3.112 \(\int \frac{(a+b \cosh ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.386817, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5866, 12, 5662, 5748, 5761, 4180, 2279, 2391, 30} \[ -\frac{i b^2 \text{PolyLog}\left (2,-i e^{\cosh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{i b^2 \text{PolyLog}\left (2,i e^{\cosh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac{b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{2 b \tan ^{-1}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d e^4}+\frac{b^2}{3 d e^4 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

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

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 12

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

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 5748

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*
d2*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*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p
])/(f*(m + 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(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, 0] && EqQ[e2 +
c*d2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegerQ[m] && IntegerQ[p + 1/2]

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 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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.956459, size = 251, normalized size = 1.35 \[ \frac{b^2 \left (-i \text{PolyLog}\left (2,-i e^{-\cosh ^{-1}(c+d x)}\right )+i \text{PolyLog}\left (2,i e^{-\cosh ^{-1}(c+d x)}\right )+\frac{1}{c+d x}-\frac{\cosh ^{-1}(c+d x)^2}{(c+d x)^3}+\frac{\sqrt{\frac{c+d x-1}{c+d x+1}} (c+d x+1) \cosh ^{-1}(c+d x)}{(c+d x)^2}-i \cosh ^{-1}(c+d x) \log \left (1-i e^{-\cosh ^{-1}(c+d x)}\right )+i \cosh ^{-1}(c+d x) \log \left (1+i e^{-\cosh ^{-1}(c+d x)}\right )\right )-\frac{a^2}{(c+d x)^3}+a b \left (\frac{\sqrt{\frac{c+d x-1}{c+d x+1}} (c+d x+1)}{(c+d x)^2}-\frac{2 \cosh ^{-1}(c+d x)}{(c+d x)^3}+2 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \cosh ^{-1}(c+d x)\right )\right )\right )}{3 d e^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

(-(a^2/(c + d*x)^3) + a*b*((Sqrt[(-1 + c + d*x)/(1 + c + d*x)]*(1 + c + d*x))/(c + d*x)^2 - (2*ArcCosh[c + d*x
])/(c + d*x)^3 + 2*ArcTan[Tanh[ArcCosh[c + d*x]/2]]) + 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*ArcCosh[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]]))/(3*d*e^4)

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Maple [A]  time = 0.086, size = 381, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}+{\frac{{b}^{2}{\rm arccosh} \left (dx+c\right )}{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{{b}^{2} \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2}}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}+{\frac{{b}^{2}}{3\,d{e}^{4} \left ( dx+c \right ) }}-{\frac{{\frac{i}{3}}{b}^{2}{\rm arccosh} \left (dx+c\right )}{d{e}^{4}}\ln \left ( 1+i \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) \right ) }+{\frac{{\frac{i}{3}}{b}^{2}{\rm arccosh} \left (dx+c\right )}{d{e}^{4}}\ln \left ( 1-i \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) \right ) }-{\frac{{\frac{i}{3}}{b}^{2}}{d{e}^{4}}{\it dilog} \left ( 1+i \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) \right ) }+{\frac{{\frac{i}{3}}{b}^{2}}{d{e}^{4}}{\it dilog} \left ( 1-i \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) \right ) }-{\frac{2\,ab{\rm arccosh} \left (dx+c\right )}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{ab}{3\,d{e}^{4}}\sqrt{dx+c-1}\sqrt{dx+c+1}\arctan \left ({\frac{1}{\sqrt{ \left ( dx+c \right ) ^{2}-1}}} \right ){\frac{1}{\sqrt{ \left ( dx+c \right ) ^{2}-1}}}}+{\frac{ab}{3\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{dx+c-1}\sqrt{dx+c+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^4,x)

[Out]

-1/3/d*a^2/e^4/(d*x+c)^3+1/3/d*b^2/e^4/(d*x+c)^2*arccosh(d*x+c)*(d*x+c+1)^(1/2)*(d*x+c-1)^(1/2)-1/3/d*b^2/e^4/
(d*x+c)^3*arccosh(d*x+c)^2+1/3*b^2/d/e^4/(d*x+c)-1/3*I/d*b^2/e^4*arccosh(d*x+c)*ln(1+I*(d*x+c+(d*x+c-1)^(1/2)*
(d*x+c+1)^(1/2)))+1/3*I/d*b^2/e^4*arccosh(d*x+c)*ln(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))-1/3*I/d*b^2/e
^4*dilog(1+I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))+1/3*I/d*b^2/e^4*dilog(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+
1)^(1/2)))-2/3/d*a*b/e^4/(d*x+c)^3*arccosh(d*x+c)-1/3/d*a*b/e^4*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/((d*x+c)^2-1)^
(1/2)*arctan(1/((d*x+c)^2-1)^(1/2))+1/3/d*a*b/e^4*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/(d*x+c)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{2} \log \left (d x + \sqrt{d x + c + 1} \sqrt{d x + c - 1} + c\right )^{2}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac{a^{2}}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} + \int \frac{2 \,{\left ({\left (3 \, a b d^{3} + b^{2} d^{3}\right )} x^{3} + 3 \,{\left (c^{3} - c\right )} a b +{\left (c^{3} - c\right )} b^{2} + 3 \,{\left (3 \, a b c d^{2} + b^{2} c d^{2}\right )} x^{2} +{\left (b^{2} c^{2} + 3 \,{\left (c^{2} - 1\right )} a b +{\left (3 \, a b d^{2} + b^{2} d^{2}\right )} x^{2} + 2 \,{\left (3 \, a b c d + b^{2} c d\right )} x\right )} \sqrt{d x + c + 1} \sqrt{d x + c - 1} +{\left (3 \,{\left (3 \, c^{2} d - d\right )} a b +{\left (3 \, c^{2} d - d\right )} b^{2}\right )} x\right )} \log \left (d x + \sqrt{d x + c + 1} \sqrt{d x + c - 1} + c\right )}{3 \,{\left (d^{7} e^{4} x^{7} + 7 \, c d^{6} e^{4} x^{6} + c^{7} e^{4} - c^{5} e^{4} +{\left (21 \, c^{2} d^{5} e^{4} - d^{5} e^{4}\right )} x^{5} + 5 \,{\left (7 \, c^{3} d^{4} e^{4} - c d^{4} e^{4}\right )} x^{4} + 5 \,{\left (7 \, c^{4} d^{3} e^{4} - 2 \, c^{2} d^{3} e^{4}\right )} x^{3} +{\left (21 \, c^{5} d^{2} e^{4} - 10 \, c^{3} d^{2} e^{4}\right )} x^{2} +{\left (d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + c^{6} e^{4} - c^{4} e^{4} +{\left (15 \, c^{2} d^{4} e^{4} - d^{4} e^{4}\right )} x^{4} + 4 \,{\left (5 \, c^{3} d^{3} e^{4} - c d^{3} e^{4}\right )} x^{3} + 3 \,{\left (5 \, c^{4} d^{2} e^{4} - 2 \, c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (3 \, c^{5} d e^{4} - 2 \, c^{3} d e^{4}\right )} x\right )} \sqrt{d x + c + 1} \sqrt{d x + c - 1} +{\left (7 \, c^{6} d e^{4} - 5 \, c^{4} d e^{4}\right )} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{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}}\, dx + \int \frac{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{2 a 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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))**2/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**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(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(2*a*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcosh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arccosh(d*x + c) + a)^2/(d*e*x + c*e)^4, x)