\(\int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx\) [135]

Optimal result

Integrand size = 13, antiderivative size = 76 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=\frac {a x}{a^2-b^2}-\frac {b \log (\cosh (x))}{a^2-b^2}-\frac {a^4 \log (a+b \tanh (x))}{b^3 \left (a^2-b^2\right )}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b} \] Output:

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
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {\log (1-\tanh (x))}{2 (a+b)}+\frac {\log (1+\tanh (x))}{2 (a-b)}-\frac {a^4 \log (a+b \tanh (x))}{b^3 \left (a^2-b^2\right )}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b} \] Input:

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

Output:

-1/2*Log[1 - Tanh[x]]/(a + b) + Log[1 + Tanh[x]]/(2*(a - b)) - (a^4*Log[a 
+ b*Tanh[x]])/(b^3*(a^2 - b^2)) + (a*Tanh[x])/b^2 - Tanh[x]^2/(2*b)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.154, Rules used = {3042, 4049, 27, 3042, 26, 4130, 25, 3042, 4110, 26, 3042, 26, 3956, 4100, 16}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (i x)^4}{a-i b \tan (i x)}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}+\frac {i \int -\frac {2 i \tanh (x) \left (-a \tanh ^2(x)+b \tanh (x)+a\right )}{a+b \tanh (x)}dx}{2 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tanh (x) \left (-a \tanh ^2(x)+b \tanh (x)+a\right )}{a+b \tanh (x)}dx}{b}-\frac {\tanh ^2(x)}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}+\frac {\int -\frac {i \tan (i x) \left (a \tan (i x)^2-i b \tan (i x)+a\right )}{a-i b \tan (i x)}dx}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \int \frac {\tan (i x) \left (a \tan (i x)^2-i b \tan (i x)+a\right )}{a-i b \tan (i x)}dx}{b}\)

\(\Big \downarrow \) 4130

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i \int -\frac {a^2-\left (a^2+b^2\right ) \tanh ^2(x)}{a+b \tanh (x)}dx}{b}+\frac {i a \tanh (x)}{b}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \int \frac {a^2-\left (a^2+b^2\right ) \tanh ^2(x)}{a+b \tanh (x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \int \frac {a^2+\left (a^2+b^2\right ) \tan (i x)^2}{a-i b \tan (i x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 4110

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (-\frac {i b^3 \int i \tanh (x)dx}{a^2-b^2}+\frac {a^4 \int \frac {1-\tanh ^2(x)}{a+b \tanh (x)}dx}{a^2-b^2}-\frac {a b^2 x}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (\frac {b^3 \int \tanh (x)dx}{a^2-b^2}+\frac {a^4 \int \frac {1-\tanh ^2(x)}{a+b \tanh (x)}dx}{a^2-b^2}-\frac {a b^2 x}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (\frac {b^3 \int -i \tan (i x)dx}{a^2-b^2}+\frac {a^4 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {a b^2 x}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (-\frac {i b^3 \int \tan (i x)dx}{a^2-b^2}+\frac {a^4 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {a b^2 x}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (\frac {a^4 \int \frac {\tan (i x)^2+1}{a-i b \tan (i x)}dx}{a^2-b^2}-\frac {a b^2 x}{a^2-b^2}+\frac {b^3 \log (\cosh (x))}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 4100

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (\frac {a^4 \int \frac {1}{a+b \tanh (x)}d(b \tanh (x))}{b \left (a^2-b^2\right )}-\frac {a b^2 x}{a^2-b^2}+\frac {b^3 \log (\cosh (x))}{a^2-b^2}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {\tanh ^2(x)}{2 b}-\frac {i \left (\frac {i a \tanh (x)}{b}-\frac {i \left (-\frac {a b^2 x}{a^2-b^2}+\frac {b^3 \log (\cosh (x))}{a^2-b^2}+\frac {a^4 \log (a+b \tanh (x))}{b \left (a^2-b^2\right )}\right )}{b}\right )}{b}\)

Input:

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

Output:

-1/2*Tanh[x]^2/b - (I*(((-I)*(-((a*b^2*x)/(a^2 - b^2)) + (b^3*Log[Cosh[x]] 
)/(a^2 - b^2) + (a^4*Log[a + b*Tanh[x]])/(b*(a^2 - b^2))))/b + (I*a*Tanh[x 
])/b))/b
 

Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 4049

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

rule 4100

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + 
 (f_.)*(x_)]^2), x_Symbol] :> Simp[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 4110

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] + (Simp[(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] 
- Simp[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 4130

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] + Simp[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] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(-\frac {\tanh \left (x \right )^{2}}{2 b}+\frac {a \tanh \left (x \right )}{b^{2}}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 b +2 a}-\frac {a^{4} \ln \left (a +b \tanh \left (x \right )\right )}{b^{3} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}\)	\(76\)
default	\(-\frac {\tanh \left (x \right )^{2}}{2 b}+\frac {a \tanh \left (x \right )}{b^{2}}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 b +2 a}-\frac {a^{4} \ln \left (a +b \tanh \left (x \right )\right )}{b^{3} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}\)	\(76\)
parallelrisch	\(-\frac {\tanh \left (x \right )^{2} a^{2} b^{2}-\tanh \left (x \right )^{2} b^{4}-2 \ln \left (1-\tanh \left (x \right )\right ) b^{4}+2 a^{4} \ln \left (a +b \tanh \left (x \right )\right )-2 a \,b^{3} x -2 b^{4} x -2 \tanh \left (x \right ) a^{3} b +2 \tanh \left (x \right ) a \,b^{3}}{2 b^{3} \left (a^{2}-b^{2}\right )}\)	\(91\)
risch	\(\frac {x}{a +b}-\frac {2 x \,a^{2}}{b^{3}}-\frac {2 x}{b}+\frac {2 x \,a^{4}}{b^{3} \left (a^{2}-b^{2}\right )}-\frac {2 \left ({\mathrm e}^{2 x} a -{\mathrm e}^{2 x} b +a \right )}{\left ({\mathrm e}^{2 x}+1\right )^{2} b^{2}}+\frac {\ln \left ({\mathrm e}^{2 x}+1\right ) a^{2}}{b^{3}}+\frac {\ln \left ({\mathrm e}^{2 x}+1\right )}{b}-\frac {a^{4} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{b^{3} \left (a^{2}-b^{2}\right )}\)	\(133\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 644 vs. \(2 (74) = 148\).

Time = 0.13 (sec) , antiderivative size = 644, normalized size of antiderivative = 8.47 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((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))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (61) = 122\).

Time = 0.40 (sec) , antiderivative size = 442, normalized size of antiderivative = 5.82 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=\begin {cases} \tilde {\infty } \left (x - \log {\left (\tanh {\left (x \right )} + 1 \right )} - \frac {\tanh ^{2}{\left (x \right )}}{2}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {x - \frac {\tanh ^{3}{\left (x \right )}}{3} - \tanh {\left (x \right )}}{a} & \text {for}\: b = 0 \\\frac {7 x \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} - \frac {7 x}{2 b \tanh {\left (x \right )} - 2 b} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} + \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 b \tanh {\left (x \right )} - 2 b} - \frac {\tanh ^{3}{\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} - \frac {\tanh ^{2}{\left (x \right )}}{2 b \tanh {\left (x \right )} - 2 b} + \frac {3}{2 b \tanh {\left (x \right )} - 2 b} & \text {for}\: a = - b \\\frac {x \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} + \frac {x}{2 b \tanh {\left (x \right )} + 2 b} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 b \tanh {\left (x \right )} + 2 b} - \frac {\tanh ^{3}{\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} + \frac {\tanh ^{2}{\left (x \right )}}{2 b \tanh {\left (x \right )} + 2 b} - \frac {3}{2 b \tanh {\left (x \right )} + 2 b} & \text {for}\: a = b \\- \frac {2 a^{4} \log {\left (\frac {a}{b} + \tanh {\left (x \right )} \right )}}{2 a^{2} b^{3} - 2 b^{5}} + \frac {2 a^{3} b \tanh {\left (x \right )}}{2 a^{2} b^{3} - 2 b^{5}} - \frac {a^{2} b^{2} \tanh ^{2}{\left (x \right )}}{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 {\left (x \right )}}{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 {\left (x \right )} + 1 \right )}}{2 a^{2} b^{3} - 2 b^{5}} + \frac {b^{4} \tanh ^{2}{\left (x \right )}}{2 a^{2} b^{3} - 2 b^{5}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*(x - log(tanh(x) + 1) - tanh(x)**2/2), Eq(a, 0) & Eq(b, 0)) 
, ((x - tanh(x)**3/3 - tanh(x))/a, 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* 
tanh(x) + 2*b), Eq(a, b)), (-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*log(tanh( 
x) + 1)/(2*a**2*b**3 - 2*b**5) + b**4*tanh(x)**2/(2*a**2*b**3 - 2*b**5), T 
rue))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.32 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {a^{4} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} b^{3} - b^{5}} + \frac {2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} + a\right )}}{2 \, b^{2} e^{\left (-2 \, x\right )} + b^{2} e^{\left (-4 \, x\right )} + b^{2}} + \frac {x}{a + b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{3}} \] Input:

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

Output:

-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
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.29 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=-\frac {a^{4} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} b^{3} - b^{5}} + \frac {x}{a - b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{3}} - \frac {2 \, {\left (a b + {\left (a b - b^{2}\right )} e^{\left (2 \, x\right )}\right )}}{b^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \] Input:

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

Output:

-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)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=\frac {x}{a+b}-\frac {{\mathrm {tanh}\left (x\right )}^2}{2\,b}+\frac {b\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )}{a^2-b^2}+\frac {a\,\mathrm {tanh}\left (x\right )}{b^2}-\frac {a^4\,\ln \left (a+b\,\mathrm {tanh}\left (x\right )\right )}{b^3\,\left (a^2-b^2\right )} \] Input:

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

Output:

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))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.51 \[ \int \frac {\tanh ^4(x)}{a+b \tanh (x)} \, dx=\frac {-2 \,\mathrm {log}\left (e^{2 x} a +e^{2 x} b +a -b \right ) b^{4}-2 \,\mathrm {log}\left (\tanh \left (x \right ) b +a \right ) a^{4}+2 \,\mathrm {log}\left (\tanh \left (x \right ) b +a \right ) b^{4}-\tanh \left (x \right )^{2} a^{2} b^{2}+\tanh \left (x \right )^{2} b^{4}+2 \tanh \left (x \right ) a^{3} b -2 \tanh \left (x \right ) a \,b^{3}+2 a \,b^{3} x +2 b^{4} x}{2 b^{3} \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

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