3.90 \(\int \frac{x}{\sinh ^{-1}(a+b x)^3} \, dx\)

Optimal. Leaf size=147 \[ -\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{\sqrt{(a+b x)^2+1} (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a \sqrt{(a+b x)^2+1}}{2 b^2 \sinh ^{-1}(a+b x)^2} \]

[Out]

(a*Sqrt[1 + (a + b*x)^2])/(2*b^2*ArcSinh[a + b*x]^2) - ((a + b*x)*Sqrt[1 + (a + b*x)^2])/(2*b^2*ArcSinh[a + b*
x]^2) - 1/(2*b^2*ArcSinh[a + b*x]) + (a*(a + b*x))/(2*b^2*ArcSinh[a + b*x]) - (a + b*x)^2/(b^2*ArcSinh[a + b*x
]) - (a*CoshIntegral[ArcSinh[a + b*x]])/(2*b^2) + SinhIntegral[2*ArcSinh[a + b*x]]/b^2

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Rubi [A]  time = 0.249732, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.2, Rules used = {5865, 5803, 5655, 5774, 5657, 3301, 5667, 5669, 5448, 12, 3298, 5675} \[ -\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{\sqrt{(a+b x)^2+1} (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a \sqrt{(a+b x)^2+1}}{2 b^2 \sinh ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a + b*x]^3,x]

[Out]

(a*Sqrt[1 + (a + b*x)^2])/(2*b^2*ArcSinh[a + b*x]^2) - ((a + b*x)*Sqrt[1 + (a + b*x)^2])/(2*b^2*ArcSinh[a + b*
x]^2) - 1/(2*b^2*ArcSinh[a + b*x]) + (a*(a + b*x))/(2*b^2*ArcSinh[a + b*x]) - (a + b*x)^2/(b^2*ArcSinh[a + b*x
]) - (a*CoshIntegral[ArcSinh[a + b*x]])/(2*b^2) + SinhIntegral[2*ArcSinh[a + b*x]]/b^2

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5774

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

Rule 5657

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cosh[a/b - x/b], x], x,
a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5667

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c
^2*x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 5675

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

Rubi steps

\begin{align*} \int \frac{x}{\sinh ^{-1}(a+b x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-\frac{a}{b}+\frac{x}{b}}{\sinh ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b \sinh ^{-1}(x)^3}+\frac{x}{b \sinh ^{-1}(x)^3}\right ) \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{\sinh ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sinh ^{-1}(x)^3} \, dx,x,a+b x\right )}{b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2} \sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}+\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}+\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac{2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{a \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{(a+b x) \sqrt{1+(a+b x)^2}}{2 b^2 \sinh ^{-1}(a+b x)^2}-\frac{1}{2 b^2 \sinh ^{-1}(a+b x)}+\frac{a (a+b x)}{2 b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2}{b^2 \sinh ^{-1}(a+b x)}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.10173, size = 117, normalized size = 0.8 \[ -\frac{b x \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2 \sinh ^{-1}(a+b x)+2 b^2 x^2 \sinh ^{-1}(a+b x)+a \sinh ^{-1}(a+b x)^2 \text{Chi}\left (\sinh ^{-1}(a+b x)\right )-2 \sinh ^{-1}(a+b x)^2 \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )+3 a b x \sinh ^{-1}(a+b x)+\sinh ^{-1}(a+b x)}{2 b^2 \sinh ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSinh[a + b*x]^3,x]

[Out]

-(b*x*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + ArcSinh[a + b*x] + a^2*ArcSinh[a + b*x] + 3*a*b*x*ArcSinh[a + b*x] +
2*b^2*x^2*ArcSinh[a + b*x] + a*ArcSinh[a + b*x]^2*CoshIntegral[ArcSinh[a + b*x]] - 2*ArcSinh[a + b*x]^2*SinhI
ntegral[2*ArcSinh[a + b*x]])/(2*b^2*ArcSinh[a + b*x]^2)

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Maple [A]  time = 0.037, size = 107, normalized size = 0.7 \begin{align*}{\frac{1}{{b}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{4\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{2\,{\it Arcsinh} \left ( bx+a \right ) }}+{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) -{\frac{a}{2\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}} \left ({\it Chi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}- \left ( bx+a \right ){\it Arcsinh} \left ( bx+a \right ) -\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arcsinh(b*x+a)^3,x)

[Out]

1/b^2*(-1/4/arcsinh(b*x+a)^2*sinh(2*arcsinh(b*x+a))-1/2/arcsinh(b*x+a)*cosh(2*arcsinh(b*x+a))+Shi(2*arcsinh(b*
x+a))-1/2*a*(Chi(arcsinh(b*x+a))*arcsinh(b*x+a)^2-(b*x+a)*arcsinh(b*x+a)-(1+(b*x+a)^2)^(1/2))/arcsinh(b*x+a)^2
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^8*x^8 + 7*a*b^7*x^7 + 3*(7*a^2*b^6 + b^6)*x^6 + 5*(7*a^3*b^5 + 3*a*b^5)*x^5 + (35*a^4*b^4 + 30*a^2*b^4
+ 3*b^4)*x^4 + 3*(7*a^5*b^3 + 10*a^3*b^3 + 3*a*b^3)*x^3 + (7*a^6*b^2 + 15*a^4*b^2 + 9*a^2*b^2 + b^2)*x^2 + (b
^5*x^5 + 4*a*b^4*x^4 + (6*a^2*b^3 + b^3)*x^3 + 2*(2*a^3*b^2 + a*b^2)*x^2 + (a^4*b + a^2*b)*x)*(b^2*x^2 + 2*a*b
*x + a^2 + 1)^(3/2) + (3*b^6*x^6 + 15*a*b^5*x^5 + 5*(6*a^2*b^4 + b^4)*x^4 + 15*(2*a^3*b^3 + a*b^3)*x^3 + (15*a
^4*b^2 + 15*a^2*b^2 + 2*b^2)*x^2 + (3*a^5*b + 5*a^3*b + 2*a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (a^7*b + 3*a
^5*b + 3*a^3*b + a*b)*x + (2*b^8*x^8 + 15*a*b^7*x^7 + a^8 + (49*a^2*b^6 + 6*b^6)*x^6 + 3*a^6 + (91*a^3*b^5 + 3
3*a*b^5)*x^5 + 3*(35*a^4*b^4 + 25*a^2*b^4 + 2*b^4)*x^4 + 3*a^4 + (77*a^5*b^3 + 90*a^3*b^3 + 21*a*b^3)*x^3 + (3
5*a^6*b^2 + 60*a^4*b^2 + 27*a^2*b^2 + 2*b^2)*x^2 + (2*b^5*x^5 + 9*a*b^4*x^4 + a^5 + 2*(8*a^2*b^3 + b^3)*x^3 +
2*a^3 + 2*(7*a^3*b^2 + 3*a*b^2)*x^2 + 6*(a^4*b + a^2*b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + (6*b^6*x^
6 + 33*a*b^5*x^5 + 3*a^6 + 5*(15*a^2*b^4 + 2*b^4)*x^4 + 7*a^4 + (90*a^3*b^3 + 37*a*b^3)*x^3 + (60*a^4*b^2 + 51
*a^2*b^2 + 5*b^2)*x^2 + 5*a^2 + (21*a^5*b + 31*a^3*b + 10*a*b)*x + 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + a^2 + 3*
(3*a^7*b + 7*a^5*b + 5*a^3*b + a*b)*x + (6*b^7*x^7 + 39*a*b^6*x^6 + 3*a^7 + 2*(54*a^2*b^5 + 7*b^5)*x^5 + 8*a^5
+ (165*a^3*b^4 + 64*a*b^4)*x^4 + (150*a^4*b^3 + 116*a^2*b^3 + 11*b^3)*x^3 + 7*a^3 + (81*a^5*b^2 + 104*a^3*b^2
+ 29*a*b^2)*x^2 + (24*a^6*b + 46*a^4*b + 25*a^2*b + 3*b)*x + 2*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*log(b*x
+ a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (3*b^7*x^7 + 18*a*b^6*x^6 + (45*a^2*b^5 + 7*b^5)*x^5 + 4*(15*a^3*b^
4 + 7*a*b^4)*x^4 + (45*a^4*b^3 + 42*a^2*b^3 + 5*b^3)*x^3 + 2*(9*a^5*b^2 + 14*a^3*b^2 + 5*a*b^2)*x^2 + (3*a^6*b
+ 7*a^4*b + 5*a^2*b + b)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((b^8*x^6 + 6*a*b^7*x^5 + a^6*b^2 + 3*a^4*b^2
+ 3*(5*a^2*b^6 + b^6)*x^4 + 3*a^2*b^2 + 4*(5*a^3*b^5 + 3*a*b^5)*x^3 + 3*(5*a^4*b^4 + 6*a^2*b^4 + b^4)*x^2 + (b
^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 3*(b^6*x^4 + 4*a*b^5*x^3 +
a^4*b^2 + a^2*b^2 + (6*a^2*b^4 + b^4)*x^2 + 2*(2*a^3*b^3 + a*b^3)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + b^2 + 6*
(a^5*b^3 + 2*a^3*b^3 + a*b^3)*x + 3*(b^7*x^5 + 5*a*b^6*x^4 + a^5*b^2 + 2*a^3*b^2 + 2*(5*a^2*b^5 + b^5)*x^3 + a
*b^2 + 2*(5*a^3*b^4 + 3*a*b^4)*x^2 + (5*a^4*b^3 + 6*a^2*b^3 + b^3)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*log(b
*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2) + integrate(1/2*(4*b^9*x^9 + 35*a*b^8*x^8 + 3*a^9 + 8*(17*a^2*b
^7 + 2*b^7)*x^7 + 12*a^7 + 4*(77*a^3*b^6 + 27*a*b^6)*x^6 + 8*(56*a^4*b^5 + 39*a^2*b^5 + 3*b^5)*x^5 + 18*a^5 +
2*(217*a^5*b^4 + 250*a^3*b^4 + 57*a*b^4)*x^4 + 8*(35*a^6*b^3 + 60*a^4*b^3 + 27*a^2*b^3 + 2*b^3)*x^3 + (4*b^5*x
^5 + 19*a*b^4*x^4 + 36*a^2*b^3*x^3 + 34*a^3*b^2*x^2 + 16*a^4*b*x + 3*a^5 - 3*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^
2 + 12*a^3 + 4*(29*a^7*b^2 + 69*a^5*b^2 + 51*a^3*b^2 + 11*a*b^2)*x^2 + (16*b^6*x^6 + 92*a*b^5*x^5 + 12*a^6 + 4
*(55*a^2*b^4 + 4*b^4)*x^4 + 12*a^4 + 20*(14*a^3*b^3 + 3*a*b^3)*x^3 + 4*(50*a^4*b^2 + 21*a^2*b^2)*x^2 - 3*a^2 +
(76*a^5*b + 52*a^3*b - 3*a*b)*x - 3)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 3*(8*b^7*x^7 + 54*a*b^6*x^6 + 6*a^
7 + 4*(39*a^2*b^5 + 4*b^5)*x^5 + 12*a^5 + 2*(125*a^3*b^4 + 38*a*b^4)*x^4 + 8*(30*a^4*b^3 + 18*a^2*b^3 + b^3)*x
^3 + 7*a^3 + (138*a^5*b^2 + 136*a^3*b^2 + 23*a*b^2)*x^2 + 2*(22*a^6*b + 32*a^4*b + 11*a^2*b)*x + a)*(b^2*x^2 +
2*a*b*x + a^2 + 1) + 4*(7*a^8*b + 22*a^6*b + 24*a^4*b + 10*a^2*b + b)*x + (16*b^8*x^8 + 124*a*b^7*x^7 + 12*a^
8 + 12*(35*a^2*b^6 + 4*b^6)*x^6 + 36*a^6 + 4*(203*a^3*b^5 + 69*a*b^5)*x^5 + 4*(245*a^4*b^4 + 165*a^2*b^4 + 12*
b^4)*x^4 + 39*a^4 + 3*(252*a^5*b^3 + 280*a^3*b^3 + 61*a*b^3)*x^3 + (364*a^6*b^2 + 600*a^4*b^2 + 261*a^2*b^2 +
19*b^2)*x^2 + 18*a^2 + (100*a^7*b + 228*a^5*b + 165*a^3*b + 37*a*b)*x + 3)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) +
3*a)/((b^9*x^8 + 8*a*b^8*x^7 + a^8*b + 4*a^6*b + 4*(7*a^2*b^7 + b^7)*x^6 + 8*(7*a^3*b^6 + 3*a*b^6)*x^5 + 6*a^
4*b + 2*(35*a^4*b^5 + 30*a^2*b^5 + 3*b^5)*x^4 + 8*(7*a^5*b^4 + 10*a^3*b^4 + 3*a*b^4)*x^3 + (b^5*x^4 + 4*a*b^4*
x^3 + 6*a^2*b^3*x^2 + 4*a^3*b^2*x + a^4*b)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 4*a^2*b + 4*(7*a^6*b^3 + 15*a^4*b
^3 + 9*a^2*b^3 + b^3)*x^2 + 4*(b^6*x^5 + 5*a*b^5*x^4 + a^5*b + a^3*b + (10*a^2*b^4 + b^4)*x^3 + (10*a^3*b^3 +
3*a*b^3)*x^2 + (5*a^4*b^2 + 3*a^2*b^2)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 6*(b^7*x^6 + 6*a*b^6*x^5 + a^6
*b + 2*a^4*b + (15*a^2*b^5 + 2*b^5)*x^4 + 4*(5*a^3*b^4 + 2*a*b^4)*x^3 + a^2*b + (15*a^4*b^3 + 12*a^2*b^3 + b^3
)*x^2 + 2*(3*a^5*b^2 + 4*a^3*b^2 + a*b^2)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 8*(a^7*b^2 + 3*a^5*b^2 + 3*a^3*b^
2 + a*b^2)*x + 4*(b^8*x^7 + 7*a*b^7*x^6 + a^7*b + 3*a^5*b + 3*(7*a^2*b^6 + b^6)*x^5 + 5*(7*a^3*b^5 + 3*a*b^5)*
x^4 + 3*a^3*b + (35*a^4*b^4 + 30*a^2*b^4 + 3*b^4)*x^3 + 3*(7*a^5*b^3 + 10*a^3*b^3 + 3*a*b^3)*x^2 + a*b + (7*a^
6*b^2 + 15*a^4*b^2 + 9*a^2*b^2 + b^2)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + b)*log(b*x + a + sqrt(b^2*x^2 + 2
*a*b*x + a^2 + 1))), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\operatorname{arsinh}\left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsinh(b*x + a)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{asinh}^{3}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/asinh(b*x+a)**3,x)

[Out]

Integral(x/asinh(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/arcsinh(b*x + a)^3, x)