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

Optimal. Leaf size=57 \[ \frac{\left (a^2-b^2\right ) \log (\cosh (x))}{a^3}-\frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{a^3}-\frac{b \text{sech}(x)}{a^2}+\frac{\text{sech}^2(x)}{2 a} \]

[Out]

((a^2 - b^2)*Log[Cosh[x]])/a^3 - ((a^2 - b^2)*Log[a + b*Cosh[x]])/a^3 - (b*Sech[x])/a^2 + Sech[x]^2/(2*a)

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Rubi [A]  time = 0.0988762, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2721, 894} \[ \frac{\left (a^2-b^2\right ) \log (\cosh (x))}{a^3}-\frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{a^3}-\frac{b \text{sech}(x)}{a^2}+\frac{\text{sech}^2(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Cosh[x]),x]

[Out]

((a^2 - b^2)*Log[Cosh[x]])/a^3 - ((a^2 - b^2)*Log[a + b*Cosh[x]])/a^3 - (b*Sech[x])/a^2 + Sech[x]^2/(2*a)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\tanh ^3(x)}{a+b \cosh (x)} \, dx &=-\operatorname{Subst}\left (\int \frac{b^2-x^2}{x^3 (a+x)} \, dx,x,b \cosh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{b^2}{a x^3}-\frac{b^2}{a^2 x^2}+\frac{-a^2+b^2}{a^3 x}+\frac{a^2-b^2}{a^3 (a+x)}\right ) \, dx,x,b \cosh (x)\right )\\ &=\frac{\left (a^2-b^2\right ) \log (\cosh (x))}{a^3}-\frac{\left (a^2-b^2\right ) \log (a+b \cosh (x))}{a^3}-\frac{b \text{sech}(x)}{a^2}+\frac{\text{sech}^2(x)}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.0925007, size = 46, normalized size = 0.81 \[ \frac{2 \left (a^2-b^2\right ) (\log (\cosh (x))-\log (a+b \cosh (x)))+a^2 \text{sech}^2(x)-2 a b \text{sech}(x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 - b^2)*(Log[Cosh[x]] - Log[a + b*Cosh[x]]) - 2*a*b*Sech[x] + a^2*Sech[x]^2)/(2*a^3)

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Maple [B]  time = 0.03, size = 140, normalized size = 2.5 \begin{align*} -{\frac{1}{a}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }+{\frac{{b}^{2}}{{a}^{3}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }+{\frac{1}{a}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-2\,{\frac{1}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{b}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{1}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)+1/a^3*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)*b^2+1/a*ln(tanh(1/2
*x)^2+1)-1/a^3*ln(tanh(1/2*x)^2+1)*b^2-2/a/(tanh(1/2*x)^2+1)-2/a^2/(tanh(1/2*x)^2+1)*b+2/a/(tanh(1/2*x)^2+1)^2

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Maxima [A]  time = 1.56926, size = 130, normalized size = 2.28 \begin{align*} -\frac{2 \,{\left (b e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} + b e^{\left (-3 \, x\right )}\right )}}{2 \, a^{2} e^{\left (-2 \, x\right )} + a^{2} e^{\left (-4 \, x\right )} + a^{2}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{3}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*e^(-x) - a*e^(-2*x) + b*e^(-3*x))/(2*a^2*e^(-2*x) + a^2*e^(-4*x) + a^2) - (a^2 - b^2)*log(2*a*e^(-x) + b
*e^(-2*x) + b)/a^3 + (a^2 - b^2)*log(e^(-2*x) + 1)/a^3

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Fricas [B]  time = 2.00553, size = 1166, normalized size = 20.46 \begin{align*} -\frac{2 \, a b \cosh \left (x\right )^{3} + 2 \, a b \sinh \left (x\right )^{3} - 2 \, a^{2} \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \,{\left (3 \, a b \cosh \left (x\right ) - a^{2}\right )} \sinh \left (x\right )^{2} +{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{4} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} + a^{2} - b^{2} + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{4} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} + a^{2} - b^{2} + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \,{\left (3 \, a b \cosh \left (x\right )^{2} - 2 \, a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )}{a^{3} \cosh \left (x\right )^{4} + 4 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right )^{3} + a^{3} \sinh \left (x\right )^{4} + 2 \, a^{3} \cosh \left (x\right )^{2} + a^{3} + 2 \,{\left (3 \, a^{3} \cosh \left (x\right )^{2} + a^{3}\right )} \sinh \left (x\right )^{2} + 4 \,{\left (a^{3} \cosh \left (x\right )^{3} + a^{3} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*b*cosh(x)^3 + 2*a*b*sinh(x)^3 - 2*a^2*cosh(x)^2 + 2*a*b*cosh(x) + 2*(3*a*b*cosh(x) - a^2)*sinh(x)^2 + ((
a^2 - b^2)*cosh(x)^4 + 4*(a^2 - b^2)*cosh(x)*sinh(x)^3 + (a^2 - b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(
3*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*s
inh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - ((a^2 - b^2)*cosh(x)^4 + 4*(a^2 - b^2)*cosh(x)*sinh(x)^3
+ (a^2 - b^2)*sinh(x)^4 + 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 -
b^2 + 4*((a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 2*(3*a*b*c
osh(x)^2 - 2*a^2*cosh(x) + a*b)*sinh(x))/(a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + 2*a^3*cosh
(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 + a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + a^3*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15605, size = 155, normalized size = 2.72 \begin{align*} \frac{{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-x\right )} + e^{x}\right )}{a^{3}} - \frac{{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{3} b} - \frac{3 \, a^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 3 \, b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 4 \, a b{\left (e^{\left (-x\right )} + e^{x}\right )} - 4 \, a^{2}}{2 \, a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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