3.601 \(\int \frac{(d+i c d x)^{3/2} (a+b \sinh ^{-1}(c x))^2}{(f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=584 \[ -\frac{32 b^2 d^4 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^4 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 d^4 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{32 b d^4 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{4 b d^4 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i d^4 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^4 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b^2 d^4 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

(8*d^4*(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)) + (d^4*(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)) + (32*b*d^4*(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)) - (32*b^
2*d^4*(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)) + (4*
b*d^4*(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)) + (((8*I)/3)*b^2*d^4*(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)) + (((8*I)/3)*d^4*(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)) - (((2*I)/3)*d^4*(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))

________________________________________________________________________________________

Rubi [A]  time = 1.2055, antiderivative size = 584, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.351, Rules used = {5712, 5833, 5675, 5831, 3318, 4186, 3767, 8, 4184, 3716, 2190, 2279, 2391} \[ -\frac{32 b^2 d^4 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^4 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 d^4 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{32 b d^4 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{4 b d^4 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i d^4 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^4 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 i b^2 d^4 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^(3/2)*(a + b*ArcSinh[c*x])^2)/(f - I*c*f*x)^(5/2),x]

[Out]

(8*d^4*(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)) + (d^4*(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)) + (32*b*d^4*(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)) - (32*b^
2*d^4*(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)) + (4*
b*d^4*(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)) + (((8*I)/3)*b^2*d^4*(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)) + (((8*I)/3)*d^4*(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)) - (((2*I)/3)*d^4*(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 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 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]

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 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 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 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 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 8

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^{3/2} \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)^4 \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{d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}}-\frac{4 d^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x)^2 \sqrt{1+c^2 x^2}}-\frac{4 i d^4 \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 (4 i d^4 \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 (d^4 \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 (4 d^4 \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{d^4 \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 (4 i d^4 \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 (4 c d^4 \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{d^4 \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 (d^4 \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 (2 d^4 \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{d^4 \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{4 b d^4 \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{4 i d^4 \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{2 i d^4 \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 (2 d^4 \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 (8 i b d^4 \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 (4 b^2 d^4 \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{4 d^4 \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{d^4 \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{4 b d^4 \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{8 i d^4 \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{2 i d^4 \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 (8 i b d^4 \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 (16 i b d^4 \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 (8 i b^2 d^4 \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{8 d^4 \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{d^4 \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 i b^2 d^4 \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{16 b d^4 \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{4 b d^4 \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{8 i d^4 \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{2 i d^4 \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^4 \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 (16 b^2 d^4 \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{8 d^4 \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{d^4 \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 i b^2 d^4 \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{32 b d^4 \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 d^4 \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{8 i d^4 \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{2 i d^4 \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 b^2 d^4 \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 (16 b^2 d^4 \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{8 d^4 \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{d^4 \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 i b^2 d^4 \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{32 b d^4 \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{16 b^2 d^4 \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{4 b d^4 \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{8 i d^4 \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{2 i d^4 \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 b^2 d^4 \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{8 d^4 \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{d^4 \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 i b^2 d^4 \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{32 b d^4 \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{32 b^2 d^4 \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{4 b d^4 \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{8 i d^4 \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{2 i d^4 \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*}

Mathematica [B]  time = 10.0731, size = 1617, normalized size = 2.77 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^(3/2)*(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)]*((((4*I)/3)*a^2*d)/(f^3*(I + c*x)^2) - (8*a^2*d)/(3*f^3*(I + c*x)
)))/c + (a^2*d^(3/2)*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*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[Coth[ArcSinh[c*x]/2]] + I*Lo
g[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)*L
og[Sqrt[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(I*ArcSinh[c*x] + (2*I)*ArcTan[Coth[ArcSinh[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[Sqrt[1 + c^2*x^2]]))*Sinh[Ar
cSinh[c*x]/2]))/(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[ArcS
inh[c*x]/2])^4) + (a*b*d*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(Cosh[Ar
cSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(Cosh[(3*ArcSinh[c*x])/2]*((14*I - 3*ArcSinh[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]]))*Sin
h[ArcSinh[c*x]/2]))/(6*c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSinh[c*x]/2] - I*Sin
h[ArcSinh[c*x]/2])^4) - ((I/3)*b^2*d*(-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^ArcSinh[c*x]] - 2*Log[-Cos[(Pi +
(2*I)*ArcSinh[c*x])/4]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) + 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[ArcSin
h[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))]*Sq
rt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^2) + (b^2*d*(-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*ArcS
inh[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^Ar
cSinh[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[Ar
cSinh[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[ArcSi
nh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^2)

________________________________________________________________________________________

Maple [F]  time = 0.266, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{{\frac{3}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x)

[Out]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} c d x - i \, b^{2} d\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 \, a b c d x - 2 i \, a b d\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 (a^{2} c d x - i \, a^{2} d\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

integral(((b^2*c*d*x - I*b^2*d)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1))^2 + (2*a*b*c
*d*x - 2*I*a*b*d)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (a^2*c*d*x - I*a^2*d)*sq
rt(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**(3/2)*(a+b*asinh(c*x))**2/(f-I*c*f*x)**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))^2/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError