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

Optimal. Leaf size=224 \[ -\frac{b^2 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{\left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 b \left (c^2 x^2+1\right )^{3/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

(x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + ((1 + c^2*x^2)^(3/2)*(a +
b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]
)*Log[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*(1 + c^2*x^2)^(3/2)*PolyLog[
2, -E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))

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Rubi [A]  time = 0.423362, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {5712, 5687, 5714, 3718, 2190, 2279, 2391} \[ -\frac{b^2 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{\left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 b \left (c^2 x^2+1\right )^{3/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + ((1 + c^2*x^2)^(3/2)*(a +
b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]
)*Log[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*(1 + c^2*x^2)^(3/2)*PolyLog[
2, -E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))

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 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]

Rubi steps

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

Mathematica [B]  time = 1.36882, size = 488, normalized size = 2.18 \[ \frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+a^2 c x-a b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )+2 a b c x \sinh ^{-1}(c x)-b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)^2-2 i \pi b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)-2 b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \pi b^2 \sqrt{c^2 x^2+1} \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-2 b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-i \pi b^2 \sqrt{c^2 x^2+1} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+4 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (e^{\sinh ^{-1}(c x)}+1\right )-i \pi b^2 \sqrt{c^2 x^2+1} \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+i \pi b^2 \sqrt{c^2 x^2+1} \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-4 i \pi b^2 \sqrt{c^2 x^2+1} \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+b^2 c x \sinh ^{-1}(c x)^2}{c d f \sqrt{d+i c d x} \sqrt{f-i c f x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*c*x + 2*a*b*c*x*ArcSinh[c*x] - (2*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + b^2*c*x*ArcSinh[c*x]^2 - b^2
*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 + I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 - I/E^ArcSinh[c*x]] - 2*b^2*Sqrt[1 + c^2*
x^2]*ArcSinh[c*x]*Log[1 - I/E^ArcSinh[c*x]] - I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + I/E^ArcSinh[c*x]] - 2*b^2*Sqr
t[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + I/E^ArcSinh[c*x]] + (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + E^ArcSinh[c*x]]
- a*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]]
- (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Cosh[ArcSinh[c*x]/2]] - I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Sin[(Pi + (2*I)*Ar
cSinh[c*x])/4]] + 2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)/E^ArcSinh[c*x]] + 2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2,
I/E^ArcSinh[c*x]])/(c*d*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

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Maple [F]  time = 0.277, 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{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)^(3/2)/(f-I*c*f*x)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{2} \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{{\left (i \, c d x + d\right )}^{\frac{3}{2}}{\left (-i \, c f x + f\right )}^{\frac{3}{2}}}\,{d x} - \frac{a b c \sqrt{\frac{1}{c^{4} d f}} \log \left (x^{2} + \frac{1}{c^{2}}\right )}{d f} + \frac{2 \, a b x \operatorname{arsinh}\left (c x\right )}{\sqrt{c^{2} d f x^{2} + d f} d f} + \frac{a^{2} x}{\sqrt{c^{2} d f x^{2} + d f} d f} \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)^(3/2)/(f-I*c*f*x)^(3/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out