3.114 \(\int \frac{\tanh ^4(x)}{a+b \text{csch}(x)} \, dx\)

Optimal. Leaf size=183 \[ \frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{a x}{a^2+b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}+\frac{b^3 \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2} \]

[Out]

(a*b^2*x)/(a^2 + b^2)^2 + (b^4*x)/(a*(a^2 + b^2)^2) + (a*x)/(a^2 + b^2) + (2*b^5*ArcTanh[(a - b*Tanh[x/2])/Sqr
t[a^2 + b^2]])/(a*(a^2 + b^2)^(5/2)) + (b^3*Sech[x])/(a^2 + b^2)^2 + (b*Sech[x])/(a^2 + b^2) - (b*Sech[x]^3)/(
3*(a^2 + b^2)) - (a*b^2*Tanh[x])/(a^2 + b^2)^2 - (a*Tanh[x])/(a^2 + b^2) - (a*Tanh[x]^3)/(3*(a^2 + b^2))

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Rubi [A]  time = 0.384148, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {3898, 2902, 2606, 3473, 8, 2735, 2660, 618, 204} \[ \frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{a x}{a^2+b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}+\frac{b^3 \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(a + b*Csch[x]),x]

[Out]

(a*b^2*x)/(a^2 + b^2)^2 + (b^4*x)/(a*(a^2 + b^2)^2) + (a*x)/(a^2 + b^2) + (2*b^5*ArcTanh[(a - b*Tanh[x/2])/Sqr
t[a^2 + b^2]])/(a*(a^2 + b^2)^(5/2)) + (b^3*Sech[x])/(a^2 + b^2)^2 + (b*Sech[x])/(a^2 + b^2) - (b*Sech[x]^3)/(
3*(a^2 + b^2)) - (a*b^2*Tanh[x])/(a^2 + b^2)^2 - (a*Tanh[x])/(a^2 + b^2) - (a*Tanh[x]^3)/(3*(a^2 + b^2))

Rule 3898

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

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

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])

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{a+b \text{csch}(x)} \, dx &=i \int \frac{\sinh (x) \tanh ^4(x)}{i b+i a \sinh (x)} \, dx\\ &=\frac{a \int \tanh ^4(x) \, dx}{a^2+b^2}-\frac{b \int \text{sech}(x) \tanh ^3(x) \, dx}{a^2+b^2}+\frac{\left (i b^2\right ) \int \frac{\sinh (x) \tanh ^2(x)}{i b+i a \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac{\left (a b^2\right ) \int \tanh ^2(x) \, dx}{\left (a^2+b^2\right )^2}-\frac{b^3 \int \text{sech}(x) \tanh (x) \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (i b^4\right ) \int \frac{\sinh (x)}{i b+i a \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{a \int \tanh ^2(x) \, dx}{a^2+b^2}-\frac{b \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\text{sech}(x)\right )}{a^2+b^2}\\ &=\frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac{\left (a b^2\right ) \int 1 \, dx}{\left (a^2+b^2\right )^2}+\frac{b^3 \operatorname{Subst}(\int 1 \, dx,x,\text{sech}(x))}{\left (a^2+b^2\right )^2}-\frac{\left (i b^5\right ) \int \frac{1}{i b+i a \sinh (x)} \, dx}{a \left (a^2+b^2\right )^2}+\frac{a \int 1 \, dx}{a^2+b^2}\\ &=\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{a x}{a^2+b^2}+\frac{b^3 \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac{\left (2 i b^5\right ) \operatorname{Subst}\left (\int \frac{1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right )^2}\\ &=\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{a x}{a^2+b^2}+\frac{b^3 \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac{\left (4 i b^5\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,2 i a-2 i b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2+b^2\right )^2}\\ &=\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{b^4 x}{a \left (a^2+b^2\right )^2}+\frac{a x}{a^2+b^2}+\frac{2 b^5 \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2}}+\frac{b^3 \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{b \text{sech}(x)}{a^2+b^2}-\frac{b \text{sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{a \tanh ^3(x)}{3 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.695776, size = 141, normalized size = 0.77 \[ \frac{1}{3} \left (-\frac{a \left (4 a^2+7 b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{b \text{sech}^3(x)}{a^2+b^2}+\frac{3 b \left (a^2+2 b^2\right ) \text{sech}(x)}{\left (a^2+b^2\right )^2}+\frac{3 \left (x-\frac{2 b^5 \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{5/2}}\right )}{a}+\frac{a \tanh (x) \text{sech}^2(x)}{a^2+b^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(a + b*Csch[x]),x]

[Out]

((3*(x - (2*b^5*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(5/2)))/a + (3*b*(a^2 + 2*b^2)*Sech[x
])/(a^2 + b^2)^2 - (b*Sech[x]^3)/(a^2 + b^2) - (a*(4*a^2 + 7*b^2)*Tanh[x])/(a^2 + b^2)^2 + (a*Sech[x]^2*Tanh[x
])/(a^2 + b^2))/3

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Maple [A]  time = 0.053, size = 207, normalized size = 1.1 \begin{align*}{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-2\,{\frac{{b}^{5}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{5}+{b}^{3} \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+ \left ( -10/3\,{a}^{3}-16/3\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}+ \left ( 2\,{a}^{2}b+4\,{b}^{3} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \tanh \left ( x/2 \right ) +2/3\,{a}^{2}b+5/3\,{b}^{3}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a+b*csch(x)),x)

[Out]

1/a*ln(tanh(1/2*x)+1)-2/a*b^5/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^
(1/2))-1/a*ln(tanh(1/2*x)-1)+2/(a^4+2*a^2*b^2+b^4)*((-a^3-2*a*b^2)*tanh(1/2*x)^5+b^3*tanh(1/2*x)^4+(-10/3*a^3-
16/3*a*b^2)*tanh(1/2*x)^3+(2*a^2*b+4*b^3)*tanh(1/2*x)^2+(-a^3-2*a*b^2)*tanh(1/2*x)+2/3*a^2*b+5/3*b^3)/(tanh(1/
2*x)^2+1)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*csch(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.85507, size = 4128, normalized size = 22.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*csch(x)),x, algorithm="fricas")

[Out]

1/3*(3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x*cosh(x)^6 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x*sinh(x)^6 + 8
*a^6 + 22*a^4*b^2 + 14*a^2*b^4 + 6*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x)^5 + 6*(a^5*b + 3*a^3*b^3 + 2*a*b^5 +
3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x*cosh(x))*sinh(x)^5 + 3*(4*a^6 + 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 + 3*a^
4*b^2 + 3*a^2*b^4 + b^6)*x)*cosh(x)^4 + 3*(4*a^6 + 10*a^4*b^2 + 6*a^2*b^4 + 15*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6)*x*cosh(x)^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x + 10*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x))*sinh(x)
^4 + 4*(a^5*b + 5*a^3*b^3 + 4*a*b^5)*cosh(x)^3 + 4*(a^5*b + 5*a^3*b^3 + 4*a*b^5 + 15*(a^6 + 3*a^4*b^2 + 3*a^2*
b^4 + b^6)*x*cosh(x)^3 + 15*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x)^2 + 3*(4*a^6 + 10*a^4*b^2 + 6*a^2*b^4 + 3*(a
^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x)*cosh(x))*sinh(x)^3 + 3*(4*a^6 + 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6)*x)*cosh(x)^2 + 3*(4*a^6 + 12*a^4*b^2 + 8*a^2*b^4 + 15*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
)*x*cosh(x)^4 + 20*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x)^3 + 6*(4*a^6 + 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 + 3*a^
4*b^2 + 3*a^2*b^4 + b^6)*x)*cosh(x)^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x + 4*(a^5*b + 5*a^3*b^3 + 4*a*b
^5)*cosh(x))*sinh(x)^2 + 3*(b^5*cosh(x)^6 + 6*b^5*cosh(x)*sinh(x)^5 + b^5*sinh(x)^6 + 3*b^5*cosh(x)^4 + 3*b^5*
cosh(x)^2 + b^5 + 3*(5*b^5*cosh(x)^2 + b^5)*sinh(x)^4 + 4*(5*b^5*cosh(x)^3 + 3*b^5*cosh(x))*sinh(x)^3 + 3*(5*b
^5*cosh(x)^4 + 6*b^5*cosh(x)^2 + b^5)*sinh(x)^2 + 6*(b^5*cosh(x)^5 + 2*b^5*cosh(x)^3 + b^5*cosh(x))*sinh(x))*s
qrt(a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x
) + 2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b
)*sinh(x) - a)) + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x + 6*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*cosh(x) + 6*(3*(a^
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x*cosh(x)^5 + a^5*b + 3*a^3*b^3 + 2*a*b^5 + 5*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*c
osh(x)^4 + 2*(4*a^6 + 10*a^4*b^2 + 6*a^2*b^4 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x)*cosh(x)^3 + 2*(a^5*b +
5*a^3*b^3 + 4*a*b^5)*cosh(x)^2 + (4*a^6 + 12*a^4*b^2 + 8*a^2*b^4 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x)*c
osh(x))*sinh(x))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^6 + 6*(a
^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)*sinh(x)^5 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*sinh(x)^6 + 3*(a
^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^4 + 3*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 5*(a^7 + 3*a^5*b^2 +
3*a^3*b^4 + a*b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^3 + 3*(a^7 + 3*a^
5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)^3 + 3*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^2 + 3*(a^7 + 3
*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 5*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^4 + 6*(a^7 + 3*a^5*b^2 + 3*a^3*
b^4 + a*b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^5 + 2*(a^7 + 3*a^5*b^2 +
3*a^3*b^4 + a*b^6)*cosh(x)^3 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\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)**4/(a+b*csch(x)),x)

[Out]

Integral(tanh(x)**4/(a + b*csch(x)), x)

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Giac [A]  time = 1.20349, size = 290, normalized size = 1.58 \begin{align*} -\frac{b^{5} \log \left (\frac{{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{x}{a} + \frac{2 \,{\left (3 \, a^{2} b e^{\left (5 \, x\right )} + 6 \, b^{3} e^{\left (5 \, x\right )} + 6 \, a^{3} e^{\left (4 \, x\right )} + 9 \, a b^{2} e^{\left (4 \, x\right )} + 2 \, a^{2} b e^{\left (3 \, x\right )} + 8 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} + 6 \, b^{3} e^{x} + 4 \, a^{3} + 7 \, a b^{2}\right )}}{3 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(a+b*csch(x)),x, algorithm="giac")

[Out]

-b^5*log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/((a^5 + 2*a^3*b^2 + a*
b^4)*sqrt(a^2 + b^2)) + x/a + 2/3*(3*a^2*b*e^(5*x) + 6*b^3*e^(5*x) + 6*a^3*e^(4*x) + 9*a*b^2*e^(4*x) + 2*a^2*b
*e^(3*x) + 8*b^3*e^(3*x) + 6*a^3*e^(2*x) + 12*a*b^2*e^(2*x) + 3*a^2*b*e^x + 6*b^3*e^x + 4*a^3 + 7*a*b^2)/((a^4
+ 2*a^2*b^2 + b^4)*(e^(2*x) + 1)^3)