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

Optimal. Leaf size=157 \[ \frac {3 b^2 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}-\frac {3 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}+\frac {3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}-\frac {3 b^3 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right )}{2 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5865, 12, 5661, 5723, 5659, 3716, 2190, 2279, 2391} \[ \frac {3 b^3 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e^3}+\frac {3 b^2 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^3}-\frac {3 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac {3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.81, size = 229, normalized size = 1.46 \[ -\frac {a \left (a \left (a+3 b (c+d x) \sqrt {c^2+2 c d x+d^2 x^2+1}\right )-6 b^2 (c+d x)^2 \log (c+d x)\right )+3 b^2 \sinh ^{-1}(c+d x)^2 \left (a+b (c+d x) \left (\sqrt {c^2+2 c d x+d^2 x^2+1}-c-d x\right )\right )+3 b \sinh ^{-1}(c+d x) \left (a \left (a+2 b (c+d x) \sqrt {c^2+2 c d x+d^2 x^2+1}\right )-2 b^2 (c+d x)^2 \log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right )\right )+3 b^3 (c+d x)^2 \text {Li}_2\left (e^{-2 \sinh ^{-1}(c+d x)}\right )+b^3 \sinh ^{-1}(c+d x)^3}{2 d e^3 (c+d x)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {arsinh}\left (d x + c\right ) + a^{3}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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)/(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.22, size = 409, normalized size = 2.61 \[ -\frac {a^{3}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {3 b^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{2 d \,e^{3} \left (d x +c \right )}-\frac {3 b^{3} \arcsinh \left (d x +c \right )^{2}}{2 d \,e^{3}}-\frac {b^{3} \arcsinh \left (d x +c \right )^{3}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {3 b^{3} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {3 b^{3} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {3 b^{3} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}+\frac {3 b^{3} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{d \,e^{3}}-\frac {3 a \,b^{2} \arcsinh \left (d x +c \right )}{d \,e^{3}}-\frac {3 a \,b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{d \,e^{3} \left (d x +c \right )}-\frac {3 a \,b^{2} \arcsinh \left (d x +c \right )^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {3 a \,b^{2} \ln \left (\left (d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )^{2}-1\right )}{d \,e^{3}}-\frac {3 a^{2} b \arcsinh \left (d x +c \right )}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {3 a^{2} b \sqrt {1+\left (d x +c \right )^{2}}}{2 d \,e^{3} \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -3 \, {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d \operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c\right )}{d e^{3}}\right )} a b^{2} - \frac {1}{2} \, {\left (\frac {\log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3}}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}} - 2 \, \int \frac {3 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (d x + c\right )} + 1\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}}{2 \, {\left (d^{5} e^{3} x^{5} + 5 \, c d^{4} e^{3} x^{4} + c^{5} e^{3} + c^{3} e^{3} + {\left (10 \, c^{2} d^{3} e^{3} + d^{3} e^{3}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{3} + 3 \, c d^{2} e^{3}\right )} x^{2} + {\left (5 \, c^{4} d e^{3} + 3 \, c^{2} d e^{3}\right )} x + {\left (d^{4} e^{3} x^{4} + 4 \, c d^{3} e^{3} x^{3} + c^{4} e^{3} + c^{2} e^{3} + {\left (6 \, c^{2} d^{2} e^{3} + d^{2} e^{3}\right )} x^{2} + 2 \, {\left (2 \, c^{3} d e^{3} + c d e^{3}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}}\,{d x}\right )} b^{3} - \frac {3}{2} \, a^{2} b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac {\operatorname {arsinh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac {3 \, a b^{2} \operatorname {arsinh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac {a^{3}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {asinh}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {asinh}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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