3.2.53 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^4}{(c e+d e x)^3} \, dx\) [153]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.27, antiderivative size = 186, normalized size of antiderivative = 1.00, 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 = {5859, 12, 5776, 5800, 5775, 3797, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {6 b^3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \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}-\frac {3 b^4 \text {Li}_3\left (e^{-2 \sinh ^{-1}(c+d x)}\right )}{d e^3} \end {gather*}

Antiderivative was successfully verified.

[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 12

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3797

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))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

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

Rule 5776

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 5800

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/(f*(m + 1)))*Simp[(
d + e*x^2)^p/(1 + c^2*x^2)^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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 5859

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 6724

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 {\text {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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] Result contains complex when optimal does not.
time = 0.86, size = 360, normalized size = 1.94 \begin {gather*} \frac {-\frac {2 a^4}{(c+d x)^2}-\frac {8 a^3 b \sqrt {1+(c+d x)^2}}{c+d x}-\frac {8 a^3 b \sinh ^{-1}(c+d x)}{(c+d x)^2}-\frac {2 b^4 \sinh ^{-1}(c+d x)^4}{(c+d x)^2}+24 a^2 b^2 \left (-\frac {\sqrt {1+(c+d x)^2} \sinh ^{-1}(c+d x)}{c+d x}-\frac {\sinh ^{-1}(c+d x)^2}{2 (c+d x)^2}+\log (c+d x)\right )+8 a b^3 \left (\sinh ^{-1}(c+d x) \left (3 \sinh ^{-1}(c+d x)-\frac {3 \sqrt {1+(c+d x)^2} \sinh ^{-1}(c+d x)}{c+d x}-\frac {\sinh ^{-1}(c+d x)^2}{(c+d x)^2}+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 (i \pi ^3-8 \sinh ^{-1}(c+d x)^3-\frac {8 \sqrt {1+(c+d x)^2} \sinh ^{-1}(c+d x)^3}{c+d x}+24 \sinh ^{-1}(c+d x)^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )+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 )\right )}{4 d e^3} \end {gather*}

Antiderivative was successfully verified.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(653\) vs. \(2(206)=412\).
time = 5.12, size = 654, normalized size = 3.52

method	result	size
derivativedivides	\(\frac {-\frac {a^{4}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3}}{e^{3}}-\frac {b^{4} \arcsinh \left (d x +c \right )^{4}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 b^{4} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 b^{4} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2}}{e^{3}}-\frac {2 a \,b^{3} \arcsinh \left (d x +c \right )^{3}}{e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right )}{e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {3 a^{2} b^{2} \arcsinh \left (d x +c \right )^{2}}{e^{3} \left (d x +c \right )^{2}}+\frac {6 a^{2} b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{e^{3}}+\frac {4 a^{3} b \left (-\frac {\arcsinh \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(654\)
default	\(\frac {-\frac {a^{4}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {2 b^{4} \arcsinh \left (d x +c \right )^{3}}{e^{3}}-\frac {b^{4} \arcsinh \left (d x +c \right )^{4}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 b^{4} \polylog \left (3, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {6 b^{4} \arcsinh \left (d x +c \right )^{2} \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \arcsinh \left (d x +c \right ) \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 b^{4} \polylog \left (3, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {6 a \,b^{3} \arcsinh \left (d x +c \right )^{2}}{e^{3}}-\frac {2 a \,b^{3} \arcsinh \left (d x +c \right )^{3}}{e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right )}{e^{3}}-\frac {6 a^{2} b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {3 a^{2} b^{2} \arcsinh \left (d x +c \right )^{2}}{e^{3} \left (d x +c \right )^{2}}+\frac {6 a^{2} b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{e^{3}}+\frac {4 a^{3} b \left (-\frac {\arcsinh \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(654\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-1/2*b^4*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4/(d^3*x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3) - 6*(sqr
t(d^2*x^2 + 2*c*d*x + c^2 + 1)*d*arcsinh(d*x + c)/(d^3*x*e^3 + c*d^2*e^3) - e^(-3)*log(d*x + c)/d)*a^2*b^2 - 2
*a^3*b*(sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*d/(d^3*x*e^3 + c*d^2*e^3) + arcsinh(d*x + c)/(d^3*x^2*e^3 + 2*c*d^2*
x*e^3 + c^2*d*e^3)) - 3*a^2*b^2*arcsinh(d*x + c)^2/(d^3*x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3) - 1/2*a^4/(d^3*x^
2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3) + integrate(2*(2*(c^3 + c)*a*b^3 + (c^3 + c)*b^4 + (2*a*b^3*d^3 + b^4*d^3)*
x^3 + 3*(2*a*b^3*c*d^2 + b^4*c*d^2)*x^2 + (2*(3*c^2*d + d)*a*b^3 + (3*c^2*d + d)*b^4)*x + (b^4*c^2 + 2*(c^2 +
1)*a*b^3 + (2*a*b^3*d^2 + b^4*d^2)*x^2 + 2*(2*a*b^3*c*d + b^4*c*d)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d
*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3/(d^6*x^6*e^3 + 6*c*d^5*x^5*e^3 + (15*c^2*d^4 + d^4)*x^4*e^3 + 4*
(5*c^3*d^3 + c*d^3)*x^3*e^3 + 3*(5*c^4*d^2 + 2*c^2*d^2)*x^2*e^3 + 2*(3*c^5*d + 2*c^3*d)*x*e^3 + (c^6 + c^4)*e^
3 + (d^5*x^5*e^3 + 5*c*d^4*x^4*e^3 + (10*c^2*d^3 + d^3)*x^3*e^3 + (10*c^3*d^2 + 3*c*d^2)*x^2*e^3 + (5*c^4*d +
3*c^2*d)*x*e^3 + (c^5 + c^3)*e^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*e^(-3)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________