3.2.21 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\) [121]

Optimal. Leaf size=297 \[ \frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \text {ArcTan}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {ArcTan}\left (\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^4}-\frac {i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \cosh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}+\frac {i b^3 \text {PolyLog}\left (3,-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^3 \text {PolyLog}\left (3,i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4} \]

[Out]

b^2*(a+b*arccosh(d*x+c))/d/e^4/(d*x+c)-1/3*(a+b*arccosh(d*x+c))^3/d/e^4/(d*x+c)^3+b*(a+b*arccosh(d*x+c))^2*arc
tan(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/d/e^4-b^3*arctan((d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))/d/e^4-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^4+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^4+I*b^3*polylog(3,-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))
)/d/e^4-I*b^3*polylog(3,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^4+1/2*b*(a+b*arccosh(d*x+c))^2*(d*x+c-1
)^(1/2)*(d*x+c+1)^(1/2)/d/e^4/(d*x+c)^2

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Rubi [A]
time = 0.40, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5996, 12, 5883, 5933, 5947, 4265, 2611, 2320, 6724, 94, 209} \begin {gather*} \frac {b \text {ArcTan}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {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^4}+\frac {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^4}+\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {c+d x-1} \sqrt {c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b^3 \text {ArcTan}\left (\sqrt {c+d x-1} \sqrt {c+d x+1}\right )}{d e^4}+\frac {i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(a + b*ArcCosh[c + d*x]))/(d*e^4*(c + d*x)) + (b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(a + b*ArcCosh[c +
d*x])^2)/(2*d*e^4*(c + d*x)^2) - (a + b*ArcCosh[c + d*x])^3/(3*d*e^4*(c + d*x)^3) + (b*(a + b*ArcCosh[c + d*x]
)^2*ArcTan[E^ArcCosh[c + d*x]])/(d*e^4) - (b^3*ArcTan[Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]])/(d*e^4) - (I*b^2*
(a + b*ArcCosh[c + d*x])*PolyLog[2, (-I)*E^ArcCosh[c + d*x]])/(d*e^4) + (I*b^2*(a + b*ArcCosh[c + d*x])*PolyLo
g[2, I*E^ArcCosh[c + d*x]])/(d*e^4) + (I*b^3*PolyLog[3, (-I)*E^ArcCosh[c + d*x]])/(d*e^4) - (I*b^3*PolyLog[3,
I*E^ArcCosh[c + d*x]])/(d*e^4)

Rule 12

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

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 4265

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 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5933

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_
))^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*((a + b*ArcCosh[c*x])^n/(d1*d
2*f*(m + 1))), x] + (Dist[c^2*((m + 2*p + 3)/(f^2*(m + 1))), Int[(f*x)^(m + 2)*(d1 + e1*x)^p*(d2 + e2*x)^p*(a
+ b*ArcCosh[c*x])^n, x], x] + Dist[b*c*(n/(f*(m + 1)))*Simp[(d1 + e1*x)^p/(1 + c*x)^p]*Simp[(d2 + e2*x)^p/(-1
+ c*x)^p], Int[(f*x)^(m + 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /;
FreeQ[{a, b, c, d1, e1, d2, e2, f, p}, x] && EqQ[e1, c*d1] && EqQ[e2, (-c)*d2] && GtQ[n, 0] && ILtQ[m, -1]

Rule 5947

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]
), x_Symbol] :> Dist[(1/c^(m + 1))*Simp[Sqrt[1 + c*x]/Sqrt[d1 + e1*x]]*Simp[Sqrt[-1 + c*x]/Sqrt[d2 + e2*x]], S
ubst[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*d
1] && EqQ[e2, (-c)*d2] && IGtQ[n, 0] && IntegerQ[m]

Rule 5996

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 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 \cosh ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{\sqrt {-1+x} x^3 \sqrt {1+x}} \, dx,x,c+d x\right )}{d e^4}\\ &=\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {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 )}{2 d e^4}-\frac {b^2 \text {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\cosh ^{-1}(c+d x)\right )}{2 d e^4}-\frac {b^3 \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} x \sqrt {1+x}} \, dx,x,c+d x\right )}{d e^4}\\ &=\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (i b^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^4}\\ &=\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tan ^{-1}\left (\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^4}-\frac {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^4}+\frac {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^4}+\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^4}-\frac {\left (i b^3\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e^4}\\ &=\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tan ^{-1}\left (\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^4}-\frac {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^4}+\frac {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^4}+\frac {\left (i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (i b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}\\ &=\frac {b^2 \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x} \left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \cosh ^{-1}(c+d x)\right )^2 \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tan ^{-1}\left (\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^4}-\frac {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^4}+\frac {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^4}+\frac {i b^3 \text {Li}_3\left (-i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^3 \text {Li}_3\left (i e^{\cosh ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(601\) vs. \(2(297)=594\).
time = 1.35, size = 601, normalized size = 2.02 \begin {gather*} \frac {-\frac {2 a^3}{(c+d x)^3}+\frac {3 a^2 b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{(c+d x)^2}-\frac {6 a^2 b \cosh ^{-1}(c+d x)}{(c+d x)^3}-3 a^2 b \text {ArcTan}\left (\frac {1}{\sqrt {-1+c+d x} \sqrt {1+c+d x}}\right )+6 a b^2 \left (\frac {1}{c+d x}+\frac {\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x) \cosh ^{-1}(c+d x)}{(c+d x)^2}-\frac {\cosh ^{-1}(c+d x)^2}{(c+d x)^3}-i \cosh ^{-1}(c+d x) \log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )+i \cosh ^{-1}(c+d x) \log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )-i \text {PolyLog}\left (2,-i e^{-\cosh ^{-1}(c+d x)}\right )+i \text {PolyLog}\left (2,i e^{-\cosh ^{-1}(c+d x)}\right )\right )+6 b^3 \left (\frac {\cosh ^{-1}(c+d x)}{c+d x}+\frac {\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x) \cosh ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac {\cosh ^{-1}(c+d x)^3}{3 (c+d x)^3}-2 \text {ArcTan}\left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )+\cosh ^{-1}(c+d x)^2 \text {ArcTan}\left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )-i \cosh ^{-1}(c+d x) \text {PolyLog}\left (2,-i \left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )\right )+i \cosh ^{-1}(c+d x) \text {PolyLog}\left (2,i \left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )\right )+i \text {PolyLog}\left (3,-i \left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )\right )-i \text {PolyLog}\left (3,i \left (c+d x+\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)\right )\right )\right )}{6 d e^4} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a^3)/(c + d*x)^3 + (3*a^2*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(c + d*x)^2 - (6*a^2*b*ArcCosh[c + d*x]
)/(c + d*x)^3 - 3*a^2*b*ArcTan[1/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])] + 6*a*b^2*((c + d*x)^(-1) + (Sqrt[(-1
+ c + d*x)/(1 + c + d*x)]*(1 + c + d*x)*ArcCosh[c + d*x])/(c + d*x)^2 - ArcCosh[c + d*x]^2/(c + d*x)^3 - I*Ar
cCosh[c + d*x]*Log[1 - I/E^ArcCosh[c + d*x]] + I*ArcCosh[c + d*x]*Log[1 + I/E^ArcCosh[c + d*x]] - I*PolyLog[2,
(-I)/E^ArcCosh[c + d*x]] + I*PolyLog[2, I/E^ArcCosh[c + d*x]]) + 6*b^3*(ArcCosh[c + d*x]/(c + d*x) + (Sqrt[(-
1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x)*ArcCosh[c + d*x]^2)/(2*(c + d*x)^2) - ArcCosh[c + d*x]^3/(3*(c + d*x
)^3) - 2*ArcTan[c + d*x + Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x)] + ArcCosh[c + d*x]^2*ArcTan[c + d*
x + Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x)] - I*ArcCosh[c + d*x]*PolyLog[2, (-I)*(c + d*x + Sqrt[(-1
+ c + d*x)/(1 + c + d*x)]*(1 + c + d*x))] + I*ArcCosh[c + d*x]*PolyLog[2, I*(c + d*x + Sqrt[(-1 + c + d*x)/(1
+ c + d*x)]*(1 + c + d*x))] + I*PolyLog[3, (-I)*(c + d*x + Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x))]
- I*PolyLog[3, I*(c + d*x + Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x))]))/(6*d*e^4)

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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {arccosh}\left (d x +c \right )\right )^{3}}{\left (d e x +c e \right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-1/3*b^3*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c)^3/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4
+ c^3*d*e^4) - 1/3*a^3/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) + integrate(((3*(c^3 - c
)*a*b^2 + (c^3 - c)*b^3 + (3*a*b^2*d^3 + b^3*d^3)*x^3 + 3*(3*a*b^2*c*d^2 + b^3*c*d^2)*x^2 + (b^3*c^2 + 3*(c^2
- 1)*a*b^2 + (3*a*b^2*d^2 + b^3*d^2)*x^2 + 2*(3*a*b^2*c*d + b^3*c*d)*x)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) +
(3*(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 + 3*(a^2*b
*d^3*x^3 + 3*a^2*b*c*d^2*x^2 + (3*c^2*d - d)*a^2*b*x + (c^3 - c)*a^2*b + (a^2*b*d^2*x^2 + 2*a^2*b*c*d*x + (c^2
- 1)*a^2*b)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1))*log(d*x + sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + c))/(d^7*x^7
*e^4 + 7*c*d^6*x^6*e^4 + (21*c^2*d^5 - d^5)*x^5*e^4 + 5*(7*c^3*d^4 - c*d^4)*x^4*e^4 + 5*(7*c^4*d^3 - 2*c^2*d^3
)*x^3*e^4 + (21*c^5*d^2 - 10*c^3*d^2)*x^2*e^4 + (7*c^6*d - 5*c^4*d)*x*e^4 + (d^6*x^6*e^4 + 6*c*d^5*x^5*e^4 + (
15*c^2*d^4 - d^4)*x^4*e^4 + 4*(5*c^3*d^3 - c*d^3)*x^3*e^4 + 3*(5*c^4*d^2 - 2*c^2*d^2)*x^2*e^4 + 2*(3*c^5*d - 2
*c^3*d)*x*e^4 + (c^6 - c^4)*e^4)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1) + (c^7 - c^5)*e^4), x)

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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*arccosh(d*x+c))^3/(d*e*x+c*e)^4,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)*e^(-4)/(d^4*x^
4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {acosh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {acosh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {acosh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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