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

Optimal. Leaf size=179 \[ -\pi ^{5/2} b \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+\pi ^{5/2} b \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )+\frac{1}{5} \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi ^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-2 \pi ^{5/2} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{25} \pi ^{5/2} b c^5 x^5-\frac{11}{45} \pi ^{5/2} b c^3 x^3-\frac{23}{15} \pi ^{5/2} b c x \]

[Out]

(-23*b*c*Pi^(5/2)*x)/15 - (11*b*c^3*Pi^(5/2)*x^3)/45 - (b*c^5*Pi^(5/2)*x^5)/25 + Pi^2*Sqrt[Pi + c^2*Pi*x^2]*(a
+ b*ArcSinh[c*x]) + (Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/3 + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*Arc
Sinh[c*x]))/5 - 2*Pi^(5/2)*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - b*Pi^(5/2)*PolyLog[2, -E^ArcSinh[c*x
]] + b*Pi^(5/2)*PolyLog[2, E^ArcSinh[c*x]]

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Rubi [A]  time = 0.429625, antiderivative size = 329, normalized size of antiderivative = 1.84, number of steps used = 13, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5744, 5742, 5760, 4182, 2279, 2391, 8, 194} \[ -\frac{\pi ^2 b \sqrt{\pi c^2 x^2+\pi } \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+\frac{\pi ^2 b \sqrt{\pi c^2 x^2+\pi } \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 x^2+1}}+\frac{1}{5} \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi ^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 \pi ^2 \sqrt{\pi c^2 x^2+\pi } \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c^5 x^5 \sqrt{\pi c^2 x^2+\pi }}{25 \sqrt{c^2 x^2+1}}-\frac{11 \pi ^2 b c^3 x^3 \sqrt{\pi c^2 x^2+\pi }}{45 \sqrt{c^2 x^2+1}}-\frac{23 \pi ^2 b c x \sqrt{\pi c^2 x^2+\pi }}{15 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*b*c*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2])/(15*Sqrt[1 + c^2*x^2]) - (11*b*c^3*Pi^2*x^3*Sqrt[Pi + c^2*Pi*x^2])/(45*
Sqrt[1 + c^2*x^2]) - (b*c^5*Pi^2*x^5*Sqrt[Pi + c^2*Pi*x^2])/(25*Sqrt[1 + c^2*x^2]) + Pi^2*Sqrt[Pi + c^2*Pi*x^2
]*(a + b*ArcSinh[c*x]) + (Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/3 + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b
*ArcSinh[c*x]))/5 - (2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Sqrt[1 + c^2*x
^2] - (b*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (b*Pi^2*Sqrt[Pi + c^2*Pi*
x^2]*PolyLog[2, E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2]

Rule 5744

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

Rule 5742

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

Rule 5760

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

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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 8

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

Rule 194

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

Rubi steps

\begin{align*} \int \frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi \int \frac{\left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\pi ^2 \int \frac{\sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \sqrt{1+c^2 x^2}}-\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=-\frac{8 b c \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (\pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{\sqrt{1+c^2 x^2}}-\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int 1 \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (\pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}+\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}+\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}\\ &=-\frac{23 b c \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{11 b c^3 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{45 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}-\frac{b \pi ^2 \sqrt{\pi +c^2 \pi x^2} \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}+\frac{b \pi ^2 \sqrt{\pi +c^2 \pi x^2} \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.349934, size = 257, normalized size = 1.44 \[ \frac{1}{225} \pi ^{5/2} \left (225 b \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-225 b \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+45 a c^4 x^4 \sqrt{c^2 x^2+1}+165 a c^2 x^2 \sqrt{c^2 x^2+1}+345 a \sqrt{c^2 x^2+1}-225 a \log \left (\pi \left (\sqrt{c^2 x^2+1}+1\right )\right )+225 a \log (x)-9 b c^5 x^5-55 b c^3 x^3+45 b c^4 x^4 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+165 b c^2 x^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+345 b \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)-345 b c x+225 b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-225 b \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Pi^(5/2)*(-345*b*c*x - 55*b*c^3*x^3 - 9*b*c^5*x^5 + 345*a*Sqrt[1 + c^2*x^2] + 165*a*c^2*x^2*Sqrt[1 + c^2*x^2]
+ 45*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 345*b*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 165*b*c^2*x^2*Sqrt[1 + c^2*x^2]*Arc
Sinh[c*x] + 45*b*c^4*x^4*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 225*b*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 225*
b*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 225*a*Log[x] - 225*a*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] + 225*b*PolyL
og[2, -E^(-ArcSinh[c*x])] - 225*b*PolyLog[2, E^(-ArcSinh[c*x])]))/225

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Maple [A]  time = 0.235, size = 284, normalized size = 1.6 \begin{align*}{\frac{a}{5} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}+{\frac{a\pi }{3} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}-a{\pi }^{{\frac{5}{2}}}{\it Artanh} \left ({\sqrt{\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}} \right ) +a{\pi }^{2}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }+b{\pi }^{{\frac{5}{2}}}{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) -b{\pi }^{{\frac{5}{2}}}{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) -b{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) +b{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) -{\frac{b{c}^{5}{\pi }^{{\frac{5}{2}}}{x}^{5}}{25}}-{\frac{11\,b{c}^{3}{\pi }^{5/2}{x}^{3}}{45}}+{\frac{23\,b{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) }{15}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{23\,bc{\pi }^{5/2}x}{15}}+{\frac{b{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ){x}^{4}{c}^{4}}{5}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{11\,b{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ){x}^{2}{c}^{2}}{15}\sqrt{{c}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(Pi*c^2*x^2+Pi)^(5/2)*a+1/3*a*Pi*(Pi*c^2*x^2+Pi)^(3/2)-a*Pi^(5/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2))+
a*Pi^2*(Pi*c^2*x^2+Pi)^(1/2)+b*Pi^(5/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))-b*Pi^(5/2)*polylog(2,-c*x-(c^2*x^2+1)
^(1/2))-b*Pi^(5/2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+b*Pi^(5/2)*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2)
)-1/25*b*c^5*Pi^(5/2)*x^5-11/45*b*c^3*Pi^(5/2)*x^3+23/15*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2)-23/15*b*c*P
i^(5/2)*x+1/5*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2)*x^4*c^4+11/15*b*arcsinh(c*x)*Pi^(5/2)*(c^2*x^2+1)^(1/2
)*x^2*c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{15} \,{\left (15 \, \pi ^{\frac{5}{2}} \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - 15 \, \pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} - 5 \, \pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} - 3 \,{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}}\right )} a + b \int \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*pi^(5/2)*arcsinh(1/(sqrt(c^2)*abs(x))) - 15*pi^2*sqrt(pi + pi*c^2*x^2) - 5*pi*(pi + pi*c^2*x^2)^(3/2
) - 3*(pi + pi*c^2*x^2)^(5/2))*a + b*integrate((pi + pi*c^2*x^2)^(5/2)*log(c*x + sqrt(c^2*x^2 + 1))/x, x)

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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 (\pi ^{2} a c^{4} x^{4} + 2 \, \pi ^{2} a c^{2} x^{2} + \pi ^{2} a +{\left (\pi ^{2} b c^{4} x^{4} + 2 \, \pi ^{2} b c^{2} x^{2} + \pi ^{2} b\right )} \operatorname{arsinh}\left (c x\right )\right )}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi^2*a*c^4*x^4 + 2*pi^2*a*c^2*x^2 + pi^2*a + (pi^2*b*c^4*x^4 + 2*pi^2*b*c^2*x^
2 + pi^2*b)*arcsinh(c*x))/x, 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((pi*c**2*x**2+pi)**(5/2)*(a+b*asinh(c*x))/x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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