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

Optimal. Leaf size=76 \[ -\frac {b x}{a^2-b^2}+\frac {b \tanh (x)}{a^2}+\frac {\left (a^2+b^2\right ) \log (\cosh (x))}{a^3}+\frac {b^4 \log (a \sinh (x)+b \cosh (x))}{a^3 \left (a^2-b^2\right )}-\frac {\tanh ^2(x)}{2 a} \]

[Out]

-b*x/(a^2-b^2)+(a^2+b^2)*ln(cosh(x))/a^3+b^4*ln(b*cosh(x)+a*sinh(x))/a^3/(a^2-b^2)+b*tanh(x)/a^2-1/2*tanh(x)^2
/a

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Rubi [A]  time = 0.33, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3569, 3649, 3652, 3530, 3475} \[ -\frac {b x}{a^2-b^2}+\frac {\left (a^2+b^2\right ) \log (\cosh (x))}{a^3}+\frac {b^4 \log (a \sinh (x)+b \cosh (x))}{a^3 \left (a^2-b^2\right )}+\frac {b \tanh (x)}{a^2}-\frac {\tanh ^2(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Coth[x]),x]

[Out]

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3652

Int[((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Simp[((a*(A*c - c*C) - b*(A*d - C*d))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[
(A*b^2 + a^2*C)/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] - Dist[(c^2*C
+ A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c,
d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rubi steps

\begin {align*} \int \frac {\tanh ^3(x)}{a+b \coth (x)} \, dx &=-\frac {\tanh ^2(x)}{2 a}-\frac {i \int \frac {\left (-2 i b+2 i a \coth (x)+2 i b \coth ^2(x)\right ) \tanh ^2(x)}{a+b \coth (x)} \, dx}{2 a}\\ &=\frac {b \tanh (x)}{a^2}-\frac {\tanh ^2(x)}{2 a}-\frac {\int \frac {\left (-2 \left (a^2+b^2\right )+2 b^2 \coth ^2(x)\right ) \tanh (x)}{a+b \coth (x)} \, dx}{2 a^2}\\ &=-\frac {b x}{a^2-b^2}+\frac {b \tanh (x)}{a^2}-\frac {\tanh ^2(x)}{2 a}+\frac {\left (i b^4\right ) \int \frac {-i b-i a \coth (x)}{a+b \coth (x)} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {\left (a^2+b^2\right ) \int \tanh (x) \, dx}{a^3}\\ &=-\frac {b x}{a^2-b^2}+\frac {\left (a^2+b^2\right ) \log (\cosh (x))}{a^3}+\frac {b^4 \log (b \cosh (x)+a \sinh (x))}{a^3 \left (a^2-b^2\right )}+\frac {b \tanh (x)}{a^2}-\frac {\tanh ^2(x)}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 88, normalized size = 1.16 \[ \frac {a^2 \left (a^2-b^2\right ) \text {sech}^2(x)+2 \left (\left (a^4-b^4\right ) \log (\cosh (x))-a^3 b x+a b \left (a^2-b^2\right ) \tanh (x)+b^4 \log (a \sinh (x)+b \cosh (x))\right )}{2 a^3 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 637, normalized size = 8.38 \[ -\frac {{\left (a^{4} + a^{3} b\right )} x \cosh \relax (x)^{4} + 4 \, {\left (a^{4} + a^{3} b\right )} x \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{4} + a^{3} b\right )} x \sinh \relax (x)^{4} + 2 \, a^{3} b - 2 \, a b^{3} - 2 \, {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3} - {\left (a^{4} + a^{3} b\right )} x\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3} - 3 \, {\left (a^{4} + a^{3} b\right )} x \cosh \relax (x)^{2} - {\left (a^{4} + a^{3} b\right )} x\right )} \sinh \relax (x)^{2} + {\left (a^{4} + a^{3} b\right )} x - {\left (b^{4} \cosh \relax (x)^{4} + 4 \, b^{4} \cosh \relax (x) \sinh \relax (x)^{3} + b^{4} \sinh \relax (x)^{4} + 2 \, b^{4} \cosh \relax (x)^{2} + b^{4} + 2 \, {\left (3 \, b^{4} \cosh \relax (x)^{2} + b^{4}\right )} \sinh \relax (x)^{2} + 4 \, {\left (b^{4} \cosh \relax (x)^{3} + b^{4} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left ({\left (a^{4} - b^{4}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{4} - b^{4}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{4} - b^{4}\right )} \sinh \relax (x)^{4} + a^{4} - b^{4} + 2 \, {\left (a^{4} - b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} - b^{4} + 3 \, {\left (a^{4} - b^{4}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 4 \, {\left ({\left (a^{4} - b^{4}\right )} \cosh \relax (x)^{3} + {\left (a^{4} - b^{4}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left ({\left (a^{4} + a^{3} b\right )} x \cosh \relax (x)^{3} - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3} - {\left (a^{4} + a^{3} b\right )} x\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{a^{5} - a^{3} b^{2} + {\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \sinh \relax (x)^{4} + 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{5} - a^{3} b^{2} + 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 4 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x)^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^4 + a^3*b)*x*cosh(x)^4 + 4*(a^4 + a^3*b)*x*cosh(x)*sinh(x)^3 + (a^4 + a^3*b)*x*sinh(x)^4 + 2*a^3*b - 2*a*
b^3 - 2*(a^4 - a^3*b - a^2*b^2 + a*b^3 - (a^4 + a^3*b)*x)*cosh(x)^2 - 2*(a^4 - a^3*b - a^2*b^2 + a*b^3 - 3*(a^
4 + a^3*b)*x*cosh(x)^2 - (a^4 + a^3*b)*x)*sinh(x)^2 + (a^4 + a^3*b)*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))*sinh(x))*log(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) - ((a^4 - b^4)*cosh(x)^4 + 4*(a^4 - b^4)*cosh
(x)*sinh(x)^3 + (a^4 - b^4)*sinh(x)^4 + a^4 - b^4 + 2*(a^4 - b^4)*cosh(x)^2 + 2*(a^4 - b^4 + 3*(a^4 - b^4)*cos
h(x)^2)*sinh(x)^2 + 4*((a^4 - b^4)*cosh(x)^3 + (a^4 - b^4)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))
) + 4*((a^4 + a^3*b)*x*cosh(x)^3 - (a^4 - a^3*b - a^2*b^2 + a*b^3 - (a^4 + a^3*b)*x)*cosh(x))*sinh(x))/(a^5 -
a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4 + 2*(a^5
- a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2 + 3*(a^5 - a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2)*cosh(x)
^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x))

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giac [A]  time = 0.12, size = 97, normalized size = 1.28 \[ \frac {b^{4} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{5} - a^{3} b^{2}} - \frac {x}{a - b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{3}} - \frac {2 \, {\left (a b - {\left (a^{2} - a b\right )} e^{\left (2 \, x\right )}\right )}}{a^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 167, normalized size = 2.20 \[ \frac {b^{4} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {32 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{32 a +32 b}-\frac {32 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{32 a -32 b}+\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 b \tanh \left (\frac {x}{2}\right )}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{a}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^4/(a+b)/(a-b)/a^3*ln(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)-32/(32*a+32*b)*ln(tanh(1/2*x)-1)-32/(32*a-32*b)*ln(t
anh(1/2*x)+1)+2/a^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*b-2/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^2+2/a^2/(tanh(1/2*
x)^2+1)^2*b*tanh(1/2*x)+1/a*ln(tanh(1/2*x)^2+1)+1/a^3*ln(tanh(1/2*x)^2+1)*b^2

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maxima [A]  time = 0.41, size = 94, normalized size = 1.24 \[ \frac {b^{4} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{5} - a^{3} b^{2}} + \frac {2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} + b\right )}}{2 \, a^{2} e^{\left (-2 \, x\right )} + a^{2} e^{\left (-4 \, x\right )} + a^{2}} + \frac {x}{a + b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.51, size = 111, normalized size = 1.46 \[ \frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )\,\left (a^2+b^2\right )}{a^3}-\frac {x}{a-b}-\frac {2}{a\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}+\frac {b^4\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^5-a^3\,b^2}+\frac {2\,\left (a^2-b^2\right )}{a^2\,\left (a+b\right )\,\left ({\mathrm {e}}^{2\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a + b*coth(x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{3}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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