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

Optimal. Leaf size=113 \[ -\frac{a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac{b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac{b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac{b^4 \log (a+b \text{csch}(x))}{a \left (a^2+b^2\right )^2}-\frac{\tanh ^2(x) (a-b \text{csch}(x))}{2 \left (a^2+b^2\right )}+\frac{\log (\sinh (x))}{a} \]

[Out]

-((b^3*ArcTan[Sinh[x]])/(a^2 + b^2)^2) - (b*ArcTan[Sinh[x]])/(2*(a^2 + b^2)) + (b^4*Log[a + b*Csch[x]])/(a*(a^
2 + b^2)^2) + Log[Sinh[x]]/a - (a*(a^2 + 2*b^2)*Log[Tanh[x]])/(a^2 + b^2)^2 - ((a - b*Csch[x])*Tanh[x]^2)/(2*(
a^2 + b^2))

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Rubi [A]  time = 0.162987, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3885, 894, 639, 203, 635, 260} \[ -\frac{a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac{b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac{b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac{b^4 \log (a+b \text{csch}(x))}{a \left (a^2+b^2\right )^2}-\frac{\tanh ^2(x) (a-b \text{csch}(x))}{2 \left (a^2+b^2\right )}+\frac{\log (\sinh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Csch[x]),x]

[Out]

-((b^3*ArcTan[Sinh[x]])/(a^2 + b^2)^2) - (b*ArcTan[Sinh[x]])/(2*(a^2 + b^2)) + (b^4*Log[a + b*Csch[x]])/(a*(a^
2 + b^2)^2) + Log[Sinh[x]]/a - (a*(a^2 + 2*b^2)*Log[Tanh[x]])/(a^2 + b^2)^2 - ((a - b*Csch[x])*Tanh[x]^2)/(2*(
a^2 + b^2))

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]))

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 203

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

Rule 635

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

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]

Rubi steps

\begin{align*} \int \frac{\tanh ^3(x)}{a+b \text{csch}(x)} \, dx &=-\left (b^4 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (-b^2-x^2\right )^2} \, dx,x,b \text{csch}(x)\right )\right )\\ &=-\left (b^4 \operatorname{Subst}\left (\int \left (\frac{1}{a b^4 x}-\frac{1}{a \left (a^2+b^2\right )^2 (a+x)}+\frac{-b^2-a x}{b^2 \left (a^2+b^2\right ) \left (b^2+x^2\right )^2}+\frac{-b^4-a \left (a^2+2 b^2\right ) x}{b^4 \left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \text{csch}(x)\right )\right )\\ &=\frac{b^4 \log (a+b \text{csch}(x))}{a \left (a^2+b^2\right )^2}+\frac{\log (\sinh (x))}{a}-\frac{\operatorname{Subst}\left (\int \frac{-b^4-a \left (a^2+2 b^2\right ) x}{b^2+x^2} \, dx,x,b \text{csch}(x)\right )}{\left (a^2+b^2\right )^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{-b^2-a x}{\left (b^2+x^2\right )^2} \, dx,x,b \text{csch}(x)\right )}{a^2+b^2}\\ &=\frac{b^4 \log (a+b \text{csch}(x))}{a \left (a^2+b^2\right )^2}+\frac{\log (\sinh (x))}{a}-\frac{(a-b \text{csch}(x)) \tanh ^2(x)}{2 \left (a^2+b^2\right )}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \text{csch}(x)\right )}{\left (a^2+b^2\right )^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \text{csch}(x)\right )}{2 \left (a^2+b^2\right )}+\frac{\left (a \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \text{csch}(x)\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac{b^3 \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}-\frac{b \tan ^{-1}(\sinh (x))}{2 \left (a^2+b^2\right )}+\frac{b^4 \log (a+b \text{csch}(x))}{a \left (a^2+b^2\right )^2}+\frac{\log (\sinh (x))}{a}-\frac{a \left (a^2+2 b^2\right ) \log (\tanh (x))}{\left (a^2+b^2\right )^2}-\frac{(a-b \text{csch}(x)) \tanh ^2(x)}{2 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.189544, size = 191, normalized size = 1.69 \[ \frac{a^2 \left (a^2+b^2\right ) \text{sech}^2(x)+2 a^2 b^2 \log (-\sinh (x)+i)+2 a^2 b^2 \log (\sinh (x)+i)+a b \left (a^2+b^2\right ) \tan ^{-1}(\sinh (x))+a b \left (a^2+b^2\right ) \tanh (x) \text{sech}(x)+i a^3 b \log (-\sinh (x)+i)-i a^3 b \log (\sinh (x)+i)+a^4 \log (-\sinh (x)+i)+a^4 \log (\sinh (x)+i)+2 i a b^3 \log (-\sinh (x)+i)-2 i a b^3 \log (\sinh (x)+i)+2 b^4 \log (a \sinh (x)+b)}{2 a \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(a^2 + b^2)*ArcTan[Sinh[x]] + a^4*Log[I - Sinh[x]] + I*a^3*b*Log[I - Sinh[x]] + 2*a^2*b^2*Log[I - Sinh[x]
] + (2*I)*a*b^3*Log[I - Sinh[x]] + a^4*Log[I + Sinh[x]] - I*a^3*b*Log[I + Sinh[x]] + 2*a^2*b^2*Log[I + Sinh[x]
] - (2*I)*a*b^3*Log[I + Sinh[x]] + 2*b^4*Log[b + a*Sinh[x]] + a^2*(a^2 + b^2)*Sech[x]^2 + a*b*(a^2 + b^2)*Sech
[x]*Tanh[x])/(2*a*(a^2 + b^2)^2)

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Maple [B]  time = 0.049, size = 324, normalized size = 2.9 \begin{align*} -{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{4}}{a \left ({a}^{2}+{b}^{2} \right ) ^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-{\frac{{b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{{b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{{a}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\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}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-3\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ){b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*ln(tanh(1/2*x)+1)+b^4/a/(a^2+b^2)^2*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)-1/a*ln(tanh(1/2*x)-1)-1/(a^2+b^
2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a^2*b-1/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*b^3-2/(a^2+b^2)^2
/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a^3-2/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2*a*b^2+1/(a^2+b^2)^2/(ta
nh(1/2*x)^2+1)^2*tanh(1/2*x)*a^2*b+1/(a^2+b^2)^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*b^3+1/(a^2+b^2)^2*ln(tanh(1/2
*x)^2+1)*a^3+2/(a^2+b^2)^2*ln(tanh(1/2*x)^2+1)*a*b^2-3/(a^2+b^2)^2*arctan(tanh(1/2*x))*b^3-1/(a^2+b^2)^2*arcta
n(tanh(1/2*x))*a^2*b

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Maxima [A]  time = 1.58664, size = 232, normalized size = 2.05 \begin{align*} \frac{b^{4} \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (a^{2} b + 3 \, b^{3}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{b e^{\left (-x\right )} + 2 \, a e^{\left (-2 \, x\right )} - b e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \,{\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} +{\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} + \frac{x}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.95952, size = 2379, normalized size = 21.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^4 + (a^4 + 2*a^2*b^2 + b^4)*x*sinh(x)^4 - (a^3*b + a*b^3)*cosh(x)^3 - (a^3
*b + a*b^3 - 4*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x))*sinh(x)^3 - 2*(a^4 + a^2*b^2 - (a^4 + 2*a^2*b^2 + b^4)*x)*co
sh(x)^2 - (2*a^4 + 2*a^2*b^2 - 6*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^2 - 2*(a^4 + 2*a^2*b^2 + b^4)*x + 3*(a^3*b
+ a*b^3)*cosh(x))*sinh(x)^2 + (a^4 + 2*a^2*b^2 + b^4)*x + ((a^3*b + 3*a*b^3)*cosh(x)^4 + 4*(a^3*b + 3*a*b^3)*c
osh(x)*sinh(x)^3 + (a^3*b + 3*a*b^3)*sinh(x)^4 + a^3*b + 3*a*b^3 + 2*(a^3*b + 3*a*b^3)*cosh(x)^2 + 2*(a^3*b +
3*a*b^3 + 3*(a^3*b + 3*a*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3*b + 3*a*b^3)*cosh(x)^3 + (a^3*b + 3*a*b^3)*cosh(x
))*sinh(x))*arctan(cosh(x) + sinh(x)) + (a^3*b + a*b^3)*cosh(x) - (b^4*cosh(x)^4 + 4*b^4*cosh(x)*sinh(x)^3 + b
^4*sinh(x)^4 + 2*b^4*cosh(x)^2 + b^4 + 2*(3*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 4*(b^4*cosh(x)^3 + b^4*cosh(x))*s
inh(x))*log(2*(a*sinh(x) + b)/(cosh(x) - sinh(x))) - ((a^4 + 2*a^2*b^2)*cosh(x)^4 + 4*(a^4 + 2*a^2*b^2)*cosh(x
)*sinh(x)^3 + (a^4 + 2*a^2*b^2)*sinh(x)^4 + a^4 + 2*a^2*b^2 + 2*(a^4 + 2*a^2*b^2)*cosh(x)^2 + 2*(a^4 + 2*a^2*b
^2 + 3*(a^4 + 2*a^2*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + 2*a^2*b^2)*cosh(x)^3 + (a^4 + 2*a^2*b^2)*cosh(x))*si
nh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + (4*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^3 + a^3*b + a*b^3 - 3*(a^3*b
+ a*b^3)*cosh(x)^2 - 4*(a^4 + a^2*b^2 - (a^4 + 2*a^2*b^2 + b^4)*x)*cosh(x))*sinh(x))/(a^5 + 2*a^3*b^2 + a*b^4
+ (a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^4 + 4*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)*sinh(x)^3 + (a^5 + 2*a^3*b^2 + a*b
^4)*sinh(x)^4 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4 + 3*(a^5 + 2*a^3*b^2 + a*b^
4)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)^3 + (a^5 + 2*a^3*b^2 + 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 ^{3}{\left (x \right )}}{a + b \operatorname{csch}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.22909, size = 316, normalized size = 2.8 \begin{align*} \frac{b^{4} \log \left ({\left | -a{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} - \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}{\left (a^{2} b + 3 \, b^{3}\right )}}{4 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{a^{3}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 2 \, a b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 2 \, a^{2} b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b^{3}{\left (e^{\left (-x\right )} - e^{x}\right )} + 4 \, a b^{2}}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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