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

Optimal. Leaf size=400 \[ \frac{5 b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt{c^2 d x^2+d}}-\frac{5 b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt{c^2 d x^2+d}}-\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt{c^2 d x^2+d}}+\frac{5 c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{2 d x^2 \left (c^2 d x^2+d\right )^{3/2}}+\frac{5 b c^3 x}{12 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{3 b c \sqrt{c^2 x^2+1}}{4 d^2 x \sqrt{c^2 d x^2+d}}+\frac{b c}{4 d^2 x \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{13 b c^2 \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.559931, antiderivative size = 400, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {5747, 5755, 5764, 5760, 4182, 2279, 2391, 203, 199, 290, 325} \[ \frac{5 b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt{c^2 d x^2+d}}-\frac{5 b c^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \sqrt{c^2 d x^2+d}}-\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \sqrt{c^2 d x^2+d}}+\frac{5 c^2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 d x^2+d}}-\frac{5 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{2 d x^2 \left (c^2 d x^2+d\right )^{3/2}}+\frac{5 b c^3 x}{12 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{3 b c \sqrt{c^2 x^2+1}}{4 d^2 x \sqrt{c^2 d x^2+d}}+\frac{b c}{4 d^2 x \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{13 b c^2 \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5764

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

Rubi steps

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

Mathematica [A]  time = 6.57009, size = 437, normalized size = 1.09 \[ \frac{b c^2 \left (-60 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+\frac{4 c x}{\sqrt{c^2 x^2+1}}-\frac{8 \sinh ^{-1}(c x)}{c^2 x^2+1}-60 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+60 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+6 \sqrt{c^2 x^2+1} \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-6 \sqrt{c^2 x^2+1} \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-3 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-3 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )+104 \sqrt{c^2 x^2+1} \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-48 \sinh ^{-1}(c x)\right )}{24 d^2 \sqrt{d \left (c^2 x^2+1\right )}}+\sqrt{d \left (c^2 x^2+1\right )} \left (-\frac{2 a c^2}{d^3 \left (c^2 x^2+1\right )}-\frac{a c^2}{3 d^3 \left (c^2 x^2+1\right )^2}-\frac{a}{2 d^3 x^2}\right )+\frac{5 a c^2 \log \left (\sqrt{d} \sqrt{d \left (c^2 x^2+1\right )}+d\right )}{2 d^{5/2}}-\frac{5 a c^2 \log (x)}{2 d^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[d*(1 + c^2*x^2)]*(-a/(2*d^3*x^2) - (a*c^2)/(3*d^3*(1 + c^2*x^2)^2) - (2*a*c^2)/(d^3*(1 + c^2*x^2))) - (5*
a*c^2*Log[x])/(2*d^(5/2)) + (5*a*c^2*Log[d + Sqrt[d]*Sqrt[d*(1 + c^2*x^2)]])/(2*d^(5/2)) + (b*c^2*((4*c*x)/Sqr
t[1 + c^2*x^2] - 48*ArcSinh[c*x] - (8*ArcSinh[c*x])/(1 + c^2*x^2) + 104*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[
c*x]/2]] - 6*Sqrt[1 + c^2*x^2]*Coth[ArcSinh[c*x]/2] - 3*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2
- 60*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 60*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + E^(
-ArcSinh[c*x])] - 60*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(-ArcSinh[c*x])] + 60*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(-Ar
cSinh[c*x])] - 3*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 6*Sqrt[1 + c^2*x^2]*Tanh[ArcSinh[c*x]
/2]))/(24*d^2*Sqrt[d*(1 + c^2*x^2)])

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Maple [A]  time = 0.204, size = 546, normalized size = 1.4 \begin{align*} -{\frac{a}{2\,d{x}^{2}} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,a{c}^{2}}{6\,d} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,a{c}^{2}}{2\,{d}^{2}}{\frac{1}{\sqrt{{c}^{2}d{x}^{2}+d}}}}+{\frac{5\,a{c}^{2}}{2}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{{c}^{2}d{x}^{2}+d} \right ) } \right ){d}^{-{\frac{5}{2}}}}-{\frac{5\,b{x}^{2}{\it Arcsinh} \left ( cx \right ){c}^{4}}{ \left ( 2\,{c}^{4}{x}^{4}+4\,{c}^{2}{x}^{2}+2 \right ){d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bx{c}^{3}}{ \left ( 3\,{c}^{4}{x}^{4}+6\,{c}^{2}{x}^{2}+3 \right ){d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{10\,b{\it Arcsinh} \left ( cx \right ){c}^{2}}{ \left ( 3\,{c}^{4}{x}^{4}+6\,{c}^{2}{x}^{2}+3 \right ){d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bc}{ \left ( 2\,{c}^{4}{x}^{4}+4\,{c}^{2}{x}^{2}+2 \right ){d}^{3}x}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{ \left ( 2\,{c}^{4}{x}^{4}+4\,{c}^{2}{x}^{2}+2 \right ){d}^{3}{x}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{13\,b{c}^{2}}{3\,{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\arctan \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,b{c}^{2}}{2\,{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it dilog} \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,b{c}^{2}}{2\,{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it dilog} \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,b{\it Arcsinh} \left ( cx \right ){c}^{2}}{2\,{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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