3.600 \(\int \frac {(d+i c d x)^{5/2} (a+b \sinh ^{-1}(c x))^2}{(f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=794 \[ -\frac {2 i a b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {112 b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (c^2 x^2+1\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

-2*I*a*b*d^5*x*(c^2*x^2+1)^(5/2)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2*I*b^2*d^5*(c^2*x^2+1)^3/c/(d+I*c*d*x)^(
5/2)/(f-I*c*f*x)^(5/2)-2*I*b^2*d^5*x*(c^2*x^2+1)^(5/2)*arcsinh(c*x)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+28/3*d
^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+I*d^5*(c^2*x^2+1)^3*(a+b*arcsi
nh(c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+5/3*d^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^3/b/c/(d+I*c*d*x
)^(5/2)/(f-I*c*f*x)^(5/2)+112/3*b*d^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*ln(1+I/(c*x+(c^2*x^2+1)^(1/2)))/c/(
d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-112/3*b^2*d^5*(c^2*x^2+1)^(5/2)*polylog(2,-I/(c*x+(c^2*x^2+1)^(1/2)))/c/(d+
I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+8/3*b*d^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*sec(1/4*Pi+1/2*I*arcsinh(c*x))
^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+16/3*I*b^2*d^5*(c^2*x^2+1)^(5/2)*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+
I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+28/3*I*d^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2*tan(1/4*Pi+1/2*I*arcsinh(c*
x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-4/3*I*d^5*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2*sec(1/4*Pi+1/2*I*ar
csinh(c*x))^2*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 1.40, antiderivative size = 794, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 16, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {5712, 5833, 5675, 5717, 5653, 261, 5831, 3318, 4186, 3767, 8, 4184, 3716, 2190, 2279, 2391} \[ -\frac {112 b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i a b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (c^2 x^2+1\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(f - I*c*f*x)^(5/2),x]

[Out]

((-2*I)*a*b*d^5*x*(1 + c^2*x^2)^(5/2))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((2*I)*b^2*d^5*(1 + c^2*x^2
)^3)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - ((2*I)*b^2*d^5*x*(1 + c^2*x^2)^(5/2)*ArcSinh[c*x])/((d + I*
c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (28*d^5*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/(3*c*(d + I*c*d*x)^(5/
2)*(f - I*c*f*x)^(5/2)) + (I*d^5*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^
(5/2)) + (5*d^5*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^3)/(3*b*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) +
(112*b*d^5*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]])/(3*c*(d + I*c*d*x)^(5/2)*(f - I
*c*f*x)^(5/2)) - (112*b^2*d^5*(1 + c^2*x^2)^(5/2)*PolyLog[2, (-I)/E^ArcSinh[c*x]])/(3*c*(d + I*c*d*x)^(5/2)*(f
- I*c*f*x)^(5/2)) + (8*b*d^5*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/(3*c*
(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (((16*I)/3)*b^2*d^5*(1 + c^2*x^2)^(5/2)*Tan[Pi/4 + (I/2)*ArcSinh[c*
x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (((28*I)/3)*d^5*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2*
Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((4*I)/3)*d^5*(1 + c^2*x^2)^(5/
2)*(a + b*ArcSinh[c*x])^2*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5
/2)*(f - I*c*f*x)^(5/2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5712

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :>
Dist[((d + e*x)^q*(f + g*x)^q)/(1 + c^2*x^2)^q, Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n,
x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 + e^2, 0] && HalfIntegerQ[p,
q] && GeQ[p - q, 0]

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5831

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{
a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 5833

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(a + b*ArcSinh[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Fr
eeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(d+i c d x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{(f-i c f x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x)^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \left (\frac {5 d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}}+\frac {i c d^5 x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}}-\frac {8 d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x)^2 \sqrt {1+c^2 x^2}}-\frac {12 i d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x) \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {\left (12 i d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x) \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (5 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x)^2 \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (12 i d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^2}{i c+c \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^2}{(i c+c \sinh (x))^2} \, dx,x,\sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \sinh ^{-1}(c x) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc ^4\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (6 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {12 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (24 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b^2 c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {12 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (16 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (48 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int 1 \, dx,x,\cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {48 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (32 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (32 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (32 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {112 b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}

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Mathematica [B]  time = 13.91, size = 2552, normalized size = 3.21 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(f - I*c*f*x)^(5/2),x]

[Out]

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((I*a^2*d^2)/f^3 + (((8*I)/3)*a^2*d^2)/(f^3*(I + c*x)^2) - (28*a^
2*d^2)/(3*f^3*(I + c*x))))/c + (5*a^2*d^(5/2)*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(
I + c*x)]])/(c*f^(5/2)) - ((I/3)*a*b*d^2*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2
*x^2))]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[C
oth[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]])) + Cosh[ArcSinh[c*x]/2]*(4*I + 3*ArcSinh[c*x] - 6*ArcTan[Coth
[ArcSinh[c*x]/2]] + (3*I)*Log[Sqrt[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(I*ArcSinh[c*x] + (2*I)*ArcTan[Coth[A
rcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]) + 2*(1 + I*ArcSinh[c*x] + (2*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[S
qrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x]/2]))/(c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[Ar
cSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) + (a*b*d^2*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-
(d*f*(1 + c^2*x^2))]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(Cosh[(3*ArcSinh[c*x])/2]*((14*I - 3*ArcS
inh[c*x])*ArcSinh[c*x] + (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] - 14*Log[Sqrt[1 + c^2*x^2]]) + Cosh[ArcSinh[c*x]/
2]*(8 + (6*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 - (84*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 42*Log[Sqrt[1 + c^2*x^2]
]) - (2*I)*(4 + (4*I)*ArcSinh[c*x] + 6*ArcSinh[c*x]^2 - (56*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 28*Log[Sqrt[1 +
c^2*x^2]] + Sqrt[1 + c^2*x^2]*(ArcSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 14
*Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x]/2]))/(3*c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*
(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) - ((I/3)*b^2*d^2*(-I + c*x)*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(
-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*((-1 - I)*ArcSinh[c*x]^2 - (2*ArcSinh[c*x]*(2*I + ArcSinh[c*x]))
/(I + c*x) - (2*I)*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] - I*Pi*(3*ArcSinh[c*x] - 4*Log[1 + E^Ar
cSinh[c*x]] - 2*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) + 4*PolyLog[2, (-I)/E^Ar
cSinh[c*x]] - (4*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^3 + (2*(
4 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(c*f^3*Sqrt[-(((-I
)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^2) + ((I/3)*b^2
*d^2*(-I + c*x)*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(((-6*I)*c*x*ArcS
inh[c*x])/Sqrt[1 + c^2*x^2] + ((13 + 13*I)*ArcSinh[c*x]^2)/Sqrt[1 + c^2*x^2] + (3*ArcSinh[c*x]^3)/Sqrt[1 + c^2
*x^2] + (2*ArcSinh[c*x]*(2*I + ArcSinh[c*x]))/((I + c*x)*Sqrt[1 + c^2*x^2]) + (3*I)*(2 + ArcSinh[c*x]^2) + ((1
3*I)*(2*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] + Pi*(3*ArcSinh[c*x] - 4*Log[1 + E^ArcSinh[c*x]] -
2*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) + (4*I)*PolyLog[2, (-I)/E^ArcSinh[c*x
]]))/Sqrt[1 + c^2*x^2] + (4*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] - I*
Sinh[ArcSinh[c*x]/2])^3) - (2*(4 + 13*ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c
*x]/2] - I*Sinh[ArcSinh[c*x]/2]))))/(c*f^3*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSinh[c*x]/2] + I*S
inh[ArcSinh[c*x]/2])^2) + (2*b^2*d^2*(-I + c*x)*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(
1 + c^2*x^2))]*(-21*Pi*ArcSinh[c*x] - (7 - 7*I)*ArcSinh[c*x]^2 + I*ArcSinh[c*x]^3 + ((2*I)*ArcSinh[c*x]*(2*I +
ArcSinh[c*x]))/(I + c*x) - 14*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] + 28*Pi*Log[1 + E^ArcSinh[c
*x]] + 14*Pi*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] - 28*Pi*Log[Cosh[ArcSinh[c*x]/2]] - (28*I)*PolyLog[2, (-I)
/E^ArcSinh[c*x]] - ((2*I)*(4 + 7*ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[
c*x]/2]) + (4*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2])^3))/(3*c*f^
3*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^2)
- ((I/6)*a*b*d^2*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(Cosh[ArcSinh[c
*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(9 - (35*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 + (52
*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 26*Log[Sqrt[1 + c^2*x^2]])) + Cosh[ArcSinh[c*x]/2]*(20 + (24*I)*ArcSinh[c*x
] + 27*ArcSinh[c*x]^2 + (156*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 78*Log[Sqrt[1 + c^2*x^2]]) - I*(3*(-I + ArcSinh
[c*x])*Cosh[(5*ArcSinh[c*x])/2] + 2*(13 + (7*I)*ArcSinh[c*x] + 18*ArcSinh[c*x]^2 + (104*I)*ArcTan[Coth[ArcSinh
[c*x]/2]] + (3*I)*(I + ArcSinh[c*x])*Cosh[2*ArcSinh[c*x]] + 52*Log[Sqrt[1 + c^2*x^2]] + Sqrt[1 + c^2*x^2]*(6 +
(38*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 + (52*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 26*Log[Sqrt[1 + c^2*x^2]]))*Si
nh[ArcSinh[c*x]/2])))/(c*f^3*(-I + c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSinh[c*x]/2] - I*Sinh
[ArcSinh[c*x]/2])^4)

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (i \, b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c d^{2} x - i \, b^{2} d^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + {\left (2 i \, a b c^{2} d^{2} x^{2} + 4 \, a b c d^{2} x - 2 i \, a b d^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (i \, a^{2} c^{2} d^{2} x^{2} + 2 \, a^{2} c d^{2} x - i \, a^{2} d^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{c^{3} f^{3} x^{3} + 3 i \, c^{2} f^{3} x^{2} - 3 \, c f^{3} x - i \, f^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

integral(((I*b^2*c^2*d^2*x^2 + 2*b^2*c*d^2*x - I*b^2*d^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(
c^2*x^2 + 1))^2 + (2*I*a*b*c^2*d^2*x^2 + 4*a*b*c*d^2*x - 2*I*a*b*d^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log
(c*x + sqrt(c^2*x^2 + 1)) + (I*a^2*c^2*d^2*x^2 + 2*a^2*c*d^2*x - I*a^2*d^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x +
f))/(c^3*f^3*x^3 + 3*I*c^2*f^3*x^2 - 3*c*f^3*x - I*f^3), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\left (i c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (-i c f x +f \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x)

[Out]

int((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(5/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))^2*(d + c*d*x*1i)^(5/2))/(f - c*f*x*1i)^(5/2),x)

[Out]

int(((a + b*asinh(c*x))^2*(d + c*d*x*1i)^(5/2))/(f - c*f*x*1i)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**(5/2)*(a+b*asinh(c*x))**2/(f-I*c*f*x)**(5/2),x)

[Out]

Timed out

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