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

Optimal. Leaf size=743 \[ -\frac{b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{f x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b f \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i b f \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b^2 f \left (c^2 x^2+1\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

((-I/3)*b^2*f*(1 + c^2*x^2)^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b^2*f*x*(1 + c^2*x^2)^2)/(3*(d +
I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b*f*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c*(d + I*c*d*x)^(5/2)
*(f - I*c*f*x)^(5/2)) - ((I/3)*b*f*x*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/((d + I*c*d*x)^(5/2)*(f - I*c*f
*x)^(5/2)) + ((I/3)*f*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (f*x
*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*f*x*(1 + c^2*x^2)^2*(a
+ b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*f*(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)) - (((2*I)/3)*b*f*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])
*ArcTan[E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (4*b*f*(1 + c^2*x^2)^(5/2)*(a + b*ArcSi
nh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b^2*f*(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)) + (b^2*f*(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)) - (2*b^2*f*(1 + c^2*x^2)^(5/2)*Po
lyLog[2, -E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.908861, antiderivative size = 743, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 14, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.378, Rules used = {5712, 5821, 5690, 5687, 5714, 3718, 2190, 2279, 2391, 5717, 191, 5693, 4180, 261} \[ -\frac{b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 b^2 f \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{f x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b f \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i b f \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b^2 f \left (c^2 x^2+1\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/3)*b^2*f*(1 + c^2*x^2)^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b^2*f*x*(1 + c^2*x^2)^2)/(3*(d +
I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b*f*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c*(d + I*c*d*x)^(5/2)
*(f - I*c*f*x)^(5/2)) - ((I/3)*b*f*x*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/((d + I*c*d*x)^(5/2)*(f - I*c*f
*x)^(5/2)) + ((I/3)*f*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (f*x
*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*f*x*(1 + c^2*x^2)^2*(a
+ b*ArcSinh[c*x])^2)/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*f*(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)) - (((2*I)/3)*b*f*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x])
*ArcTan[E^ArcSinh[c*x]])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (4*b*f*(1 + c^2*x^2)^(5/2)*(a + b*ArcSi
nh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b^2*f*(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)) + (b^2*f*(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)) - (2*b^2*f*(1 + c^2*x^2)^(5/2)*Po
lyLog[2, -E^(2*ArcSinh[c*x])])/(3*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 5821

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

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5714

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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]

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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 5693

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

Rule 4180

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

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{(f-i c f x) \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{f \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}}-\frac{i c f x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{\left (f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{\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 (i c f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x \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{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 i b f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 b c f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (i b f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b^2 f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (4 b c f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (i b^2 c f \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (i b f \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{sech}(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 f \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int (a+b x) \tanh (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{i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \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{2 i b f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (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 f \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 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 (b^2 f \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 (b^2 f \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{i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \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{2 i b f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (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 f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b^2 f \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{\left (b^2 f \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{\left (4 b^2 f \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 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{i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \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{2 i b f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (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 f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f \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{b^2 f \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{\left (2 b^2 f \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b f \left (1+c^2 x^2\right )^{3/2} \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 b f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i f \left (1+c^2 x^2\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{f x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 f \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{2 i b f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (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 f \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 f \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{b^2 f \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{2 b^2 f \left (1+c^2 x^2\right )^{5/2} \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 8.25687, size = 754, normalized size = 1.01 \[ \frac{i b^2 \sqrt{c^2 x^2+1} \sqrt{i (c d x-i d)} \sqrt{-i (c f x+i f)} \left (6 i \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+10 i \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )-(1+4 i) \sinh ^{-1}(c x)^2+\frac{\left (2+i \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)}{c x-i}+7 \pi \sinh ^{-1}(c x)-5 \left (\pi +2 i \sinh ^{-1}(c x)\right ) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+3 \left (\pi -2 i \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-16 \pi \log \left (e^{\sinh ^{-1}(c x)}+1\right )+\frac{3 i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)^2}{\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )}+\frac{2 i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c 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 )^3}+\frac{\left (5 \sinh ^{-1}(c x)^2-4\right ) \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )}{\sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-i \cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )}+5 \pi \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-3 \pi \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+16 \pi \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{6 c d^2 f \sqrt{-(c d x-i d) (c f x+i f)} \sqrt{-d f \left (c^2 x^2+1\right )}}+\frac{\sqrt{i d (c x-i)} \sqrt{-i f (c x+i)} \left (\frac{5 a^2}{12 d^3 f^2 (c x-i)}+\frac{a^2}{4 d^3 f^2 (c x+i)}-\frac{i a^2}{6 d^3 f^2 (c x-i)^2}\right )}{c}+\frac{i a b \sqrt{i (c d x-i d)} \sqrt{-i (c f x+i f)} \left (\sqrt{c^2 x^2+1} \left (-4 \log \left (\sqrt{c^2 x^2+1}\right )+2 c x \left (\tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-2 i \log \left (\sqrt{c^2 x^2+1}\right )\right )-2 i \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+1\right )+4 c x \sinh ^{-1}(c x)+2 i \sinh ^{-1}(c x) \cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{3 c d^2 f (c x-i) \sqrt{-(c d x-i d) (c f x+i f)} \sqrt{-d f \left (c^2 x^2+1\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*(((-I/6)*a^2)/(d^3*f^2*(-I + c*x)^2) + (5*a^2)/(12*d^3*f^2*(-I +
c*x)) + a^2/(4*d^3*f^2*(I + c*x))))/c + ((I/3)*a*b*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*(4*c*x*Ar
cSinh[c*x] + (2*I)*ArcSinh[c*x]*Cosh[2*ArcSinh[c*x]] + Sqrt[1 + c^2*x^2]*(1 - (2*I)*ArcTan[Tanh[ArcSinh[c*x]/2
]] + 2*c*x*(ArcTan[Tanh[ArcSinh[c*x]/2]] - (2*I)*Log[Sqrt[1 + c^2*x^2]]) - 4*Log[Sqrt[1 + c^2*x^2]])))/(c*d^2*
f*(-I + c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[-(d*f*(1 + c^2*x^2))]) + ((I/6)*b^2*Sqrt[I*((-I)*d +
c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[1 + c^2*x^2]*(7*Pi*ArcSinh[c*x] + ((2 + I*ArcSinh[c*x])*ArcSinh[c*x])/(
-I + c*x) - (1 + 4*I)*ArcSinh[c*x]^2 - 5*(Pi + (2*I)*ArcSinh[c*x])*Log[1 - I/E^ArcSinh[c*x]] + 3*(Pi - (2*I)*A
rcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] - 16*Pi*Log[1 + E^ArcSinh[c*x]] - 3*Pi*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x
])/4]] + 16*Pi*Log[Cosh[ArcSinh[c*x]/2]] + 5*Pi*Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]] + (6*I)*PolyLog[2, (-I)/
E^ArcSinh[c*x]] + (10*I)*PolyLog[2, I/E^ArcSinh[c*x]] + ((3*I)*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcS
inh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2]) + ((2*I)*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] + I*
Sinh[ArcSinh[c*x]/2])^3 + ((-4 + 5*ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[Arc
Sinh[c*x]/2])))/(c*d^2*f*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[-(d*f*(1 + c^2*x^2))])

________________________________________________________________________________________

Maple [F]  time = 0.276, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{-{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2 \, b^{2} c^{2} x^{2} - 2 i \, b^{2} c x + b^{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 (3 \, c^{4} d^{3} f^{2} x^{3} - 3 i \, c^{3} d^{3} f^{2} x^{2} + 3 \, c^{2} d^{3} f^{2} x - 3 i \, c d^{3} f^{2}\right )}{\rm integral}\left (\frac{-3 i \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a^{2} - 2 \,{\left (3 i \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a b +{\left (2 \, b^{2} c^{2} x^{2} - 2 i \, b^{2} c x + b^{2}\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{3 \, c^{5} d^{3} f^{2} x^{5} - 3 i \, c^{4} d^{3} f^{2} x^{4} + 6 \, c^{3} d^{3} f^{2} x^{3} - 6 i \, c^{2} d^{3} f^{2} x^{2} + 3 \, c d^{3} f^{2} x - 3 i \, d^{3} f^{2}}, x\right )}{3 \, c^{4} d^{3} f^{2} x^{3} - 3 i \, c^{3} d^{3} f^{2} x^{2} + 3 \, c^{2} d^{3} f^{2} x - 3 i \, c d^{3} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b^2*c^2*x^2 - 2*I*b^2*c*x + 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^4*d^3*f^2*x^3 - 3*I*c^3*d^3*f^2*x^2 + 3*c^2*d^3*f^2*x - 3*I*c*d^3*f^2)*integral((-3*I*sqrt(I*c*d*x + d)*sqrt
(-I*c*f*x + f)*a^2 - 2*(3*I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a*b + (2*b^2*c^2*x^2 - 2*I*b^2*c*x + b^2)*sqr
t(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))*log(c*x + sqrt(c^2*x^2 + 1)))/(3*c^5*d^3*f^2*x^5 - 3*I*c^
4*d^3*f^2*x^4 + 6*c^3*d^3*f^2*x^3 - 6*I*c^2*d^3*f^2*x^2 + 3*c*d^3*f^2*x - 3*I*d^3*f^2), x))/(3*c^4*d^3*f^2*x^3
- 3*I*c^3*d^3*f^2*x^2 + 3*c^2*d^3*f^2*x - 3*I*c*d^3*f^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Exception raised: AttributeError