3.153 \(\int \frac{(a+b \sinh ^{-1}(c+d x))^4}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=186 \[ -\frac{6 b^3 \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b^4 \text{PolyLog}\left (3,e^{-2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^2 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^3}-\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}+\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]

[Out]

(2*b*(a + b*ArcSinh[c + d*x])^3)/(d*e^3) - (2*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/(d*e^3*(c +
d*x)) - (a + b*ArcSinh[c + d*x])^4/(2*d*e^3*(c + d*x)^2) + (6*b^2*(a + b*ArcSinh[c + d*x])^2*Log[1 - E^(-2*Arc
Sinh[c + d*x])])/(d*e^3) - (6*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[2, E^(-2*ArcSinh[c + d*x])])/(d*e^3) - (3*b
^4*PolyLog[3, E^(-2*ArcSinh[c + d*x])])/(d*e^3)

________________________________________________________________________________________

Rubi [A]  time = 0.32809, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {5865, 12, 5661, 5723, 5659, 3716, 2190, 2531, 2282, 6589} \[ \frac{6 b^3 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b^4 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e^3}-\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcSinh[c + d*x])^4/(c*e + d*e*x)^3,x]

[Out]

(-2*b*(a + b*ArcSinh[c + d*x])^3)/(d*e^3) - (2*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/(d*e^3*(c +
d*x)) - (a + b*ArcSinh[c + d*x])^4/(2*d*e^3*(c + d*x)^2) + (6*b^2*(a + b*ArcSinh[c + d*x])^2*Log[1 - E^(2*Arc
Sinh[c + d*x])])/(d*e^3) + (6*b^3*(a + b*ArcSinh[c + d*x])*PolyLog[2, E^(2*ArcSinh[c + d*x])])/(d*e^3) - (3*b^
4*PolyLog[3, E^(2*ArcSinh[c + d*x])])/(d*e^3)

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 5661

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

Rule 5723

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[(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*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 5659

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

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{(c e+d e x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^4}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{x^2 \sqrt{1+x^2}} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \coth (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)^2}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac{\left (12 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac{\left (6 b^4\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}\\ &=-\frac{2 b \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^3 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}-\frac{3 b^4 \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{d e^3}\\ \end{align*}

Mathematica [C]  time = 1.23193, size = 360, normalized size = 1.94 \[ \frac{8 a b^3 \left (\sinh ^{-1}(c+d x) \left (-\frac{\sinh ^{-1}(c+d x)^2}{(c+d x)^2}-\frac{3 \sqrt{(c+d x)^2+1} \sinh ^{-1}(c+d x)}{c+d x}+3 \sinh ^{-1}(c+d x)+6 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right )\right )-3 \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c+d x)}\right )\right )+b^4 \left (24 \sinh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right )-12 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right )-\frac{8 \sqrt{(c+d x)^2+1} \sinh ^{-1}(c+d x)^3}{c+d x}-8 \sinh ^{-1}(c+d x)^3+24 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )+i \pi ^3\right )+24 a^2 b^2 \left (\log (c+d x)-\frac{\sinh ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac{\sqrt{(c+d x)^2+1} \sinh ^{-1}(c+d x)}{c+d x}\right )-\frac{8 a^3 b \sqrt{(c+d x)^2+1}}{c+d x}-\frac{8 a^3 b \sinh ^{-1}(c+d x)}{(c+d x)^2}-\frac{2 a^4}{(c+d x)^2}-\frac{2 b^4 \sinh ^{-1}(c+d x)^4}{(c+d x)^2}}{4 d e^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^4/(c*e + d*e*x)^3,x]

[Out]

((-2*a^4)/(c + d*x)^2 - (8*a^3*b*Sqrt[1 + (c + d*x)^2])/(c + d*x) - (8*a^3*b*ArcSinh[c + d*x])/(c + d*x)^2 - (
2*b^4*ArcSinh[c + d*x]^4)/(c + d*x)^2 + 24*a^2*b^2*(-((Sqrt[1 + (c + d*x)^2]*ArcSinh[c + d*x])/(c + d*x)) - Ar
cSinh[c + d*x]^2/(2*(c + d*x)^2) + Log[c + d*x]) + 8*a*b^3*(ArcSinh[c + d*x]*(3*ArcSinh[c + d*x] - (3*Sqrt[1 +
(c + d*x)^2]*ArcSinh[c + d*x])/(c + d*x) - ArcSinh[c + d*x]^2/(c + d*x)^2 + 6*Log[1 - E^(-2*ArcSinh[c + d*x])
]) - 3*PolyLog[2, E^(-2*ArcSinh[c + d*x])]) + b^4*(I*Pi^3 - 8*ArcSinh[c + d*x]^3 - (8*Sqrt[1 + (c + d*x)^2]*Ar
cSinh[c + d*x]^3)/(c + d*x) + 24*ArcSinh[c + d*x]^2*Log[1 - E^(2*ArcSinh[c + d*x])] + 24*ArcSinh[c + d*x]*Poly
Log[2, E^(2*ArcSinh[c + d*x])] - 12*PolyLog[3, E^(2*ArcSinh[c + d*x])]))/(4*d*e^3)

________________________________________________________________________________________

Maple [B]  time = 0.066, size = 723, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^4/(d*e*x+c*e)^3,x)

[Out]

-1/2/d*a^4/e^3/(d*x+c)^2-2/d*b^4/e^3*arcsinh(d*x+c)^3/(d*x+c)*(1+(d*x+c)^2)^(1/2)-2/d*b^4/e^3*arcsinh(d*x+c)^3
-1/2/d*b^4/e^3*arcsinh(d*x+c)^4/(d*x+c)^2+6/d*b^4/e^3*arcsinh(d*x+c)^2*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+12/d*b^
4/e^3*arcsinh(d*x+c)*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-12/d*b^4/e^3*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))+
6/d*b^4/e^3*arcsinh(d*x+c)^2*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+12/d*b^4/e^3*arcsinh(d*x+c)*polylog(2,d*x+c+(1+(d
*x+c)^2)^(1/2))-12/d*b^4/e^3*polylog(3,d*x+c+(1+(d*x+c)^2)^(1/2))-6/d*a*b^3/e^3*arcsinh(d*x+c)^2/(d*x+c)*(1+(d
*x+c)^2)^(1/2)-6/d*a*b^3/e^3*arcsinh(d*x+c)^2-2/d*a*b^3/e^3*arcsinh(d*x+c)^3/(d*x+c)^2+12/d*a*b^3/e^3*arcsinh(
d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+12/d*a*b^3/e^3*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))+12/d*a*b^3/e^3*arc
sinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+12/d*a*b^3/e^3*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-6/d*a^2*b^2/e^
3*arcsinh(d*x+c)-6/d*a^2*b^2/e^3*arcsinh(d*x+c)/(d*x+c)*(1+(d*x+c)^2)^(1/2)-3/d*a^2*b^2/e^3*arcsinh(d*x+c)^2/(
d*x+c)^2+6/d*a^2*b^2/e^3*ln((d*x+c+(1+(d*x+c)^2)^(1/2))^2-1)-2/d*a^3*b/e^3/(d*x+c)^2*arcsinh(d*x+c)-2/d*a^3*b/
e^3/(d*x+c)*(1+(d*x+c)^2)^(1/2)

________________________________________________________________________________________

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(d*x+c))^4/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^4/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

integral((b^4*arcsinh(d*x + c)^4 + 4*a*b^3*arcsinh(d*x + c)^3 + 6*a^2*b^2*arcsinh(d*x + c)^2 + 4*a^3*b*arcsinh
(d*x + c) + a^4)/(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), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{4}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{b^{4} \operatorname{asinh}^{4}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{4 a b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{6 a^{2} b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac{4 a^{3} b \operatorname{asinh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))**4/(d*e*x+c*e)**3,x)

[Out]

(Integral(a**4/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(b**4*asinh(c + d*x)**4/(c**3 + 3
*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(4*a*b**3*asinh(c + d*x)**3/(c**3 + 3*c**2*d*x + 3*c*d**2
*x**2 + d**3*x**3), x) + Integral(6*a**2*b**2*asinh(c + d*x)**2/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3
), x) + Integral(4*a**3*b*asinh(c + d*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x))/e**3

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^4/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^4/(d*e*x + c*e)^3, x)