3.256 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{b^2 \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{\pi ^{3/2} c}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{\pi \sqrt{\pi c^2 x^2+\pi }}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\pi ^{3/2} c}-\frac{2 b \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} c} \]

[Out]

(a + b*ArcSinh[c*x])^2/(c*Pi^(3/2)) + (x*(a + b*ArcSinh[c*x])^2)/(Pi*Sqrt[Pi + c^2*Pi*x^2]) - (2*b*(a + b*ArcS
inh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c*Pi^(3/2)) - (b^2*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c*Pi^(3/2))

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Rubi [A]  time = 0.176843, antiderivative size = 179, normalized size of antiderivative = 1.72, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {5687, 5714, 3718, 2190, 2279, 2391} \[ -\frac{b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{\pi c \sqrt{\pi c^2 x^2+\pi }}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{\pi \sqrt{\pi c^2 x^2+\pi }}+\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{\pi c \sqrt{\pi c^2 x^2+\pi }}-\frac{2 b \sqrt{c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi c \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

(x*(a + b*ArcSinh[c*x])^2)/(Pi*Sqrt[Pi + c^2*Pi*x^2]) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(c*Pi*Sqrt[
Pi + c^2*Pi*x^2]) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c*Pi*Sqrt[Pi + c
^2*Pi*x^2]) - (b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c*Pi*Sqrt[Pi + c^2*Pi*x^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]

Rubi steps

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

Mathematica [A]  time = 0.387552, size = 153, normalized size = 1.47 \[ \frac{b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+a \left (a c x-b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )\right )+2 b \sinh ^{-1}(c x) \left (a c x-b \sqrt{c^2 x^2+1} \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )\right )+b^2 \left (-\left (\sqrt{c^2 x^2+1}-c x\right )\right ) \sinh ^{-1}(c x)^2}{\pi ^{3/2} c \sqrt{c^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.129, size = 306, normalized size = 2.9 \begin{align*}{\frac{{a}^{2}x}{\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}c{x}^{2}}{{\pi }^{{\frac{3}{2}}} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}x}{{\pi }^{{\frac{3}{2}}}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{\pi }^{{\frac{3}{2}}}c \left ({c}^{2}{x}^{2}+1 \right ) }}+2\,{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{\pi }^{3/2}c}}-2\,{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{{\pi }^{3/2}c}}-{\frac{{b}^{2}}{{\pi }^{{\frac{3}{2}}}c}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+4\,{\frac{ab{\it Arcsinh} \left ( cx \right ) }{{\pi }^{3/2}c}}-2\,{\frac{ab{\it Arcsinh} \left ( cx \right ) c{x}^{2}}{{\pi }^{3/2} \left ({c}^{2}{x}^{2}+1 \right ) }}+2\,{\frac{ab{\it Arcsinh} \left ( cx \right ) x}{{\pi }^{3/2}\sqrt{{c}^{2}{x}^{2}+1}}}-2\,{\frac{ab{\it Arcsinh} \left ( cx \right ) }{{\pi }^{3/2}c \left ({c}^{2}{x}^{2}+1 \right ) }}-2\,{\frac{ab\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{{\pi }^{3/2}c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

a^2/Pi*x/(Pi*c^2*x^2+Pi)^(1/2)-b^2/Pi^(3/2)*arcsinh(c*x)^2*c/(c^2*x^2+1)*x^2+b^2/Pi^(3/2)*arcsinh(c*x)^2/(c^2*
x^2+1)^(1/2)*x-b^2/Pi^(3/2)*arcsinh(c*x)^2/c/(c^2*x^2+1)+2*b^2/c/Pi^(3/2)*arcsinh(c*x)^2-2*b^2/c/Pi^(3/2)*arcs
inh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-b^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/Pi^(3/2)+4*a*b/c/Pi^(3/2)
*arcsinh(c*x)-2*a*b/Pi^(3/2)*arcsinh(c*x)*c/(c^2*x^2+1)*x^2+2*a*b/Pi^(3/2)*arcsinh(c*x)/(c^2*x^2+1)^(1/2)*x-2*
a*b/Pi^(3/2)*arcsinh(c*x)/c/(c^2*x^2+1)-2*a*b/c/Pi^(3/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\pi + \pi c^{2} x^{2}}{\left (b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}\right )}}{\pi ^{2} c^{4} x^{4} + 2 \, \pi ^{2} c^{2} x^{2} + \pi ^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(pi^2*c^4*x^4 + 2*pi^2*c^2*x^2
+ pi^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a**2/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b**2*asinh(c*x)**2/(c**2*x
**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(2*a*b*asinh(c*x)/(c**2*x**2*sqrt(c**2*x**2 + 1)
+ sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/(pi + pi*c^2*x^2)^(3/2), x)