3.2.12 \(\int \frac {(a+b \cosh ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\) [112]

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

[Out]

1/3*b^2/d/e^4/(d*x+c)-1/3*(a+b*arccosh(d*x+c))^2/d/e^4/(d*x+c)^3+2/3*b*(a+b*arccosh(d*x+c))*arctan(d*x+c+(d*x+
c-1)^(1/2)*(d*x+c+1)^(1/2))/d/e^4-1/3*I*b^2*polylog(2,-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^4+1/3*I*
b^2*polylog(2,I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))/d/e^4+1/3*b*(a+b*arccosh(d*x+c))*(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 = 0.25, antiderivative size = 186, normalized size of antiderivative = 1.00, 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 = {5996, 12, 5883, 5933, 5947, 4265, 2317, 2438, 30} \begin {gather*} \frac {2 b \text {ArcTan}\left (e^{\cosh ^{-1}(c+d x)}\right ) \left (a+b \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 {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}+\frac {b^2}{3 d e^4 (c+d x)} \end {gather*}

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 12

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

Rule 30

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

Rule 2317

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 2438

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 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
]

Rubi steps

\begin {align*} \int \frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \cosh ^{-1}(x)\right )^2}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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) \text {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 \text {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 \text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.69, size = 251, normalized size = 1.35 \begin {gather*} \frac {-\frac {a^2}{(c+d x)^3}+a b \left (\frac {\sqrt {\frac {-1+c+d x}{1+c+d x}} (1+c+d x)}{(c+d x)^2}-\frac {2 \cosh ^{-1}(c+d x)}{(c+d x)^3}+2 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \cosh ^{-1}(c+d x)\right )\right )\right )+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 )}{3 d e^4} \end {gather*}

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 = 16.74, size = 352, normalized size = 1.89

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \mathrm {arccosh}\left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{2}}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2}}{3 e^{4} \left (d x +c \right )}-\frac {i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}+\frac {i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}+\frac {i b^{2} \dilog \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}-\frac {2 a b \,\mathrm {arccosh}\left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{3}}-\frac {a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{3 e^{4} \sqrt {\left (d x +c \right )^{2}-1}}+\frac {a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}}{3 e^{4} \left (d x +c \right )^{2}}}{d}\)	\(352\)
default	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \mathrm {arccosh}\left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{2}}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2}}{3 e^{4} \left (d x +c \right )}-\frac {i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}+\frac {i b^{2} \mathrm {arccosh}\left (d x +c \right ) \ln \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (1+i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}+\frac {i b^{2} \dilog \left (1-i \left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )\right )}{3 e^{4}}-\frac {2 a b \,\mathrm {arccosh}\left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{3}}-\frac {a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{3 e^{4} \sqrt {\left (d x +c \right )^{2}-1}}+\frac {a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}}{3 e^{4} \left (d x +c \right )^{2}}}{d}\)	\(352\)

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,method=_RETURNVERBOSE)

[Out]

1/d*(-1/3*a^2/e^4/(d*x+c)^3+1/3*b^2/e^4/(d*x+c)^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*arccosh(d*x+c)-1/3*b^2/e^4/(
d*x+c)^3*arccosh(d*x+c)^2+1/3*b^2/e^4/(d*x+c)-1/3*I*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*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*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*b^2/e^4*dilog(1-I*(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)))-
2/3*a*b/e^4/(d*x+c)^3*arccosh(d*x+c)-1/3*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*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 = 0.00, 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))^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*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^2/(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(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*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))^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)*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^{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 {gather*}

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 = 0.00, 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))^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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acosh}\left (c+d\,x\right )\right )}^2}{{\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))^2/(c*e + d*e*x)^4,x)

[Out]

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

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