3.117 \(\int \frac{\tanh ^3(x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=35 \[ \frac{\left (1-\frac{a^2}{b^2}\right ) \log (a+b \text{sech}(x))}{a}+\frac{\log (\cosh (x))}{a}+\frac{\text{sech}(x)}{b} \]

[Out]

Log[Cosh[x]]/a + ((1 - a^2/b^2)*Log[a + b*Sech[x]])/a + Sech[x]/b

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Rubi [A]  time = 0.0750012, antiderivative size = 35, 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 = {3885, 894} \[ \frac{\left (1-\frac{a^2}{b^2}\right ) \log (a+b \text{sech}(x))}{a}+\frac{\log (\cosh (x))}{a}+\frac{\text{sech}(x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Sech[x]),x]

[Out]

Log[Cosh[x]]/a + ((1 - a^2/b^2)*Log[a + b*Sech[x]])/a + Sech[x]/b

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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 \text{sech}(x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{b^2-x^2}{x (a+x)} \, dx,x,b \text{sech}(x)\right )}{b^2}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{b^2}{a x}+\frac{a^2-b^2}{a (a+x)}\right ) \, dx,x,b \text{sech}(x)\right )}{b^2}\\ &=\frac{\log (\cosh (x))}{a}+\frac{\left (1-\frac{a^2}{b^2}\right ) \log (a+b \text{sech}(x))}{a}+\frac{\text{sech}(x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0840565, size = 37, normalized size = 1.06 \[ \frac{\left (b^2-a^2\right ) \log (a \cosh (x)+b)+a^2 \log (\cosh (x))+a b \text{sech}(x)}{a b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.03, size = 107, normalized size = 3.1 \begin{align*} -{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{a}{{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{1}{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-{\frac{a}{{b}^{2}}\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 ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56517, size = 90, normalized size = 2.57 \begin{align*} \frac{x}{a} + \frac{2 \, e^{\left (-x\right )}}{b e^{\left (-2 \, x\right )} + b} + \frac{a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{2}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.5721, size = 543, normalized size = 15.51 \begin{align*} -\frac{b^{2} x \cosh \left (x\right )^{2} + b^{2} x \sinh \left (x\right )^{2} + b^{2} x - 2 \, a b \cosh \left (x\right ) +{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} + a^{2} - b^{2}\right )} \log \left (\frac{2 \,{\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} + a^{2}\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \,{\left (b^{2} x \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )}{a b^{2} \cosh \left (x\right )^{2} + 2 \, a b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a b^{2} \sinh \left (x\right )^{2} + a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13766, size = 99, normalized size = 2.83 \begin{align*} \frac{a \log \left (e^{\left (-x\right )} + e^{x}\right )}{b^{2}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left ({\left | a{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a b^{2}} - \frac{a{\left (e^{\left (-x\right )} + e^{x}\right )} - 2 \, b}{b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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