3.257 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=204 \[ \frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} c \left (c^2 x^2+1\right )}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^{5/2} c}-\frac {4 b \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} c}-\frac {b^2 x}{3 \pi ^{5/2} \sqrt {c^2 x^2+1}}-\frac {2 b^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 \pi ^{5/2} c} \]

[Out]

1/3*b*(a+b*arcsinh(c*x))/c/Pi^(5/2)/(c^2*x^2+1)+2/3*(a+b*arcsinh(c*x))^2/c/Pi^(5/2)+1/3*x*(a+b*arcsinh(c*x))^2
/Pi/(Pi*c^2*x^2+Pi)^(3/2)-4/3*b*(a+b*arcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c/Pi^(5/2)-2/3*b^2*polylog(
2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/Pi^(5/2)-1/3*b^2*x/Pi^(5/2)/(c^2*x^2+1)^(1/2)+2/3*x*(a+b*arcsinh(c*x))^2/Pi^2/
(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 292, normalized size of antiderivative = 1.43, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5690, 5687, 5714, 3718, 2190, 2279, 2391, 5717, 191} \[ -\frac {2 b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 \pi ^2 c \sqrt {\pi c^2 x^2+\pi }}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 c \sqrt {\pi c^2 x^2+\pi }}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {4 b \sqrt {c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c \sqrt {\pi c^2 x^2+\pi }}-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

-(b^2*x)/(3*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (b*(a + b*ArcSinh[c*x]))/(3*c*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*P
i*x^2]) + (x*(a + b*ArcSinh[c*x])^2)/(3*Pi*(Pi + c^2*Pi*x^2)^(3/2)) + (2*x*(a + b*ArcSinh[c*x])^2)/(3*Pi^2*Sqr
t[Pi + c^2*Pi*x^2]) + (2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(3*c*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) - (4*b*Sqr
t[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(3*c*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) - (2*b^2*Sqr
t[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*x])])/(3*c*Pi^2*Sqrt[Pi + c^2*Pi*x^2])

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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx}{3 \pi }-\frac {\left (2 b c \sqrt {1+c^2 x^2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (b^2 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (4 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}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (4 b \sqrt {1+c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (8 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {4 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\left (4 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {4 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\left (2 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b^2 x}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{3 c \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {4 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 )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {2 b^2 \sqrt {1+c^2 x^2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 293, normalized size = 1.44 \[ \frac {2 a^2 c^3 x^3+3 a^2 c x+a b \sqrt {c^2 x^2+1}-2 a b c^2 x^2 \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )-2 a b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )-b \sinh ^{-1}(c x) \left (-4 a c^3 x^3-6 a c x-b \sqrt {c^2 x^2+1}+4 b \left (c^2 x^2+1\right )^{3/2} \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )\right )-b^2 c^3 x^3+2 b^2 \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (-e^{-2 \sinh ^{-1}(c x)}\right )-b^2 \left (-2 c^3 x^3+2 c^2 x^2 \sqrt {c^2 x^2+1}+2 \sqrt {c^2 x^2+1}-3 c x\right ) \sinh ^{-1}(c x)^2-b^2 c x}{3 \pi ^{5/2} c \left (c^2 x^2+1\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

(3*a^2*c*x - b^2*c*x + 2*a^2*c^3*x^3 - b^2*c^3*x^3 + a*b*Sqrt[1 + c^2*x^2] - b^2*(-3*c*x - 2*c^3*x^3 + 2*Sqrt[
1 + c^2*x^2] + 2*c^2*x^2*Sqrt[1 + c^2*x^2])*ArcSinh[c*x]^2 - b*ArcSinh[c*x]*(-6*a*c*x - 4*a*c^3*x^3 - b*Sqrt[1
+ c^2*x^2] + 4*b*(1 + c^2*x^2)^(3/2)*Log[1 + E^(-2*ArcSinh[c*x])]) - 2*a*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2]
- 2*a*b*c^2*x^2*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + 2*b^2*(1 + c^2*x^2)^(3/2)*PolyLog[2, -E^(-2*ArcSinh[c*x]
)])/(3*c*Pi^(5/2)*(1 + c^2*x^2)^(3/2))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}\right )}}{\pi ^{3} c^{6} x^{6} + 3 \, \pi ^{3} c^{4} x^{4} + 3 \, \pi ^{3} c^{2} x^{2} + \pi ^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(5/2),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^3*c^6*x^6 + 3*pi^3*c^4*x^4
+ 3*pi^3*c^2*x^2 + pi^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/(pi + pi*c^2*x^2)^(5/2), x)

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maple [B]  time = 0.36, size = 1730, normalized size = 8.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

-2/3*b^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/Pi^(5/2)+4/3*b^2/c/Pi^(5/2)*arcsinh(c*x)^2+1/3*a^2/Pi*x/(Pi*c
^2*x^2+Pi)^(3/2)+2/3*a^2/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)-22/3*b^2/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(
c*x)^2*x^2+2/3*b^2/Pi^(5/2)*c^7/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^8+10/3*b^2/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1
)^2*x^6+6*b^2/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^4+14/3*b^2/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^2
-8/3*b^2/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)^2+4/3*b^2/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*
arcsinh(c*x)-2/3*b^2/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)*x^6-b^2/Pi^(5/2)*c^4/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/
2)*x^5+4*b^2/Pi^(5/2)/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)^2*x-b^2/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1
)*x^2-5/3*b^2/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)*x^4-7/3*b^2/Pi^(5/2)*c^2/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*
x^3+4/3*a*b/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2+4*a*b/Pi^(5/2)*c^4/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(
c*x)*x^5+34/3*a*b/Pi^(5/2)*c^2/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^3-40/3*a*b/Pi^(5/2)*c^3/(3*c^2*x
^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^4-44/3*a*b/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^2-4*a*b/Pi
^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^6+16/3*a*b/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^2-16
/3*a*b/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)-4/3*b^2/Pi^(5/2)/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*x-
4*a*b/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)*x^4+16/3*b^2/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x
)*x^6-20/3*b^2/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)^2*x^4+4/3*b^2/Pi^(5/2)*c^7/(3*c^2*x^2+4)/
(c^2*x^2+1)^2*arcsinh(c*x)*x^8-2*b^2/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)^2*x^6+16/3*b^2/Pi^(
5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^2+8*b^2/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x
)*x^4+2*b^2/Pi^(5/2)*c^4/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)^2*x^5-4*b^2/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(
c^2*x^2+1)*arcsinh(c*x)*x^4+17/3*b^2/Pi^(5/2)*c^2/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)^2*x^3-3*b^2/Pi^
(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)*arcsinh(c*x)*x^2-4/3*b^2/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)*arcsinh(c*x)
*x^6+4/3*b^2/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2-4/3*b^2/c/Pi^(5/2)*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2
))^2)+8/3*a*b/c/Pi^(5/2)*arcsinh(c*x)-4/3*a*b/c/Pi^(5/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+8*a*b/Pi^(5/2)/(3*c^2
*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x-3*a*b/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)*x^2+8*a*b/Pi^(5/2)*c^3/(3*
c^2*x^2+4)/(c^2*x^2+1)^2*x^4+4/3*a*b/Pi^(5/2)*c^7/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^8+16/3*a*b/Pi^(5/2)*c^5/(3*c^2
*x^2+4)/(c^2*x^2+1)^2*x^6-4/3*a*b/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)*x^6

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a b c {\left (\frac {1}{\pi ^{\frac {5}{2}} c^{4} x^{2} + \pi ^{\frac {5}{2}} c^{2}} - \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac {5}{2}} c^{2}}\right )} + \frac {2}{3} \, a b {\left (\frac {x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a^{2} {\left (\frac {x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}}\right )} + b^{2} \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/(pi*c^2*x^2+pi)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*b*c*(1/(pi^(5/2)*c^4*x^2 + pi^(5/2)*c^2) - 2*log(c^2*x^2 + 1)/(pi^(5/2)*c^2)) + 2/3*a*b*(x/(pi*(pi + pi*
c^2*x^2)^(3/2)) + 2*x/(pi^2*sqrt(pi + pi*c^2*x^2)))*arcsinh(c*x) + 1/3*a^2*(x/(pi*(pi + pi*c^2*x^2)^(3/2)) + 2
*x/(pi^2*sqrt(pi + pi*c^2*x^2))) + b^2*integrate(log(c*x + sqrt(c^2*x^2 + 1))^2/(pi + pi*c^2*x^2)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2/(Pi + Pi*c^2*x^2)^(5/2),x)

[Out]

int((a + b*asinh(c*x))^2/(Pi + Pi*c^2*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(pi*c**2*x**2+pi)**(5/2),x)

[Out]

(Integral(a**2/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + I
ntegral(b**2*asinh(c*x)**2/(c**4*x**4*sqrt(c**2*x**2 + 1) + 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**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c
**2*x**2 + 1)), x))/pi**(5/2)

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