3.6.75 \(\int \frac {\sqrt {f-i c f x} (a+b \sinh ^{-1}(c x))^2}{(d+i c d x)^{5/2}} \, dx\) [575]

Optimal. Leaf size=518 \[ -\frac {f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i b^2 f^3 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b^2 f^3 \left (1+c^2 x^2\right )^{5/2} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

-1/3*f^3*(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)-4/3*I*b^2*f^3*(c^2*x^2+1
)^(5/2)*cot(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*I*f^3*(c^2*x^2+1)^(5/2)*(a+b*
arcsinh(c*x))^2*cot(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*b*f^3*(c^2*x^2+1)^(5/
2)*(a+b*arcsinh(c*x))*csc(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)+1/3*I*f^3*(c^2*x^
2+1)^(5/2)*(a+b*arcsinh(c*x))^2*cot(1/4*Pi+1/2*I*arcsinh(c*x))*csc(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)+4/3*b*f^3*(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)+4/3*b^2*f^3*(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)

________________________________________________________________________________________

Rubi [A]
time = 0.76, antiderivative size = 518, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {5796, 5844, 5843, 3399, 4271, 3852, 8, 4269, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {f^3 \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 {4 b f^3 \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 {i f^3 \left (c^2 x^2+1\right )^{5/2} \cot \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 b f^3 \left (c^2 x^2+1\right )^{5/2} \csc ^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 {i f^3 \left (c^2 x^2+1\right )^{5/2} \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^2\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 b^2 f^3 \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 {4 i b^2 f^3 \left (c^2 x^2+1\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(f^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((4*I)/3)
*b^2*f^3*(1 + c^2*x^2)^(5/2)*Cot[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - ((I
/3)*f^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2*Cot[Pi/4 + (I/2)*ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f -
I*c*f*x)^(5/2)) + (2*b*f^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])*Csc[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/(3*c*(d
+ I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((I/3)*f^3*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2*Cot[Pi/4 + (I/2
)*ArcSinh[c*x]]*Csc[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (4*b*f^3*(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)) +
(4*b^2*f^3*(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))

Rule 8

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3797

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))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3852

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 4269

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 4271

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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5796

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 5843

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 5844

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 {\sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(f-i c f x)^3 \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 {2 f^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{(-i+c x)^2 \sqrt {1+c^2 x^2}}+\frac {i f^3 \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 (i f^3 \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 (2 f^3 \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 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 c f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {\left (f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 )}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 )}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {\left (f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (2 i b f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (2 b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {f^3 \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 f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {\left (4 i b f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (4 i b f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (4 i b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {f^3 \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 {4 i b^2 f^3 \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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {4 b f^3 \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 {\left (8 i b f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (4 b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {f^3 \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 {4 i b^2 f^3 \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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {4 b f^3 \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 {\left (8 b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 (4 b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {f^3 \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 {4 i b^2 f^3 \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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {4 b f^3 \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 {4 b^2 f^3 \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 {\left (8 b^2 f^3 \left (1+c^2 x^2\right )^{5/2}\right ) \text {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 {f^3 \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 {4 i b^2 f^3 \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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^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 {2 b f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \csc ^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 {i f^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \csc ^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 {4 b f^3 \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 {4 b^2 f^3 \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}}\\ \end {align*}

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Mathematica [A]
time = 7.30, size = 783, normalized size = 1.51 \begin {gather*} \frac {\sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)} \left (-\frac {2 i a^2}{3 d^3 (-i+c x)^2}-\frac {a^2}{3 d^3 (-i+c x)}\right )}{c}+\frac {i a b \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (-i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\sinh ^{-1}(c x)-2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-i \log \left (\sqrt {1+c^2 x^2}\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (4+3 i \sinh ^{-1}(c x)-6 i \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+3 \log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (\sqrt {1+c^2 x^2} \left (\sinh ^{-1}(c x)+2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (i+\sinh ^{-1}(c x)+2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (\sqrt {1+c^2 x^2}\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{3 c d^3 (i+c x) \sqrt {-((-i d+c d x) (i f+c f x))} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^4}+\frac {i b^2 (i+c x) \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left ((-1+i) \sinh ^{-1}(c x)^2-\frac {2 \sinh ^{-1}(c x) \left (-2 i+\sinh ^{-1}(c x)\right )}{-i+c x}+2 i \left (\pi +2 i \sinh ^{-1}(c x)\right ) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-i \pi \left (\sinh ^{-1}(c x)-4 \log \left (1+e^{\sinh ^{-1}(c x)}\right )+4 \log \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+2 \log \left (\sin \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )\right )+4 \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )-\frac {4 \sinh ^{-1}(c x)^2 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{\left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^3}+\frac {2 \left (4+\sinh ^{-1}(c x)^2\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}\right )}{3 c d^3 \sqrt {-((-i d+c d x) (i f+c f x))} \sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((-2*I)/3)*a^2)/(d^3*(-I + c*x)^2) - a^2/(3*d^3*(-I + c*x))))/c
+ ((I/3)*a*b*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])*((-I)*Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2]] - I
*Log[Sqrt[1 + c^2*x^2]]) + Cosh[ArcSinh[c*x]/2]*(4 + (3*I)*ArcSinh[c*x] - (6*I)*ArcTan[Coth[ArcSinh[c*x]/2]] +
3*Log[Sqrt[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(ArcSinh[c*x] + 2*ArcTan[Coth[ArcSinh[c*x]/2]] + I*Log[Sqrt[
1 + c^2*x^2]]) + 2*(I + ArcSinh[c*x] + 2*ArcTan[Coth[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSin
h[c*x]/2]))/(c*d^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) + ((I/3)*b^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*(ArcSinh[c*x] - 4*Log[1 + E^ArcSinh[c*x]] + 4*Log[Cosh[ArcSinh[c*x]/2]] + 2*
Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]]) + 4*PolyLog[2, I/E^ArcSinh[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[Ar
cSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])))/(c*d^3*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*x^2]*(Co
sh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^2)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2} \sqrt {-i c f x +f}}{\left (i c d x +d \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2*(f-I*c*f*x)**(1/2)/(d+I*c*d*x)**(5/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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