3.2.41 \(\int \frac {\tanh ^3(x)}{a+b \coth (x)} \, dx\) [141]

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

[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 = 0.22, 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 = {3650, 3730, 3733, 3611, 3556} \begin {gather*} -\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} \end {gather*}

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 3556

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

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], 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 3650

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 3730

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*Ta
n[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 3733

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 = 0.23, size = 88, normalized size = 1.16 \begin {gather*} \frac {a^2 \left (a^2-b^2\right ) \text {sech}^2(x)+2 \left (-a^3 b x+\left (a^4-b^4\right ) \log (\cosh (x))+b^4 \log (b \cosh (x)+a \sinh (x))+a b \left (a^2-b^2\right ) \tanh (x)\right )}{2 a^3 (a-b) (a+b)} \end {gather*}

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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Maple [A]
time = 0.66, size = 138, normalized size = 1.82

method result size
risch \(\frac {x}{a +b}-\frac {2 x}{a}-\frac {2 x \,b^{2}}{a^{3}}-\frac {2 x \,b^{4}}{a^{3} \left (a^{2}-b^{2}\right )}+\frac {2 \,{\mathrm e}^{2 x} a -2 b \,{\mathrm e}^{2 x}-2 b}{\left (1+{\mathrm e}^{2 x}\right )^{2} a^{2}}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{a}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right ) b^{2}}{a^{3}}+\frac {b^{4} \ln \left ({\mathrm e}^{2 x}-\frac {a -b}{a +b}\right )}{a^{3} \left (a^{2}-b^{2}\right )}\) \(135\)
default \(-\frac {32 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{32 a -32 b}-\frac {32 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{32 a +32 b}+\frac {b^{4} \ln \left (b \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}+\frac {\frac {2 \left (a b \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )-a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\tanh \left (\frac {x}{2}\right ) b a \right )}{\left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\left (a^{2}+b^{2}\right ) \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{a^{3}}\) \(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 94, normalized size = 1.24 \begin {gather*} \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}} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 637 vs. \(2 (74) = 148\).
time = 0.40, size = 637, normalized size = 8.38 \begin {gather*} -\frac {{\left (a^{4} + a^{3} b\right )} x \cosh \left (x\right )^{4} + 4 \, {\left (a^{4} + a^{3} b\right )} x \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{4} + a^{3} b\right )} x \sinh \left (x\right )^{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 \left (x\right )^{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 \left (x\right )^{2} - {\left (a^{4} + a^{3} b\right )} x\right )} \sinh \left (x\right )^{2} + {\left (a^{4} + a^{3} b\right )} x - {\left (b^{4} \cosh \left (x\right )^{4} + 4 \, b^{4} \cosh \left (x\right ) \sinh \left (x\right )^{3} + b^{4} \sinh \left (x\right )^{4} + 2 \, b^{4} \cosh \left (x\right )^{2} + b^{4} + 2 \, {\left (3 \, b^{4} \cosh \left (x\right )^{2} + b^{4}\right )} \sinh \left (x\right )^{2} + 4 \, {\left (b^{4} \cosh \left (x\right )^{3} + b^{4} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left ({\left (a^{4} - b^{4}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{4} - b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{4} - b^{4}\right )} \sinh \left (x\right )^{4} + a^{4} - b^{4} + 2 \, {\left (a^{4} - b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} - b^{4} + 3 \, {\left (a^{4} - b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (a^{4} - b^{4}\right )} \cosh \left (x\right )^{3} + {\left (a^{4} - b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left ({\left (a^{4} + a^{3} b\right )} x \cosh \left (x\right )^{3} - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3} - {\left (a^{4} + a^{3} b\right )} x\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{a^{5} - a^{3} b^{2} + {\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \sinh \left (x\right )^{4} + 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{5} - a^{3} b^{2} + 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right )^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{3}{\left (x \right )}}{a + b \coth {\left (x \right )}}\, dx \end {gather*}

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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Giac [A]
time = 0.41, size = 97, normalized size = 1.28 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 1.51, size = 111, normalized size = 1.46 \begin {gather*} \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 )} \end {gather*}

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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