3.77 \(\int \frac{(\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=205 \[ -\frac{5}{2} \pi ^{5/2} b c^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+\frac{5}{2} \pi ^{5/2} b c^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )+\frac{5}{6} \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{2} \pi ^2 c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-5 \pi ^{5/2} c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{9} \pi ^{5/2} b c^5 x^3-\frac{7}{3} \pi ^{5/2} b c^3 x-\frac{\pi ^{5/2} b c}{2 x} \]

[Out]

-(b*c*Pi^(5/2))/(2*x) - (7*b*c^3*Pi^(5/2)*x)/3 - (b*c^5*Pi^(5/2)*x^3)/9 + (5*c^2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*(a
+ b*ArcSinh[c*x]))/2 + (5*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/6 - ((Pi + c^2*Pi*x^2)^(5/2)*(
a + b*ArcSinh[c*x]))/(2*x^2) - 5*c^2*Pi^(5/2)*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - (5*b*c^2*Pi^(5/2)
*PolyLog[2, -E^ArcSinh[c*x]])/2 + (5*b*c^2*Pi^(5/2)*PolyLog[2, E^ArcSinh[c*x]])/2

________________________________________________________________________________________

Rubi [A]  time = 0.430901, antiderivative size = 355, normalized size of antiderivative = 1.73, number of steps used = 13, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {5739, 5744, 5742, 5760, 4182, 2279, 2391, 8, 270} \[ -\frac{5 \pi ^2 b c^2 \sqrt{\pi c^2 x^2+\pi } \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt{c^2 x^2+1}}+\frac{5 \pi ^2 b c^2 \sqrt{\pi c^2 x^2+\pi } \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt{c^2 x^2+1}}+\frac{5}{6} \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{2} \pi ^2 c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{5 \pi ^2 c^2 \sqrt{\pi c^2 x^2+\pi } \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c^5 x^3 \sqrt{\pi c^2 x^2+\pi }}{9 \sqrt{c^2 x^2+1}}-\frac{7 \pi ^2 b c^3 x \sqrt{\pi c^2 x^2+\pi }}{3 \sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c \sqrt{\pi c^2 x^2+\pi }}{2 x \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^3,x]

[Out]

-(b*c*Pi^2*Sqrt[Pi + c^2*Pi*x^2])/(2*x*Sqrt[1 + c^2*x^2]) - (7*b*c^3*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2])/(3*Sqrt[1 +
c^2*x^2]) - (b*c^5*Pi^2*x^3*Sqrt[Pi + c^2*Pi*x^2])/(9*Sqrt[1 + c^2*x^2]) + (5*c^2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*
(a + b*ArcSinh[c*x]))/2 + (5*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/6 - ((Pi + c^2*Pi*x^2)^(5/2)
*(a + b*ArcSinh[c*x]))/(2*x^2) - (5*c^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] - (5*b*c^2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*PolyLog[2, -E^ArcSinh[c*x]])/(2*Sqrt[1 + c^2*x^2])
+ (5*b*c^2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*PolyLog[2, E^ArcSinh[c*x]])/(2*Sqrt[1 + c^2*x^2])

Rule 5739

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 + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x
)^(m + 2)*(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 + 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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && 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^3} \, dx &=-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{1}{2} \left (5 c^2 \pi \right ) \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 \frac{\left (1+c^2 x^2\right )^2}{x^2} \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{1}{2} \left (5 c^2 \pi ^2\right ) \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 (2 c^2+\frac{1}{x^2}+c^4 x^2\right ) \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right ) \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{2 x \sqrt{1+c^2 x^2}}+\frac{b c^3 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{6 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{\left (5 c^2 \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}{2 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int 1 \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{2 x \sqrt{1+c^2 x^2}}-\frac{7 b c^3 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{3 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{\left (5 c^2 \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 )}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{2 x \sqrt{1+c^2 x^2}}-\frac{7 b c^3 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{3 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{5 c^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 (5 b c^2 \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 )}{2 \sqrt{1+c^2 x^2}}+\frac{\left (5 b c^2 \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 )}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{2 x \sqrt{1+c^2 x^2}}-\frac{7 b c^3 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{3 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{5 c^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 (5 b c^2 \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 )}{2 \sqrt{1+c^2 x^2}}+\frac{\left (5 b c^2 \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 )}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}{2 x \sqrt{1+c^2 x^2}}-\frac{7 b c^3 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{3 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{9 \sqrt{1+c^2 x^2}}+\frac{5}{2} c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{6} c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{5 c^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{5 b c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt{1+c^2 x^2}}+\frac{5 b c^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.89382, size = 349, normalized size = 1.7 \[ \frac{\pi ^{5/2} \left (180 b c^2 x^2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-180 b c^2 x^2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+24 a c^4 x^4 \sqrt{c^2 x^2+1}+168 a c^2 x^2 \sqrt{c^2 x^2+1}-36 a \sqrt{c^2 x^2+1}+180 a c^2 x^2 \log (x)-180 a c^2 x^2 \log \left (\pi \left (\sqrt{c^2 x^2+1}+1\right )\right )-8 b c^5 x^5-168 b c^3 x^3+24 b c^4 x^4 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+168 b c^2 x^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+180 b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-180 b c^2 x^2 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-9 b c^3 x^3 \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-9 b c^2 x^2 \sinh ^{-1}(c x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )+36 b c x \sinh ^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-36 b \sinh ^{-1}(c x) \sinh ^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )}{72 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/x^3,x]

[Out]

(Pi^(5/2)*(-168*b*c^3*x^3 - 8*b*c^5*x^5 - 36*a*Sqrt[1 + c^2*x^2] + 168*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 24*a*c^4*
x^4*Sqrt[1 + c^2*x^2] + 168*b*c^2*x^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 24*b*c^4*x^4*Sqrt[1 + c^2*x^2]*ArcSinh[
c*x] - 9*b*c^3*x^3*Csch[ArcSinh[c*x]/2]^2 - 9*b*c^2*x^2*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 + 180*b*c^2*x^2*Ar
cSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 180*b*c^2*x^2*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 180*a*c^2*x^2*
Log[x] - 180*a*c^2*x^2*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] + 180*b*c^2*x^2*PolyLog[2, -E^(-ArcSinh[c*x])] - 180*b*
c^2*x^2*PolyLog[2, E^(-ArcSinh[c*x])] + 36*b*c*x*Sinh[ArcSinh[c*x]/2]^2 - 36*b*ArcSinh[c*x]*Sinh[ArcSinh[c*x]/
2]^2))/(72*x^2)

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Maple [A]  time = 0.421, size = 356, normalized size = 1.7 \begin{align*} -{\frac{a}{2\,\pi \,{x}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{7}{2}}}}+{\frac{a{c}^{2}}{2} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}+{\frac{5\,a{c}^{2}\pi }{6} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}-{\frac{5\,a{c}^{2}{\pi }^{5/2}}{2}{\it Artanh} \left ({\sqrt{\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}} \right ) }+{\frac{5\,a{c}^{2}{\pi }^{2}}{2}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{5\,b{c}^{2}{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) }{2}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{5\,b{c}^{2}{\pi }^{5/2}}{2}{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{5\,b{c}^{2}{\pi }^{5/2}}{2}{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{b{c}^{5}{\pi }^{{\frac{5}{2}}}{x}^{3}}{9}}-{\frac{7\,b{c}^{3}{\pi }^{5/2}x}{3}}+{\frac{5\,b{c}^{2}{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) }{2}\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{b{c}^{2}{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ) }{2}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ){x}^{2}{c}^{4}}{3}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{\pi }^{{\frac{5}{2}}}c}{2\,x}}-{\frac{b{\pi }^{{\frac{5}{2}}}{\it Arcsinh} \left ( cx \right ) }{2\,{x}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{7\,b{c}^{2}{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) }{3}\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^3,x)

[Out]

-1/2*a/Pi/x^2*(Pi*c^2*x^2+Pi)^(7/2)+1/2*a*c^2*(Pi*c^2*x^2+Pi)^(5/2)+5/6*a*c^2*Pi*(Pi*c^2*x^2+Pi)^(3/2)-5/2*a*c
^2*Pi^(5/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2))+5/2*a*c^2*Pi^2*(Pi*c^2*x^2+Pi)^(1/2)-5/2*b*c^2*Pi^(5/2)*ar
csinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-5/2*b*c^2*Pi^(5/2)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+5/2*b*c^2*Pi^(5/2)
*polylog(2,c*x+(c^2*x^2+1)^(1/2))-1/9*b*c^5*Pi^(5/2)*x^3-7/3*b*c^3*Pi^(5/2)*x+5/2*b*c^2*Pi^(5/2)*arcsinh(c*x)*
ln(1-c*x-(c^2*x^2+1)^(1/2))-1/2*b*Pi^(5/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^2+1/3*b*arcsinh(c*x)*Pi^(5/2)*(c^2
*x^2+1)^(1/2)*x^2*c^4-1/2*b*c*Pi^(5/2)/x-1/2*b*Pi^(5/2)/(c^2*x^2+1)^(1/2)/x^2*arcsinh(c*x)+7/3*b*arcsinh(c*x)*
Pi^(5/2)*(c^2*x^2+1)^(1/2)*c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \,{\left (15 \, \pi ^{\frac{5}{2}} c^{2} \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - 15 \, \pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} c^{2} - 5 \, \pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2} - 3 \,{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}} c^{2} + \frac{3 \,{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{7}{2}}}{\pi x^{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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

-1/6*(15*pi^(5/2)*c^2*arcsinh(1/(sqrt(c^2)*abs(x))) - 15*pi^2*sqrt(pi + pi*c^2*x^2)*c^2 - 5*pi*(pi + pi*c^2*x^
2)^(3/2)*c^2 - 3*(pi + pi*c^2*x^2)^(5/2)*c^2 + 3*(pi + pi*c^2*x^2)^(7/2)/(pi*x^2))*a + b*integrate((pi + pi*c^
2*x^2)^(5/2)*log(c*x + sqrt(c^2*x^2 + 1))/x^3, 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^{3}}, 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^3,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^3, 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**3,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^{3}}\,{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^3,x, algorithm="giac")

[Out]

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