3.133 \(\int \frac {\tanh ^5(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac {b x}{a^2-b^2}+\frac {a \log (\cosh (x))}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a^5 \log (a+b \tanh (x))}{b^4 \left (a^2-b^2\right )}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3566, 3647, 3648, 3626, 3617, 31, 3475} \[ -\frac {b x}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a^5 \log (a+b \tanh (x))}{b^4 \left (a^2-b^2\right )}+\frac {a \log (\cosh (x))}{a^2-b^2}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^5/(a + b*Tanh[x]),x]

[Out]

-((b*x)/(a^2 - b^2)) + (a*Log[Cosh[x]])/(a^2 - b^2) + (a^5*Log[a + b*Tanh[x]])/(b^4*(a^2 - b^2)) - ((a^2 + b^2
)*Tanh[x])/b^3 + (a*Tanh[x]^2)/(2*b^2) - Tanh[x]^3/(3*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]

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])))

Rule 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (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 - b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {\tanh ^5(x)}{a+b \tanh (x)} \, dx &=-\frac {\tanh ^3(x)}{3 b}-\frac {\int \frac {\tanh ^2(x) \left (-3 a-3 b \tanh (x)+3 a \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{3 b}\\ &=\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b}-\frac {\int \frac {\tanh (x) \left (6 a^2-6 \left (a^2+b^2\right ) \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{6 b^2}\\ &=-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b}-\frac {\int \frac {-6 a \left (a^2+b^2\right )-6 b^3 \tanh (x)+6 a \left (a^2+b^2\right ) \tanh ^2(x)}{a+b \tanh (x)} \, dx}{6 b^3}\\ &=-\frac {b x}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b}+\frac {a \int \tanh (x) \, dx}{a^2-b^2}+\frac {a^5 \int \frac {1-\tanh ^2(x)}{a+b \tanh (x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=-\frac {b x}{a^2-b^2}+\frac {a \log (\cosh (x))}{a^2-b^2}-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b}+\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tanh (x)\right )}{b^4 \left (a^2-b^2\right )}\\ &=-\frac {b x}{a^2-b^2}+\frac {a \log (\cosh (x))}{a^2-b^2}+\frac {a^5 \log (a+b \tanh (x))}{b^4 \left (a^2-b^2\right )}-\frac {\left (a^2+b^2\right ) \tanh (x)}{b^3}+\frac {a \tanh ^2(x)}{2 b^2}-\frac {\tanh ^3(x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 105, normalized size = 1.12 \[ -\frac {-6 a^5 \log (a \cosh (x)+b \sinh (x))+6 a \left (a^4-b^4\right ) \log (\cosh (x))+b^2 \left (b^2-a^2\right ) \text {sech}^2(x) (2 b \tanh (x)-3 a)+2 b \left (3 a^4+a^2 b^2-4 b^4\right ) \tanh (x)+6 b^5 x}{6 b^4 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^5/(a + b*Tanh[x]),x]

[Out]

-1/6*(6*b^5*x + 6*a*(a^4 - b^4)*Log[Cosh[x]] - 6*a^5*Log[a*Cosh[x] + b*Sinh[x]] + 2*b*(3*a^4 + a^2*b^2 - 4*b^4
)*Tanh[x] + b^2*(-a^2 + b^2)*Sech[x]^2*(-3*a + 2*b*Tanh[x]))/((a - b)*b^4*(a + b))

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fricas [B]  time = 0.67, size = 1296, normalized size = 13.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

-1/3*(3*(a*b^4 + b^5)*x*cosh(x)^6 + 18*(a*b^4 + b^5)*x*cosh(x)*sinh(x)^5 + 3*(a*b^4 + b^5)*x*sinh(x)^6 - 6*a^4
*b - 2*a^2*b^3 + 8*b^5 - 3*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^4 -
3*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 15*(a*b^4 + b^5)*x*cosh(x)^2 - 3*(a*b^4 + b^5)*x)*sinh
(x)^4 + 12*(5*(a*b^4 + b^5)*x*cosh(x)^3 - (2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)
*x)*cosh(x))*sinh(x)^3 - 3*(4*a^4*b - 2*a^3*b^2 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^2 + 3*(15*(a*b^
4 + b^5)*x*cosh(x)^4 - 4*a^4*b + 2*a^3*b^2 - 2*a*b^4 + 4*b^5 - 6*(2*a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 -
4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^2 + 3*(a*b^4 + b^5)*x)*sinh(x)^2 + 3*(a*b^4 + b^5)*x - 3*(a^5*cosh(x)^6 + 6
*a^5*cosh(x)*sinh(x)^5 + a^5*sinh(x)^6 + 3*a^5*cosh(x)^4 + 3*a^5*cosh(x)^2 + a^5 + 3*(5*a^5*cosh(x)^2 + a^5)*s
inh(x)^4 + 4*(5*a^5*cosh(x)^3 + 3*a^5*cosh(x))*sinh(x)^3 + 3*(5*a^5*cosh(x)^4 + 6*a^5*cosh(x)^2 + a^5)*sinh(x)
^2 + 6*(a^5*cosh(x)^5 + 2*a^5*cosh(x)^3 + a^5*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(
x))) + 3*((a^5 - a*b^4)*cosh(x)^6 + 6*(a^5 - a*b^4)*cosh(x)*sinh(x)^5 + (a^5 - a*b^4)*sinh(x)^6 + a^5 - a*b^4
+ 3*(a^5 - a*b^4)*cosh(x)^4 + 3*(a^5 - a*b^4 + 5*(a^5 - a*b^4)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a*b^4)*cosh(
x)^3 + 3*(a^5 - a*b^4)*cosh(x))*sinh(x)^3 + 3*(a^5 - a*b^4)*cosh(x)^2 + 3*(a^5 - a*b^4 + 5*(a^5 - a*b^4)*cosh(
x)^4 + 6*(a^5 - a*b^4)*cosh(x)^2)*sinh(x)^2 + 6*((a^5 - a*b^4)*cosh(x)^5 + 2*(a^5 - a*b^4)*cosh(x)^3 + (a^5 -
a*b^4)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 6*(3*(a*b^4 + b^5)*x*cosh(x)^5 - 2*(2*a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + 2*a*b^4 - 4*b^5 - 3*(a*b^4 + b^5)*x)*cosh(x)^3 - (4*a^4*b - 2*a^3*b^2 + 2*a*b^4 - 4*b^5
- 3*(a*b^4 + b^5)*x)*cosh(x))*sinh(x))/((a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^2
*b^4 - b^6)*sinh(x)^6 + a^2*b^4 - b^6 + 3*(a^2*b^4 - b^6)*cosh(x)^4 + 3*(a^2*b^4 - b^6 + 5*(a^2*b^4 - b^6)*cos
h(x)^2)*sinh(x)^4 + 4*(5*(a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^2*b^4 - b^6)*cosh(x))*sinh(x)^3 + 3*(a^2*b^4 - b^6)*
cosh(x)^2 + 3*(a^2*b^4 - b^6 + 5*(a^2*b^4 - b^6)*cosh(x)^4 + 6*(a^2*b^4 - b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^2*
b^4 - b^6)*cosh(x)^5 + 2*(a^2*b^4 - b^6)*cosh(x)^3 + (a^2*b^4 - b^6)*cosh(x))*sinh(x))

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giac [A]  time = 0.14, size = 142, normalized size = 1.51 \[ \frac {a^{5} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b^{4} - b^{6}} - \frac {x}{a - b} - \frac {{\left (a^{3} + a b^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{4}} + \frac {2 \, {\left (3 \, a^{2} b + 4 \, b^{3} + 3 \, {\left (a^{2} b - a b^{2} + 2 \, b^{3}\right )} e^{\left (4 \, x\right )} + 3 \, {\left (2 \, a^{2} b - a b^{2} + 2 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )}}{3 \, b^{4} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*tanh(x)),x, algorithm="giac")

[Out]

a^5*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^2*b^4 - b^6) - x/(a - b) - (a^3 + a*b^2)*log(e^(2*x) + 1)/b^4 +
2/3*(3*a^2*b + 4*b^3 + 3*(a^2*b - a*b^2 + 2*b^3)*e^(4*x) + 3*(2*a^2*b - a*b^2 + 2*b^3)*e^(2*x))/(b^4*(e^(2*x)
+ 1)^3)

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maple [A]  time = 0.06, size = 96, normalized size = 1.02 \[ -\frac {\tanh ^{3}\relax (x )}{3 b}+\frac {a \left (\tanh ^{2}\relax (x )\right )}{2 b^{2}}-\frac {a^{2} \tanh \relax (x )}{b^{3}}-\frac {\tanh \relax (x )}{b}-\frac {\ln \left (\tanh \relax (x )-1\right )}{2 b +2 a}-\frac {\ln \left (1+\tanh \relax (x )\right )}{2 a -2 b}+\frac {a^{5} \ln \left (a +b \tanh \relax (x )\right )}{b^{4} \left (a +b \right ) \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(a+b*tanh(x)),x)

[Out]

-1/3*tanh(x)^3/b+1/2*a*tanh(x)^2/b^2-1/b^3*a^2*tanh(x)-tanh(x)/b-1/(2*b+2*a)*ln(tanh(x)-1)-1/(2*a-2*b)*ln(1+ta
nh(x))+1/b^4*a^5/(a+b)/(a-b)*ln(a+b*tanh(x))

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maxima [A]  time = 0.41, size = 150, normalized size = 1.60 \[ \frac {a^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b^{4} - b^{6}} - \frac {2 \, {\left (3 \, a^{2} + 4 \, b^{2} + 3 \, {\left (2 \, a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-4 \, x\right )}\right )}}{3 \, {\left (3 \, b^{3} e^{\left (-2 \, x\right )} + 3 \, b^{3} e^{\left (-4 \, x\right )} + b^{3} e^{\left (-6 \, x\right )} + b^{3}\right )}} + \frac {x}{a + b} - \frac {{\left (a^{3} + a b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^5/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

a^5*log(-(a - b)*e^(-2*x) - a - b)/(a^2*b^4 - b^6) - 2/3*(3*a^2 + 4*b^2 + 3*(2*a^2 + a*b + 2*b^2)*e^(-2*x) + 3
*(a^2 + a*b + 2*b^2)*e^(-4*x))/(3*b^3*e^(-2*x) + 3*b^3*e^(-4*x) + b^3*e^(-6*x) + b^3) + x/(a + b) - (a^3 + a*b
^2)*log(e^(-2*x) + 1)/b^4

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mupad [B]  time = 0.23, size = 85, normalized size = 0.90 \[ \frac {x}{a+b}-\frac {{\mathrm {tanh}\relax (x)}^3}{3\,b}-\frac {a\,\ln \left (\mathrm {tanh}\relax (x)+1\right )}{a^2-b^2}+\frac {a\,{\mathrm {tanh}\relax (x)}^2}{2\,b^2}-\frac {\mathrm {tanh}\relax (x)\,\left (a^2+b^2\right )}{b^3}+\frac {a^5\,\ln \left (a+b\,\mathrm {tanh}\relax (x)\right )}{b^4\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^5/(a + b*tanh(x)),x)

[Out]

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

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sympy [A]  time = 1.57, size = 546, normalized size = 5.81 \[ \begin {cases} \tilde {\infty } \left (x - \frac {\tanh ^{3}{\relax (x )}}{3} - \tanh {\relax (x )}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {27 x \tanh {\relax (x )}}{6 b \tanh {\relax (x )} - 6 b} - \frac {27 x}{6 b \tanh {\relax (x )} - 6 b} - \frac {12 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{6 b \tanh {\relax (x )} - 6 b} + \frac {12 \log {\left (\tanh {\relax (x )} + 1 \right )}}{6 b \tanh {\relax (x )} - 6 b} - \frac {2 \tanh ^{4}{\relax (x )}}{6 b \tanh {\relax (x )} - 6 b} - \frac {\tanh ^{3}{\relax (x )}}{6 b \tanh {\relax (x )} - 6 b} - \frac {9 \tanh ^{2}{\relax (x )}}{6 b \tanh {\relax (x )} - 6 b} + \frac {15}{6 b \tanh {\relax (x )} - 6 b} & \text {for}\: a = - b \\\frac {3 x \tanh {\relax (x )}}{6 b \tanh {\relax (x )} + 6 b} + \frac {3 x}{6 b \tanh {\relax (x )} + 6 b} + \frac {12 \log {\left (\tanh {\relax (x )} + 1 \right )} \tanh {\relax (x )}}{6 b \tanh {\relax (x )} + 6 b} + \frac {12 \log {\left (\tanh {\relax (x )} + 1 \right )}}{6 b \tanh {\relax (x )} + 6 b} - \frac {2 \tanh ^{4}{\relax (x )}}{6 b \tanh {\relax (x )} + 6 b} + \frac {\tanh ^{3}{\relax (x )}}{6 b \tanh {\relax (x )} + 6 b} - \frac {9 \tanh ^{2}{\relax (x )}}{6 b \tanh {\relax (x )} + 6 b} + \frac {15}{6 b \tanh {\relax (x )} + 6 b} & \text {for}\: a = b \\\frac {x - \log {\left (\tanh {\relax (x )} + 1 \right )} - \frac {\tanh ^{4}{\relax (x )}}{4} - \frac {\tanh ^{2}{\relax (x )}}{2}}{a} & \text {for}\: b = 0 \\\frac {6 a^{5} \log {\left (\frac {a}{b} + \tanh {\relax (x )} \right )}}{6 a^{2} b^{4} - 6 b^{6}} - \frac {6 a^{4} b \tanh {\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} + \frac {3 a^{3} b^{2} \tanh ^{2}{\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} - \frac {2 a^{2} b^{3} \tanh ^{3}{\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} + \frac {6 a b^{4} x}{6 a^{2} b^{4} - 6 b^{6}} - \frac {6 a b^{4} \log {\left (\tanh {\relax (x )} + 1 \right )}}{6 a^{2} b^{4} - 6 b^{6}} - \frac {3 a b^{4} \tanh ^{2}{\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} - \frac {6 b^{5} x}{6 a^{2} b^{4} - 6 b^{6}} + \frac {2 b^{5} \tanh ^{3}{\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} + \frac {6 b^{5} \tanh {\relax (x )}}{6 a^{2} b^{4} - 6 b^{6}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**5/(a+b*tanh(x)),x)

[Out]

Piecewise((zoo*(x - tanh(x)**3/3 - tanh(x)), Eq(a, 0) & Eq(b, 0)), (27*x*tanh(x)/(6*b*tanh(x) - 6*b) - 27*x/(6
*b*tanh(x) - 6*b) - 12*log(tanh(x) + 1)*tanh(x)/(6*b*tanh(x) - 6*b) + 12*log(tanh(x) + 1)/(6*b*tanh(x) - 6*b)
- 2*tanh(x)**4/(6*b*tanh(x) - 6*b) - tanh(x)**3/(6*b*tanh(x) - 6*b) - 9*tanh(x)**2/(6*b*tanh(x) - 6*b) + 15/(6
*b*tanh(x) - 6*b), Eq(a, -b)), (3*x*tanh(x)/(6*b*tanh(x) + 6*b) + 3*x/(6*b*tanh(x) + 6*b) + 12*log(tanh(x) + 1
)*tanh(x)/(6*b*tanh(x) + 6*b) + 12*log(tanh(x) + 1)/(6*b*tanh(x) + 6*b) - 2*tanh(x)**4/(6*b*tanh(x) + 6*b) + t
anh(x)**3/(6*b*tanh(x) + 6*b) - 9*tanh(x)**2/(6*b*tanh(x) + 6*b) + 15/(6*b*tanh(x) + 6*b), Eq(a, b)), ((x - lo
g(tanh(x) + 1) - tanh(x)**4/4 - tanh(x)**2/2)/a, Eq(b, 0)), (6*a**5*log(a/b + tanh(x))/(6*a**2*b**4 - 6*b**6)
- 6*a**4*b*tanh(x)/(6*a**2*b**4 - 6*b**6) + 3*a**3*b**2*tanh(x)**2/(6*a**2*b**4 - 6*b**6) - 2*a**2*b**3*tanh(x
)**3/(6*a**2*b**4 - 6*b**6) + 6*a*b**4*x/(6*a**2*b**4 - 6*b**6) - 6*a*b**4*log(tanh(x) + 1)/(6*a**2*b**4 - 6*b
**6) - 3*a*b**4*tanh(x)**2/(6*a**2*b**4 - 6*b**6) - 6*b**5*x/(6*a**2*b**4 - 6*b**6) + 2*b**5*tanh(x)**3/(6*a**
2*b**4 - 6*b**6) + 6*b**5*tanh(x)/(6*a**2*b**4 - 6*b**6), True))

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