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

Optimal. Leaf size=247 \[ -\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {5 c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2}}+\frac {5 b c^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}-\frac {5 b c^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}+\frac {b c}{4 \pi ^{5/2} x \left (c^2 x^2+1\right )}+\frac {13 b c^2 \tan ^{-1}(c x)}{6 \pi ^{5/2}}+\frac {5 b c^3 x}{12 \pi ^{5/2} \left (c^2 x^2+1\right )}-\frac {3 b c}{4 \pi ^{5/2} x} \]

[Out]

-3/4*b*c/Pi^(5/2)/x+1/4*b*c/Pi^(5/2)/x/(c^2*x^2+1)+5/12*b*c^3*x/Pi^(5/2)/(c^2*x^2+1)-5/6*c^2*(a+b*arcsinh(c*x)
)/Pi/(Pi*c^2*x^2+Pi)^(3/2)+1/2*(-a-b*arcsinh(c*x))/Pi/x^2/(Pi*c^2*x^2+Pi)^(3/2)+13/6*b*c^2*arctan(c*x)/Pi^(5/2
)+5*c^2*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/Pi^(5/2)+5/2*b*c^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))
/Pi^(5/2)-5/2*b*c^2*polylog(2,c*x+(c^2*x^2+1)^(1/2))/Pi^(5/2)-5/2*c^2*(a+b*arcsinh(c*x))/Pi^2/(Pi*c^2*x^2+Pi)^
(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 325, normalized size of antiderivative = 1.32, number of steps used = 15, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5747, 5755, 5760, 4182, 2279, 2391, 203, 199, 290, 325} \[ \frac {5 b c^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}-\frac {5 b c^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {5 c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}-\frac {3 b c \sqrt {c^2 x^2+1}}{4 \pi ^2 x \sqrt {\pi c^2 x^2+\pi }}+\frac {b c}{4 \pi ^2 x \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}+\frac {13 b c^2 \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(4*Pi^2*x*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*Pi*x^2]) + (5*b*c^3*x)/(12*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^
2*Pi*x^2]) - (3*b*c*Sqrt[1 + c^2*x^2])/(4*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]) - (5*c^2*(a + b*ArcSinh[c*x]))/(6*Pi*(
Pi + c^2*Pi*x^2)^(3/2)) - (a + b*ArcSinh[c*x])/(2*Pi*x^2*(Pi + c^2*Pi*x^2)^(3/2)) - (5*c^2*(a + b*ArcSinh[c*x]
))/(2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (13*b*c^2*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(6*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) +
(5*c^2*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Pi^(5/2) + (5*b*c^2*PolyLog[2, -E^ArcSinh[c*x]])/(2*Pi^(5
/2)) - (5*b*c^2*PolyLog[2, E^ArcSinh[c*x]])/(2*Pi^(5/2))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 5747

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 + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(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*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5755

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 + 1)*(a + b*ArcSinh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(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)^F
racPart[p])/(2*f*(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] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || 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]

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {1}{2} \left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx+\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )^2} \, dx}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {\left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx}{2 \pi }+\frac {\left (3 b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (5 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx}{2 \pi ^2}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{12 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (3 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\left (5 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (5 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{5/2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {\left (5 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{5/2}}-\frac {\left (5 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{5/2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {\left (5 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}-\frac {\left (5 b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}\\ &=\frac {b c}{4 \pi ^2 x \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b c^3 x}{12 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}-\frac {3 b c \sqrt {1+c^2 x^2}}{4 \pi ^2 x \sqrt {\pi +c^2 \pi x^2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \sinh ^{-1}(c x)}{2 \pi x^2 \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {13 b c^2 \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {5 b c^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}-\frac {5 b c^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 6.31, size = 331, normalized size = 1.34 \[ \frac {-\frac {48 a c^2}{\sqrt {c^2 x^2+1}}-\frac {8 a c^2}{\left (c^2 x^2+1\right )^{3/2}}-\frac {12 a \sqrt {c^2 x^2+1}}{x^2}+60 a c^2 \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )-60 a c^2 \log (x)-60 b c^2 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )+60 b c^2 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )-\frac {56 b c^2 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}-60 b c^2 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+60 b c^2 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+6 b c^2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-6 b c^2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-3 b c^2 \sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+104 b c^2 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\frac {48 b c^4 x^2 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+\frac {4 b c^3 x}{c^2 x^2+1}-\frac {6 b c \sinh ^{-1}(c x) \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{x}}{24 \pi ^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-8*a*c^2)/(1 + c^2*x^2)^(3/2) + (4*b*c^3*x)/(1 + c^2*x^2) - (48*a*c^2)/Sqrt[1 + c^2*x^2] - (12*a*Sqrt[1 + c^
2*x^2])/x^2 - (56*b*c^2*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) - (48*b*c^4*x^2*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) +
104*b*c^2*ArcTan[Tanh[ArcSinh[c*x]/2]] - 6*b*c^2*Coth[ArcSinh[c*x]/2] - 3*b*c^2*ArcSinh[c*x]*Csch[ArcSinh[c*x]
/2]^2 - 60*b*c^2*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 60*b*c^2*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] -
60*a*c^2*Log[x] + 60*a*c^2*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] - 60*b*c^2*PolyLog[2, -E^(-ArcSinh[c*x])] + 60*b*c^
2*PolyLog[2, E^(-ArcSinh[c*x])] + 6*b*c^2*Tanh[ArcSinh[c*x]/2] - (6*b*c*ArcSinh[c*x]*Tanh[ArcSinh[c*x]/2])/x)/
(24*Pi^(5/2))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\pi ^{3} c^{6} x^{9} + 3 \, \pi ^{3} c^{4} x^{7} + 3 \, \pi ^{3} c^{2} x^{5} + \pi ^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^3*c^6*x^9 + 3*pi^3*c^4*x^7 + 3*pi^3*c^2*x^5 + pi^3*x^3
), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.34, size = 314, normalized size = 1.27 \[ -\frac {a}{2 \pi \,x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{6 \pi \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {5 a \,c^{2}}{2 \pi ^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {5 a \,c^{2} \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{2 \pi ^{\frac {5}{2}}}-\frac {5 b \,x^{2} \arcsinh \left (c x \right ) c^{4}}{2 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {b \,c^{3} x}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}-\frac {10 b \arcsinh \left (c x \right ) c^{2}}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {b c}{2 \pi ^{\frac {5}{2}} x \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{2 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} x^{2}}+\frac {13 b \,c^{2} \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{3 \pi ^{\frac {5}{2}}}+\frac {5 b \,c^{2} \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{2 \pi ^{\frac {5}{2}}}+\frac {5 b \,c^{2} \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{2 \pi ^{\frac {5}{2}}}+\frac {5 b \,c^{2} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{2 \pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^3/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

-1/2*a/Pi/x^2/(Pi*c^2*x^2+Pi)^(3/2)-5/6*a*c^2/Pi/(Pi*c^2*x^2+Pi)^(3/2)-5/2*a*c^2/Pi^2/(Pi*c^2*x^2+Pi)^(1/2)+5/
2*a*c^2/Pi^(5/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2))-5/2*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)*x^2*arcsinh(c*x)*c^4
-1/3*b*c^3*x/Pi^(5/2)/(c^2*x^2+1)-10/3*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*c^2-1/2*b*c/Pi^(5/2)/x/(c^2*x
^2+1)-1/2*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/x^2*arcsinh(c*x)+13/3*b*c^2/Pi^(5/2)*arctan(c*x+(c^2*x^2+1)^(1/2))+5/2*
b*c^2/Pi^(5/2)*dilog(c*x+(c^2*x^2+1)^(1/2))+5/2*b*c^2/Pi^(5/2)*dilog(1+c*x+(c^2*x^2+1)^(1/2))+5/2*b*c^2/Pi^(5/
2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {15 \, c^{2} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{\pi ^{\frac {5}{2}}} - \frac {5 \, c^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} - \frac {15 \, c^{2}}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}} - \frac {3}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{2}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*(15*c^2*arcsinh(1/(c*abs(x)))/pi^(5/2) - 5*c^2/(pi*(pi + pi*c^2*x^2)^(3/2)) - 15*c^2/(pi^2*sqrt(pi + pi*
c^2*x^2)) - 3/(pi*(pi + pi*c^2*x^2)^(3/2)*x^2)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/((pi + pi*c^2*x^2)^
(5/2)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^3\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x^3*(Pi + Pi*c^2*x^2)^(5/2)),x)

[Out]

int((a + b*asinh(c*x))/(x^3*(Pi + Pi*c^2*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{4} x^{7} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{5} \sqrt {c^{2} x^{2} + 1} + x^{3} \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{7} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{5} \sqrt {c^{2} x^{2} + 1} + x^{3} \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**7*sqrt(c**2*x**2 + 1) + 2*c**2*x**5*sqrt(c**2*x**2 + 1) + x**3*sqrt(c**2*x**2 + 1)), x) +
Integral(b*asinh(c*x)/(c**4*x**7*sqrt(c**2*x**2 + 1) + 2*c**2*x**5*sqrt(c**2*x**2 + 1) + x**3*sqrt(c**2*x**2
+ 1)), x))/pi**(5/2)

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