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

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

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3566, 3647, 3627, 3617, 31, 3475} \[ \frac {a x}{a^2-b^2}-\frac {a^4 \log (a+b \tanh (x))}{b^3 \left (a^2-b^2\right )}-\frac {b \log (\cosh (x))}{a^2-b^2}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(a + b*Tanh[x]),x]

[Out]

(a*x)/(a^2 - b^2) - (b*Log[Cosh[x]])/(a^2 - b^2) - (a^4*Log[a + b*Tanh[x]])/(b^3*(a^2 - b^2)) + (a*Tanh[x])/b^
2 - Tanh[x]^2/(2*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 3627

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

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 88, normalized size = 1.16 \[ \frac {b^2 \left (a^2-b^2\right ) \text {sech}^2(x)+2 \left (\left (a^4-b^4\right ) \log (\cosh (x))+a^4 (-\log (a \cosh (x)+b \sinh (x)))+a b \left (a^2-b^2\right ) \tanh (x)+a b^3 x\right )}{2 b^3 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.72, size = 644, normalized size = 8.47 \[ \frac {{\left (a b^{3} + b^{4}\right )} x \cosh \relax (x)^{4} + 4 \, {\left (a b^{3} + b^{4}\right )} x \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a b^{3} + b^{4}\right )} x \sinh \relax (x)^{4} - 2 \, a^{3} b + 2 \, a b^{3} - 2 \, {\left (a^{3} b - a^{2} b^{2} - a b^{3} + b^{4} - {\left (a b^{3} + b^{4}\right )} x\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} b - a^{2} b^{2} - a b^{3} + b^{4} - 3 \, {\left (a b^{3} + b^{4}\right )} x \cosh \relax (x)^{2} - {\left (a b^{3} + b^{4}\right )} x\right )} \sinh \relax (x)^{2} + {\left (a b^{3} + b^{4}\right )} x - {\left (a^{4} \cosh \relax (x)^{4} + 4 \, a^{4} \cosh \relax (x) \sinh \relax (x)^{3} + a^{4} \sinh \relax (x)^{4} + 2 \, a^{4} \cosh \relax (x)^{2} + a^{4} + 2 \, {\left (3 \, a^{4} \cosh \relax (x)^{2} + a^{4}\right )} \sinh \relax (x)^{2} + 4 \, {\left (a^{4} \cosh \relax (x)^{3} + a^{4} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \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 b^{3} + b^{4}\right )} x \cosh \relax (x)^{3} - {\left (a^{3} b - a^{2} b^{2} - a b^{3} + b^{4} - {\left (a b^{3} + b^{4}\right )} x\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{2} b^{3} - b^{5}\right )} \sinh \relax (x)^{4} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} b^{3} - b^{5} + 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 4 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{3} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 98, normalized size = 1.29 \[ -\frac {a^{4} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b^{3} - b^{5}} + \frac {x}{a - b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{3}} - \frac {2 \, {\left (a b + {\left (a b - b^{2}\right )} e^{\left (2 \, x\right )}\right )}}{b^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 76, normalized size = 1.00 \[ -\frac {\tanh ^{2}\relax (x )}{2 b}+\frac {a \tanh \relax (x )}{b^{2}}-\frac {\ln \left (\tanh \relax (x )-1\right )}{2 b +2 a}+\frac {\ln \left (1+\tanh \relax (x )\right )}{2 a -2 b}-\frac {a^{4} \ln \left (a +b \tanh \relax (x )\right )}{b^{3} \left (a +b \right ) \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 100, normalized size = 1.32 \[ -\frac {a^{4} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b^{3} - b^{5}} + \frac {2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} + a\right )}}{2 \, b^{2} e^{\left (-2 \, x\right )} + b^{2} e^{\left (-4 \, x\right )} + b^{2}} + \frac {x}{a + b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 68, normalized size = 0.89 \[ \frac {x}{a+b}-\frac {{\mathrm {tanh}\relax (x)}^2}{2\,b}+\frac {b\,\ln \left (\mathrm {tanh}\relax (x)+1\right )}{a^2-b^2}+\frac {a\,\mathrm {tanh}\relax (x)}{b^2}-\frac {a^4\,\ln \left (a+b\,\mathrm {tanh}\relax (x)\right )}{b^3\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(a + b*tanh(x)),x)

[Out]

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

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sympy [A]  time = 1.20, size = 442, normalized size = 5.82 \[ \begin {cases} \tilde {\infty } \left (x - \log {\left (\tanh {\relax (x )} + 1 \right )} - \frac {\tanh ^{2}{\relax (x )}}{2}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {7 x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {7 x}{2 b \tanh {\relax (x )} - 2 b} - \frac {4 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {4 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {\tanh ^{3}{\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} - \frac {\tanh ^{2}{\relax (x )}}{2 b \tanh {\relax (x )} - 2 b} + \frac {3}{2 b \tanh {\relax (x )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {x}{2 b \tanh {\relax (x )} + 2 b} - \frac {4 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {4 \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {\tanh ^{3}{\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} + \frac {\tanh ^{2}{\relax (x )}}{2 b \tanh {\relax (x )} + 2 b} - \frac {3}{2 b \tanh {\relax (x )} + 2 b} & \text {for}\: a = b \\\frac {x - \frac {\tanh ^{3}{\relax (x )}}{3} - \tanh {\relax (x )}}{a} & \text {for}\: b = 0 \\- \frac {2 a^{4} \log {\left (\frac {a}{b} + \tanh {\relax (x )} \right )}}{2 a^{2} b^{3} - 2 b^{5}} + \frac {2 a^{3} b \tanh {\relax (x )}}{2 a^{2} b^{3} - 2 b^{5}} - \frac {a^{2} b^{2} \tanh ^{2}{\relax (x )}}{2 a^{2} b^{3} - 2 b^{5}} + \frac {2 a b^{3} x}{2 a^{2} b^{3} - 2 b^{5}} - \frac {2 a b^{3} \tanh {\relax (x )}}{2 a^{2} b^{3} - 2 b^{5}} - \frac {2 b^{4} x}{2 a^{2} b^{3} - 2 b^{5}} + \frac {2 b^{4} \log {\left (\tanh {\relax (x )} + 1 \right )}}{2 a^{2} b^{3} - 2 b^{5}} + \frac {b^{4} \tanh ^{2}{\relax (x )}}{2 a^{2} b^{3} - 2 b^{5}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(a+b*tanh(x)),x)

[Out]

Piecewise((zoo*(x - log(tanh(x) + 1) - tanh(x)**2/2), Eq(a, 0) & Eq(b, 0)), (7*x*tanh(x)/(2*b*tanh(x) - 2*b) -
7*x/(2*b*tanh(x) - 2*b) - 4*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) - 2*b) + 4*log(tanh(x) + 1)/(2*b*tanh(x) -
2*b) - tanh(x)**3/(2*b*tanh(x) - 2*b) - 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) - 4*log(tanh(x) + 1)*tanh(x)/(2*b*tanh(x) + 2*b) - 4*log(
tanh(x) + 1)/(2*b*tanh(x) + 2*b) - tanh(x)**3/(2*b*tanh(x) + 2*b) + tanh(x)**2/(2*b*tanh(x) + 2*b) - 3/(2*b*ta
nh(x) + 2*b), Eq(a, b)), ((x - tanh(x)**3/3 - tanh(x))/a, Eq(b, 0)), (-2*a**4*log(a/b + tanh(x))/(2*a**2*b**3
- 2*b**5) + 2*a**3*b*tanh(x)/(2*a**2*b**3 - 2*b**5) - a**2*b**2*tanh(x)**2/(2*a**2*b**3 - 2*b**5) + 2*a*b**3*x
/(2*a**2*b**3 - 2*b**5) - 2*a*b**3*tanh(x)/(2*a**2*b**3 - 2*b**5) - 2*b**4*x/(2*a**2*b**3 - 2*b**5) + 2*b**4*l
og(tanh(x) + 1)/(2*a**2*b**3 - 2*b**5) + b**4*tanh(x)**2/(2*a**2*b**3 - 2*b**5), True))

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