3.119 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=186 \[ -\frac {6 i b^2 \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {6 i b^2 \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2}+\frac {6 i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arccosh(d*x+c))^3/d/e^2/(d*x+c)+6*b*(a+b*arccosh(d*x+c))^2*arctan(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))
/d/e^2-6*I*b^2*(a+b*arccosh(d*x+c))*polylog(2,-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2+6*I*b^2*(a+b*a
rccosh(d*x+c))*polylog(2,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2+6*I*b^3*polylog(3,-I*(d*x+c+(d*x+c-1
)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2-6*I*b^3*polylog(3,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^2

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Rubi [A]  time = 0.36, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5866, 12, 5662, 5761, 4180, 2531, 2282, 6589} \[ -\frac {6 i b^2 \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {6 i b^2 \text {PolyLog}\left (2,i e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {6 i b^3 \text {PolyLog}\left (3,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {PolyLog}\left (3,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])^3/(c*e + d*e*x)^2,x]

[Out]

-((a + b*ArcCosh[c + d*x])^3/(d*e^2*(c + d*x))) + (6*b*(a + b*ArcCosh[c + d*x])^2*ArcTan[E^ArcCosh[c + d*x]])/
(d*e^2) - ((6*I)*b^2*(a + b*ArcCosh[c + d*x])*PolyLog[2, (-I)*E^ArcCosh[c + d*x]])/(d*e^2) + ((6*I)*b^2*(a + b
*ArcCosh[c + d*x])*PolyLog[2, I*E^ArcCosh[c + d*x]])/(d*e^2) + ((6*I)*b^3*PolyLog[3, (-I)*E^ArcCosh[c + d*x]])
/(d*e^2) - ((6*I)*b^3*PolyLog[3, I*E^ArcCosh[c + d*x]])/(d*e^2)

Rule 12

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

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 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 4180

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

Rule 5662

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

Rule 5761

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]
), x_Symbol] :> Dist[1/(c^(m + 1)*Sqrt[-(d1*d2)]), Subst[Int[(a + b*x)^n*Cosh[x]^m, x], x, ArcCosh[c*x]], x] /
; FreeQ[{a, b, c, d1, e1, d2, e2}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IGtQ[n, 0] && GtQ[d1, 0] &&
LtQ[d2, 0] && IntegerQ[m]

Rule 5866

Int[((a_.) + ArcCosh[(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*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, 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 \cosh ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt {-1+x} x \sqrt {1+x}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}+\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {\left (6 i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {6 b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {Li}_2\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}+\frac {6 i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}-\frac {6 i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^2}\\ \end {align*}

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Mathematica [A]  time = 1.18, size = 327, normalized size = 1.76 \[ -\frac {\frac {a^3}{c+d x}+3 a^2 b \tan ^{-1}\left (\frac {1}{\sqrt {c+d x-1} \sqrt {c+d x+1}}\right )+\frac {3 a^2 b \cosh ^{-1}(c+d x)}{c+d x}+3 i a b^2 \left (2 \text {Li}_2\left (-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \text {Li}_2\left (i e^{-\cosh ^{-1}(c+d x)}\right )+\cosh ^{-1}(c+d x) \left (-\frac {i \cosh ^{-1}(c+d x)}{c+d x}+2 \log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )-2 \log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )\right )+b^3 \left (\frac {\cosh ^{-1}(c+d x)^3}{c+d x}-3 i \left (-2 \cosh ^{-1}(c+d x) \left (\text {Li}_2\left (-i e^{-\cosh ^{-1}(c+d x)}\right )-\text {Li}_2\left (i e^{-\cosh ^{-1}(c+d x)}\right )\right )-2 \text {Li}_3\left (-i e^{-\cosh ^{-1}(c+d x)}\right )+2 \text {Li}_3\left (i e^{-\cosh ^{-1}(c+d x)}\right )-\left (\cosh ^{-1}(c+d x)^2 \left (\log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )-\log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )\right )\right )\right )}{d e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c + d*x])^3/(c*e + d*e*x)^2,x]

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arcosh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {arcosh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {arcosh}\left (d x + c\right ) + a^{3}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.7sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad A
rgument Value

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{3}}{\left (d e x +c e \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))^3/(d*e*x+c*e)^2,x)

[Out]

int((a+b*arccosh(d*x+c))^3/(d*e*x+c*e)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="maxima")

[Out]

-b^3*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^3/(d^2*e^2*x + c*d*e^2) - 3*a^2*b*(arccosh(d*x + c)/(d
^2*e^2*x + c*d*e^2) + arcsin(d*e^2/abs(d^2*e^2*x + c*d*e^2))/(d*e^2)) - a^3/(d^2*e^2*x + c*d*e^2) + integrate(
3*((c^3 - c)*a*b^2 + (c^3 - c)*b^3 + (a*b^2*d^3 + b^3*d^3)*x^3 + 3*(a*b^2*c*d^2 + b^3*c*d^2)*x^2 + (b^3*c^2 +
(c^2 - 1)*a*b^2 + (a*b^2*d^2 + b^3*d^2)*x^2 + 2*(a*b^2*c*d + b^3*c*d)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) +
((3*c^2*d - d)*a*b^2 + (3*c^2*d - d)*b^3)*x)*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^2/(d^5*e^2*x^
5 + 5*c*d^4*e^2*x^4 + c^5*e^2 - c^3*e^2 + (10*c^2*d^3*e^2 - d^3*e^2)*x^3 + (10*c^3*d^2*e^2 - 3*c*d^2*e^2)*x^2
+ (d^4*e^2*x^4 + 4*c*d^3*e^2*x^3 + c^4*e^2 - c^2*e^2 + (6*c^2*d^2*e^2 - d^2*e^2)*x^2 + 2*(2*c^3*d*e^2 - c*d*e^
2)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + (5*c^4*d*e^2 - 3*c^2*d*e^2)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c + d*x))^3/(c*e + d*e*x)^2,x)

[Out]

int((a + b*acosh(c + d*x))^3/(c*e + d*e*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))**3/(d*e*x+c*e)**2,x)

[Out]

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

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