3.169 \(\int \frac{(c e+d e x)^3}{(a+b \sinh ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=247 \[ \frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}-\frac{e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(2*b*d*(a + b*ArcSinh[c + d*x])^2) - (3*e^3*(c + d*x)^2)/(2*b^2*d*(a
+ b*ArcSinh[c + d*x])) - (2*e^3*(c + d*x)^4)/(b^2*d*(a + b*ArcSinh[c + d*x])) + (e^3*CoshIntegral[(2*(a + b*Ar
cSinh[c + d*x]))/b]*Sinh[(2*a)/b])/(2*b^3*d) - (e^3*CoshIntegral[(4*(a + b*ArcSinh[c + d*x]))/b]*Sinh[(4*a)/b]
)/(b^3*d) - (e^3*Cosh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c + d*x]))/b])/(2*b^3*d) + (e^3*Cosh[(4*a)/b]*Si
nhIntegral[(4*(a + b*ArcSinh[c + d*x]))/b])/(b^3*d)

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Rubi [A]  time = 0.681912, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3303, 3298, 3301} \[ \frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^3 d}-\frac{e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{b^3 d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^3 d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{b^3 d}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(2*b*d*(a + b*ArcSinh[c + d*x])^2) - (3*e^3*(c + d*x)^2)/(2*b^2*d*(a
+ b*ArcSinh[c + d*x])) - (2*e^3*(c + d*x)^4)/(b^2*d*(a + b*ArcSinh[c + d*x])) + (e^3*CoshIntegral[(2*a)/b + 2*
ArcSinh[c + d*x]]*Sinh[(2*a)/b])/(2*b^3*d) - (e^3*CoshIntegral[(4*a)/b + 4*ArcSinh[c + d*x]]*Sinh[(4*a)/b])/(b
^3*d) - (e^3*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c + d*x]])/(2*b^3*d) + (e^3*Cosh[(4*a)/b]*SinhInte
gral[(4*a)/b + 4*ArcSinh[c + d*x]])/(b^3*d)

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 12

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

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 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 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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

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 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]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}+\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 (a+b x)} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \left (-\frac{\sinh (2 x)}{4 (a+b x)}+\frac{\sinh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (3 e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (2 e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (e^3 \cosh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (3 e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (2 e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (e^3 \sinh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^3 \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{2 a}{b}\right )}{2 b^3 d}-\frac{e^3 \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{4 a}{b}\right )}{b^3 d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{2 b^3 d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.660613, size = 179, normalized size = 0.72 \[ \frac{e^3 \left (-\frac{b^2 \sqrt{(c+d x)^2+1} (c+d x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-2 \sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+2 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\frac{b \left (-4 (c+d x)^4-3 (c+d x)^2\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-((b^2*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^2) + (b*(-3*(c + d*x)^2 - 4*(c + d*x)
^4))/(a + b*ArcSinh[c + d*x]) + CoshIntegral[2*(a/b + ArcSinh[c + d*x])]*Sinh[(2*a)/b] - 2*CoshIntegral[4*(a/b
+ ArcSinh[c + d*x])]*Sinh[(4*a)/b] - Cosh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c + d*x])] + 2*Cosh[(4*a)/b]
*SinhIntegral[4*(a/b + ArcSinh[c + d*x])]))/(2*b^3*d)

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Maple [B]  time = 0.127, size = 579, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/32*(8*(d*x+c)^4-8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+8*(d*x+c)^2-4*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)*e^3*(4*b*
arcsinh(d*x+c)+4*a-b)/b^2/(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)+a^2)+1/2*e^3/b^3*exp(4*a/b)*Ei(1,4*arcsin
h(d*x+c)+4*a/b)+1/16*(2*(d*x+c)^2-2*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)*e^3*(2*b*arcsinh(d*x+c)+2*a-b)/b^2/(b^2*arc
sinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)+a^2)-1/4*e^3/b^3*exp(2*a/b)*Ei(1,2*arcsinh(d*x+c)+2*a/b)+1/16*e^3/b*(2*(d*x
+c)^2+1+2*(d*x+c)*(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^2+1/8*e^3/b^2*(2*(d*x+c)^2+1+2*(d*x+c)*(1+(d*x+c)^
2)^(1/2))/(a+b*arcsinh(d*x+c))+1/4*e^3/b^3*exp(-2*a/b)*Ei(1,-2*arcsinh(d*x+c)-2*a/b)-1/32*e^3/b*(8*(d*x+c)^4+8
*(d*x+c)^2+8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)/(a+b*arcsinh(d*x+c))^2-1/8*e^3/b^2
*(8*(d*x+c)^4+8*(d*x+c)^2+8*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+4*(d*x+c)*(1+(d*x+c)^2)^(1/2)+1)/(a+b*arcsinh(d*x+c)
)-1/2*e^3/b^3*exp(-4*a/b)*Ei(1,-4*arcsinh(d*x+c)-4*a/b))

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Maxima [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((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3)/(b^3*arcsinh(d*x + c)^3 + 3*a*b^2*arcsinh(d
*x + c)^2 + 3*a^2*b*arcsinh(d*x + c) + a^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**3*(Integral(c**3/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)**3), x)
+ Integral(d**3*x**3/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)**3),
x) + Integral(3*c*d**2*x**2/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d*x)
**3), x) + Integral(3*c**2*d*x/(a**3 + 3*a**2*b*asinh(c + d*x) + 3*a*b**2*asinh(c + d*x)**2 + b**3*asinh(c + d
*x)**3), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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