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

Optimal. Leaf size=64 \[ -\frac {b x}{a^2-b^2}+\frac {a \log (\cosh (x))}{a^2-b^2}+\frac {a^3 \log (a+b \tanh (x))}{b^2 \left (a^2-b^2\right )}-\frac {\tanh (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 64, 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 = {3566, 3626, 3617, 31, 3475} \[ -\frac {b x}{a^2-b^2}+\frac {a^3 \log (a+b \tanh (x))}{b^2 \left (a^2-b^2\right )}+\frac {a \log (\cosh (x))}{a^2-b^2}-\frac {\tanh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(a + b*Tanh[x]),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rule 3566

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 - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 65, normalized size = 1.02 \[ \frac {\left (a b^2-a^3\right ) \log (\cosh (x))+a^3 \log (a \cosh (x)+b \sinh (x))+\left (b^3-a^2 b\right ) \tanh (x)-b^3 x}{b^2 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.11, size = 264, normalized size = 4.12 \[ -\frac {{\left (a b^{2} + b^{3}\right )} x \cosh \relax (x)^{2} + 2 \, {\left (a b^{2} + b^{3}\right )} x \cosh \relax (x) \sinh \relax (x) + {\left (a b^{2} + b^{3}\right )} x \sinh \relax (x)^{2} - 2 \, a^{2} b + 2 \, b^{3} + {\left (a b^{2} + b^{3}\right )} x - {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) \sinh \relax (x) + a^{3} \sinh \relax (x)^{2} + a^{3}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (a^{3} - a b^{2} + {\left (a^{3} - a b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{3} - a b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{3} - a b^{2}\right )} \sinh \relax (x)^{2}\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b^{2} - b^{4} + {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 75, normalized size = 1.17 \[ \frac {a^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {x}{a - b} - \frac {a \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{2}} + \frac {2}{b {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 67, normalized size = 1.05 \[ -\frac {\tanh \relax (x )}{b}-\frac {\ln \left (\tanh \relax (x )-1\right )}{2 b +2 a}-\frac {\ln \left (1+\tanh \relax (x )\right )}{2 a -2 b}+\frac {a^{3} \ln \left (a +b \tanh \relax (x )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 71, normalized size = 1.11 \[ \frac {a^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b^{2} - b^{4}} + \frac {x}{a + b} - \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{2}} - \frac {2}{b e^{\left (-2 \, x\right )} + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.09, size = 59, normalized size = 0.92 \[ \frac {x}{a+b}-\frac {\mathrm {tanh}\relax (x)}{b}-\frac {a\,\ln \left (\mathrm {tanh}\relax (x)+1\right )}{a^2-b^2}+\frac {a^3\,\ln \left (a+b\,\mathrm {tanh}\relax (x)\right )}{b^2\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(a + b*tanh(x)),x)

[Out]

x/(a + b) - tanh(x)/b - (a*log(tanh(x) + 1))/(a^2 - b^2) + (a^3*log(a + b*tanh(x)))/(b^2*(a^2 - b^2))

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sympy [A]  time = 0.84, size = 330, normalized size = 5.16 \[ \begin {cases} \tilde {\infty } \left (x - \tanh {\relax (x )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {5 x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {5 x}{2 b \tanh {\relax (x )} - 2 b} - \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {2 \tanh ^{2}{\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {3}{2 b \tanh {\relax (x )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {x}{2 b \tanh {\relax (x )} + 2 b} + \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {2 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {2 \tanh ^{2}{\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {3}{2 b \tanh {\relax (x )} + 2 b} & \text {for}\: a = b \\\frac {x - \log {\left (\tanh {\relax (x )} + 1 \right )} - \frac {\tanh ^{2}{\relax (x )}}{2}}{a} & \text {for}\: b = 0 \\\frac {a^{3} \log {\left (\frac {a}{b} + \tanh {\relax (x )} \right )}}{a^{2} b^{2} - b^{4}} - \frac {a^{2} b \tanh {\relax (x )}}{a^{2} b^{2} - b^{4}} + \frac {a b^{2} x}{a^{2} b^{2} - b^{4}} - \frac {a b^{2} \log {\left (\tanh {\relax (x )} + 1 \right )}}{a^{2} b^{2} - b^{4}} - \frac {b^{3} x}{a^{2} b^{2} - b^{4}} + \frac {b^{3} \tanh {\relax (x )}}{a^{2} b^{2} - b^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(x - tanh(x)), Eq(a, 0) & Eq(b, 0)), (5*x*tanh(x)/(2*b*tanh(x) - 2*b) - 5*x/(2*b*tanh(x) - 2*b)
- 2*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) - 2*b) + 2*log(tanh(x) + 1)/(2*b*tanh(x) - 2*b) - 2*tanh(x)**2/(2*b
*tanh(x) - 2*b) + 3/(2*b*tanh(x) - 2*b), Eq(a, -b)), (x*tanh(x)/(2*b*tanh(x) + 2*b) + x/(2*b*tanh(x) + 2*b) +
2*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) + 2*b) + 2*log(tanh(x) + 1)/(2*b*tanh(x) + 2*b) - 2*tanh(x)**2/(2*b*ta
nh(x) + 2*b) + 3/(2*b*tanh(x) + 2*b), Eq(a, b)), ((x - log(tanh(x) + 1) - tanh(x)**2/2)/a, Eq(b, 0)), (a**3*lo
g(a/b + tanh(x))/(a**2*b**2 - b**4) - a**2*b*tanh(x)/(a**2*b**2 - b**4) + a*b**2*x/(a**2*b**2 - b**4) - a*b**2
*log(tanh(x) + 1)/(a**2*b**2 - b**4) - b**3*x/(a**2*b**2 - b**4) + b**3*tanh(x)/(a**2*b**2 - b**4), True))

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