3.115 \(\int \frac{\tanh ^5(x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=72 \[ -\frac{\left (a^2-2 b^2\right ) \text{sech}(x)}{b^3}+\frac{\left (a^2-b^2\right )^2 \log (a+b \text{sech}(x))}{a b^4}+\frac{a \text{sech}^2(x)}{2 b^2}+\frac{\log (\cosh (x))}{a}-\frac{\text{sech}^3(x)}{3 b} \]

[Out]

Log[Cosh[x]]/a + ((a^2 - b^2)^2*Log[a + b*Sech[x]])/(a*b^4) - ((a^2 - 2*b^2)*Sech[x])/b^3 + (a*Sech[x]^2)/(2*b
^2) - Sech[x]^3/(3*b)

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Rubi [A]  time = 0.0970659, antiderivative size = 72, 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} \[ -\frac{\left (a^2-2 b^2\right ) \text{sech}(x)}{b^3}+\frac{\left (a^2-b^2\right )^2 \log (a+b \text{sech}(x))}{a b^4}+\frac{a \text{sech}^2(x)}{2 b^2}+\frac{\log (\cosh (x))}{a}-\frac{\text{sech}^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^5/(a + b*Sech[x]),x]

[Out]

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

Mathematica [A]  time = 0.170961, size = 85, normalized size = 1.18 \[ \frac{3 a^2 b^2 \text{sech}^2(x)-6 a b \left (a^2-2 b^2\right ) \text{sech}(x)-6 a^2 \left (a^2-2 b^2\right ) \log (\cosh (x))+6 \left (a^2-b^2\right )^2 \log (a \cosh (x)+b)-2 a b^3 \text{sech}^3(x)}{6 a b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^5/(a + b*Sech[x]),x]

[Out]

(-6*a^2*(a^2 - 2*b^2)*Log[Cosh[x]] + 6*(a^2 - b^2)^2*Log[b + a*Cosh[x]] - 6*a*b*(a^2 - 2*b^2)*Sech[x] + 3*a^2*
b^2*Sech[x]^2 - 2*a*b^3*Sech[x]^3)/(6*a*b^4)

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Maple [B]  time = 0.04, size = 233, normalized size = 3.2 \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 ) }+2\,{\frac{a}{{b}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+4\,{\frac{1}{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) a}{{b}^{2}}}-{\frac{8}{3\,b} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-2\,{\frac{{a}^{2}}{{b}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{a}{{b}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{1}{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+{\frac{{a}^{3}}{{b}^{4}}\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 ) }-2\,{\frac{a\ln \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+a+b \right ) }{{b}^{2}}}+{\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)^5/(a+b*sech(x)),x)

[Out]

-1/a*ln(tanh(1/2*x)+1)-1/a*ln(tanh(1/2*x)-1)+2/b^2/(tanh(1/2*x)^2+1)^2*a+4/b/(tanh(1/2*x)^2+1)^2-1/b^4*ln(tanh
(1/2*x)^2+1)*a^3+2/b^2*ln(tanh(1/2*x)^2+1)*a-8/3/b/(tanh(1/2*x)^2+1)^3-2/b^3/(tanh(1/2*x)^2+1)*a^2-2/b^2/(tanh
(1/2*x)^2+1)*a+2/b/(tanh(1/2*x)^2+1)+a^3/b^4*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b+a+b)-2*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 [B]  time = 1.66744, size = 221, normalized size = 3.07 \begin{align*} \frac{2 \,{\left (3 \, a b e^{\left (-2 \, x\right )} + 3 \, a b e^{\left (-4 \, x\right )} - 3 \,{\left (a^{2} - 2 \, b^{2}\right )} e^{\left (-x\right )} - 2 \,{\left (3 \, a^{2} - 4 \, b^{2}\right )} e^{\left (-3 \, x\right )} - 3 \,{\left (a^{2} - 2 \, b^{2}\right )} e^{\left (-5 \, x\right )}\right )}}{3 \,{\left (3 \, b^{3} e^{\left (-2 \, x\right )} + 3 \, b^{3} e^{\left (-4 \, x\right )} + b^{3} e^{\left (-6 \, x\right )} + b^{3}\right )}} + \frac{x}{a} - \frac{{\left (a^{3} - 2 \, a b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{4}} + \frac{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*sech(x)),x, algorithm="maxima")

[Out]

2/3*(3*a*b*e^(-2*x) + 3*a*b*e^(-4*x) - 3*(a^2 - 2*b^2)*e^(-x) - 2*(3*a^2 - 4*b^2)*e^(-3*x) - 3*(a^2 - 2*b^2)*e
^(-5*x))/(3*b^3*e^(-2*x) + 3*b^3*e^(-4*x) + b^3*e^(-6*x) + b^3) + x/a - (a^3 - 2*a*b^2)*log(e^(-2*x) + 1)/b^4
+ (a^4 - 2*a^2*b^2 + b^4)*log(2*b*e^(-x) + a*e^(-2*x) + a)/(a*b^4)

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Fricas [B]  time = 3.19241, size = 3216, normalized size = 44.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*sech(x)),x, algorithm="fricas")

[Out]

-1/3*(3*b^4*x*cosh(x)^6 + 3*b^4*x*sinh(x)^6 + 6*(a^3*b - 2*a*b^3)*cosh(x)^5 + 6*(3*b^4*x*cosh(x) + a^3*b - 2*a
*b^3)*sinh(x)^5 + 3*b^4*x + 3*(3*b^4*x - 2*a^2*b^2)*cosh(x)^4 + 3*(15*b^4*x*cosh(x)^2 + 3*b^4*x - 2*a^2*b^2 +
10*(a^3*b - 2*a*b^3)*cosh(x))*sinh(x)^4 + 4*(3*a^3*b - 4*a*b^3)*cosh(x)^3 + 4*(15*b^4*x*cosh(x)^3 + 3*a^3*b -
4*a*b^3 + 15*(a^3*b - 2*a*b^3)*cosh(x)^2 + 3*(3*b^4*x - 2*a^2*b^2)*cosh(x))*sinh(x)^3 + 3*(3*b^4*x - 2*a^2*b^2
)*cosh(x)^2 + 3*(15*b^4*x*cosh(x)^4 + 3*b^4*x - 2*a^2*b^2 + 20*(a^3*b - 2*a*b^3)*cosh(x)^3 + 6*(3*b^4*x - 2*a^
2*b^2)*cosh(x)^2 + 4*(3*a^3*b - 4*a*b^3)*cosh(x))*sinh(x)^2 + 6*(a^3*b - 2*a*b^3)*cosh(x) - 3*((a^4 - 2*a^2*b^
2 + b^4)*cosh(x)^6 + 6*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^5 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^6 + 3*(a^4
- 2*a^2*b^2 + b^4)*cosh(x)^4 + 3*(a^4 - 2*a^2*b^2 + b^4 + 5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^4 + a^4
- 2*a^2*b^2 + b^4 + 4*(5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x)^3 + 3
*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 3*(5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + a^4 - 2*a^2*b^2 + b^4 + 6*(a^4 -
2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 6*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^5 + 2*(a^4 - 2*a^2*b^2 + b^4)*cosh
(x)^3 + (a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) + 3*((a^4 - 2*a^2
*b^2)*cosh(x)^6 + 6*(a^4 - 2*a^2*b^2)*cosh(x)*sinh(x)^5 + (a^4 - 2*a^2*b^2)*sinh(x)^6 + 3*(a^4 - 2*a^2*b^2)*co
sh(x)^4 + 3*(a^4 - 2*a^2*b^2 + 5*(a^4 - 2*a^2*b^2)*cosh(x)^2)*sinh(x)^4 + a^4 - 2*a^2*b^2 + 4*(5*(a^4 - 2*a^2*
b^2)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2)*cosh(x))*sinh(x)^3 + 3*(a^4 - 2*a^2*b^2)*cosh(x)^2 + 3*(5*(a^4 - 2*a^2*b^
2)*cosh(x)^4 + a^4 - 2*a^2*b^2 + 6*(a^4 - 2*a^2*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^4 - 2*a^2*b^2)*cosh(x)^5 + 2
*(a^4 - 2*a^2*b^2)*cosh(x)^3 + (a^4 - 2*a^2*b^2)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 6*(3*b
^4*x*cosh(x)^5 + 5*(a^3*b - 2*a*b^3)*cosh(x)^4 + a^3*b - 2*a*b^3 + 2*(3*b^4*x - 2*a^2*b^2)*cosh(x)^3 + 2*(3*a^
3*b - 4*a*b^3)*cosh(x)^2 + (3*b^4*x - 2*a^2*b^2)*cosh(x))*sinh(x))/(a*b^4*cosh(x)^6 + 6*a*b^4*cosh(x)*sinh(x)^
5 + a*b^4*sinh(x)^6 + 3*a*b^4*cosh(x)^4 + 3*a*b^4*cosh(x)^2 + a*b^4 + 3*(5*a*b^4*cosh(x)^2 + a*b^4)*sinh(x)^4
+ 4*(5*a*b^4*cosh(x)^3 + 3*a*b^4*cosh(x))*sinh(x)^3 + 3*(5*a*b^4*cosh(x)^4 + 6*a*b^4*cosh(x)^2 + a*b^4)*sinh(x
)^2 + 6*(a*b^4*cosh(x)^5 + 2*a*b^4*cosh(x)^3 + a*b^4*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{5}{\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)**5/(a+b*sech(x)),x)

[Out]

Integral(tanh(x)**5/(a + b*sech(x)), x)

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Giac [B]  time = 1.15139, size = 205, normalized size = 2.85 \begin{align*} -\frac{{\left (a^{3} - 2 \, a b^{2}\right )} \log \left (e^{\left (-x\right )} + e^{x}\right )}{b^{4}} + \frac{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | a{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a b^{4}} + \frac{11 \, a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 22 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 12 \, a^{2} b{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 24 \, b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 12 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )} - 16 \, b^{3}}{6 \, b^{4}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*sech(x)),x, algorithm="giac")

[Out]

-(a^3 - 2*a*b^2)*log(e^(-x) + e^x)/b^4 + (a^4 - 2*a^2*b^2 + b^4)*log(abs(a*(e^(-x) + e^x) + 2*b))/(a*b^4) + 1/
6*(11*a^3*(e^(-x) + e^x)^3 - 22*a*b^2*(e^(-x) + e^x)^3 - 12*a^2*b*(e^(-x) + e^x)^2 + 24*b^3*(e^(-x) + e^x)^2 +
12*a*b^2*(e^(-x) + e^x) - 16*b^3)/(b^4*(e^(-x) + e^x)^3)