3.81 \(\int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx\)

Optimal. Leaf size=153 \[ -\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cosh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}+\frac {\log (\cosh (x))}{a} \]

[Out]

ln(cosh(x))/a+1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*cosh(x))/a^(5/3)-1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*cosh(x)+b
^(2/3)*cosh(x)^2)/a^(5/3)-1/3*ln(a+b*cosh(x)^3)/a+1/2*sech(x)^2/a-1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*co
sh(x))/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3230, 1834, 1871, 200, 31, 634, 617, 204, 628, 260} \[ -\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cosh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}+\frac {\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Cosh[x]^3),x]

[Out]

-((b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Cosh[x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3))) + Log[Cosh[x]]/a + (b^(
2/3)*Log[a^(1/3) + b^(1/3)*Cosh[x]])/(3*a^(5/3)) - (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Cosh[x] + b^(2/3)*Co
sh[x]^2])/(6*a^(5/3)) - Log[a + b*Cosh[x]^3]/(3*a) + Sech[x]^2/(2*a)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 204

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx &=-\operatorname {Subst}\left (\int \frac {1-x^2}{x^3 \left (a+b x^3\right )} \, dx,x,\cosh (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {1}{a x^3}-\frac {1}{a x}+\frac {b \left (-1+x^2\right )}{a \left (a+b x^3\right )}\right ) \, dx,x,\cosh (x)\right )\\ &=\frac {\log (\cosh (x))}{a}+\frac {\text {sech}^2(x)}{2 a}-\frac {b \operatorname {Subst}\left (\int \frac {-1+x^2}{a+b x^3} \, dx,x,\cosh (x)\right )}{a}\\ &=\frac {\log (\cosh (x))}{a}+\frac {\text {sech}^2(x)}{2 a}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a+b x^3} \, dx,x,\cosh (x)\right )}{a}-\frac {b \operatorname {Subst}\left (\int \frac {x^2}{a+b x^3} \, dx,x,\cosh (x)\right )}{a}\\ &=\frac {\log (\cosh (x))}{a}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\cosh (x)\right )}{3 a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\cosh (x)\right )}{3 a^{5/3}}\\ &=\frac {\log (\cosh (x))}{a}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}-\frac {b^{2/3} \operatorname {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\cosh (x)\right )}{6 a^{5/3}}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\cosh (x)\right )}{2 a^{4/3}}\\ &=\frac {\log (\cosh (x))}{a}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}+\frac {b^{2/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \cosh (x)}{\sqrt [3]{a}}\right )}{a^{5/3}}\\ &=-\frac {b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \cosh (x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{5/3}}+\frac {\log (\cosh (x))}{a}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}\\ \end {align*}

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Mathematica [C]  time = 1.42, size = 145, normalized size = 0.95 \[ \frac {-2 \text {RootSum}\left [\text {$\#$1}^6 b+3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b+b\& ,\frac {-3 \text {$\#$1}^4 b x+3 \text {$\#$1}^4 b \log \left (e^x-\text {$\#$1}\right )-4 \text {$\#$1}^3 a x+4 \text {$\#$1}^3 a \log \left (e^x-\text {$\#$1}\right )+b \log \left (e^x-\text {$\#$1}\right )-b x}{\text {$\#$1}^4 b+4 \text {$\#$1}^3 a+2 \text {$\#$1}^2 b+b}\& \right ]-6 x+3 \text {sech}^2(x)+6 \log (\cosh (x))}{6 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^3/(a + b*Cosh[x]^3),x]

[Out]

(-6*x + 6*Log[Cosh[x]] - 2*RootSum[b + 3*b*#1^2 + 8*a*#1^3 + 3*b*#1^4 + b*#1^6 & , (-(b*x) + b*Log[E^x - #1] -
4*a*x*#1^3 + 4*a*Log[E^x - #1]*#1^3 - 3*b*x*#1^4 + 3*b*Log[E^x - #1]*#1^4)/(b + 2*b*#1^2 + 4*a*#1^3 + b*#1^4)
& ] + 3*Sech[x]^2)/(6*a)

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fricas [C]  time = 1.44, size = 1138, normalized size = 7.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*cosh(x)^3),x, algorithm="fricas")

[Out]

-1/12*(12*sqrt(1/3)*(a*e^(4*x) + 2*a*e^(2*x) + a)*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 -
b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a
)*a + 4)/a^2)*arctan(-1/8*(2*sqrt(1/3)*sqrt(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(
1/3) + 2/a)^2*a^4*e^(2*x) + b^2*e^(4*x) - 2*a*b*e^(3*x) - 2*a*b*e^x + (a^2*b*e^(3*x) - 4*a^3*e^(2*x) + a^2*b*e
^x)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) + b^2 + 2*(2*a^2 + b^2)*e^(2
*x))*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a^3 - 2*a^2)*sqrt((((1/2)^
(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*
(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2) - sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3
+ b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^5*e^x - 4*a^2*b*e^(2*x) + 4*a^3*e^x - 4*a^2*b + 2*(a^3*b*e^(2*x)
- 2*a^4*e^x + a^3*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a))*sqrt((((1
/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) +
1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2))*e^(-x)/b^2) + 2*(a*e^(4*x) + 2*a*e^(2*x) + a
)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*log(-((1/2)^(1/3)*(I*sqrt(3) +
1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a^2*e^x + b*e^(2*x) + 2*a*e^x + b) - ((a*e^(4*x) + 2*a*e^
(2*x) + a)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) - 6*e^(4*x) - 12*e^(2
*x) - 6)*log(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^4*e^(2*x) + b^2
*e^(4*x) - 2*a*b*e^(3*x) - 2*a*b*e^x + (a^2*b*e^(3*x) - 4*a^3*e^(2*x) + a^2*b*e^x)*((1/2)^(1/3)*(I*sqrt(3) + 1
)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) + b^2 + 2*(2*a^2 + b^2)*e^(2*x)) - 12*(e^(4*x) + 2*e^(2*x)
+ 1)*log(e^(2*x) + 1) - 24*e^(2*x))/(a*e^(4*x) + 2*a*e^(2*x) + a)

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giac [A]  time = 8.02, size = 191, normalized size = 1.25 \[ -\frac {b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} + e^{\left (-x\right )} + e^{x} \right |}\right )}{3 \, a^{2}} + \frac {\log \left (e^{\left (-x\right )} + e^{x}\right )}{a} - \frac {\log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 8 \, a \right |}\right )}{3 \, a} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + e^{\left (-x\right )} + e^{x}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} {\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2}} - \frac {3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4}{2 \, a {\left (e^{\left (-x\right )} + e^{x}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*cosh(x)^3),x, algorithm="giac")

[Out]

-1/3*b*(-a/b)^(1/3)*log(abs(-2*(-a/b)^(1/3) + e^(-x) + e^x))/a^2 + log(e^(-x) + e^x)/a - 1/3*log(abs(b*(e^(-x)
+ e^x)^3 + 8*a))/a + 1/3*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) + e^(-x) + e^x)/(-a/b)^(1/3)
)/a^2 + 1/6*(-a*b^2)^(1/3)*log((e^(-x) + e^x)^2 + 2*(-a/b)^(1/3)*(e^(-x) + e^x) + 4*(-a/b)^(2/3))/a^2 - 1/2*(3
*(e^(-x) + e^x)^2 - 4)/(a*(e^(-x) + e^x)^2)

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maple [C]  time = 0.14, size = 150, normalized size = 0.98 \[ -\frac {\munderset {\textit {\_R} =\RootOf \left (\left (a -b \right ) \textit {\_Z}^{3}+\left (-3 a -3 b \right ) \textit {\_Z}^{2}+\left (3 a -3 b \right ) \textit {\_Z} -a -b \right )}{\sum }\frac {\left (\textit {\_R}^{2} a -\textit {\_R}^{2} b -2 \textit {\_R} a -4 \textit {\_R} b +a +b \right ) \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{2} a -\textit {\_R}^{2} b -2 \textit {\_R} a -2 \textit {\_R} b +a -b}}{3 a}-\frac {2}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}+\frac {2}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a+b*cosh(x)^3),x)

[Out]

-1/3/a*sum((_R^2*a-_R^2*b-2*_R*a-4*_R*b+a+b)/(_R^2*a-_R^2*b-2*_R*a-2*_R*b+a-b)*ln(tanh(1/2*x)^2-_R),_R=RootOf(
(a-b)*_Z^3+(-3*a-3*b)*_Z^2+(3*a-3*b)*_Z-a-b))-2/a/(tanh(1/2*x)^2+1)+2/a/(tanh(1/2*x)^2+1)^2+1/a*ln(tanh(1/2*x)
^2+1)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^3/(a+b*cosh(x)^3),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 0.92, size = 1173, normalized size = 7.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a + b*cosh(x)^3),x)

[Out]

2/(a + a*exp(2*x)) - 2/(a + 2*a*exp(2*x) + a*exp(4*x)) + symsum(log(-(50331648*a^6*exp(2*x) - 786432*b^6*exp(2
*x) + 452984832*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^7 + 50331648*a^6 - 786432*b^6 + 13
58954496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^8 + 1358954496*root(27*a^5*z^3 + 27*a^4
*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^9 + 50593792*a^2*b^4 - 102498304*a^4*b^2 + 1358954496*root(27*a^5*z^3 +
27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^8*exp(2*x) + 1358954496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a
^2 - b^2, z, k)^3*a^9*exp(2*x) + 50593792*a^2*b^4*exp(2*x) - 102498304*a^4*b^2*exp(2*x) + 7602176*root(27*a^5*
z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^3*b^4 - 465305600*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2
- b^2, z, k)*a^5*b^2 + 524288*a*b^5*exp(x) + 24379392*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z,
k)^2*a^4*b^4 - 1383333888*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^6*b^2 + 18874368*root(
27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^5*b^4 - 1370750976*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a
^3*z + a^2 - b^2, z, k)^3*a^7*b^2 + 452984832*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^7*ex
p(2*x) - 5242880*a^3*b^3*exp(x) - 524288*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^2*b^5*exp
(x) - 8912896*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^4*b^3*exp(x) + 7602176*root(27*a^5*z
^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^3*b^4*exp(2*x) - 465305600*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3
*z + a^2 - b^2, z, k)*a^5*b^2*exp(2*x) + 14155776*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*
a^6*b^3*exp(x) + 24379392*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^4*b^4*exp(2*x) - 13833
33888*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^6*b^2*exp(2*x) + 18874368*root(27*a^5*z^3
+ 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^5*b^4*exp(2*x) - 1370750976*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3
*z + a^2 - b^2, z, k)^3*a^7*b^2*exp(2*x))/(3*a^6*b^6))*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z,
k), k, 1, 3) + log(3221225472*a^6*exp(2*x) - 786432*b^6*exp(2*x) + 3221225472*a^6 - 786432*b^6 + 101449728*a^2
*b^4 - 3321888768*a^4*b^2 + 101449728*a^2*b^4*exp(2*x) - 3321888768*a^4*b^2*exp(2*x))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}{\relax (x )}}{a + b \cosh ^{3}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**3/(a+b*cosh(x)**3),x)

[Out]

Integral(tanh(x)**3/(a + b*cosh(x)**3), x)

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