\(\int \frac {\tanh ^4(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [149]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {x}{a^2}+\frac {(a-2 b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} d}-\frac {(a+b) \tanh (c+d x)}{2 a b d \left (a+b-b \tanh ^2(c+d x)\right )} \] Output:

x/a^2+1/2*(a-2*b)*(a+b)^(1/2)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2 
/b^(3/2)/d-1/2*(a+b)*tanh(d*x+c)/a/b/d/(a+b-b*tanh(d*x+c)^2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(228\) vs. \(2(91)=182\).

Time = 1.85 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.51 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^4(c+d x) \left (2 x (a+2 b+a \cosh (2 (c+d x)))+\frac {\left (a^2-a b-2 b^2\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (a+2 b+a \cosh (2 (c+d x))) (\cosh (2 c)-\sinh (2 c))}{b \sqrt {a+b} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {(a+b) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{b d}\right )}{8 a^2 \left (a+b \text {sech}^2(c+d x)\right )^2} \] Input:

Integrate[Tanh[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]
 

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^4*(2*x*(a + 2*b + a*Cosh[2* 
(c + d*x)]) + ((a^2 - a*b - 2*b^2)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2* 
c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh 
[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])*(Cosh[2*c] - Sinh[2*c] 
))/(b*Sqrt[a + b]*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + ((a + b)*Sech[2*c]*(( 
a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(b*d)))/(8*a^2*(a + b*Sech[c + d*x]^2 
)^2)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4629, 2075, 372, 397, 219, 221}

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.

\(\displaystyle \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (i c+i d x)^4}{\left (a+b \sec (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 372

\(\displaystyle \frac {\frac {\int \frac {-\left ((a-b) \tanh ^2(c+d x)\right )+a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a b}-\frac {(a+b) \tanh (c+d x)}{2 a b \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {2 b \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a}+\frac {(a-2 b) (a+b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a b}-\frac {(a+b) \tanh (c+d x)}{2 a b \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {(a-2 b) (a+b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}+\frac {2 b \text {arctanh}(\tanh (c+d x))}{a}}{2 a b}-\frac {(a+b) \tanh (c+d x)}{2 a b \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 b \text {arctanh}(\tanh (c+d x))}{a}+\frac {(a-2 b) \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {b}}}{2 a b}-\frac {(a+b) \tanh (c+d x)}{2 a b \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\)

Input:

Int[Tanh[c + d*x]^4/(a + b*Sech[c + d*x]^2)^2,x]
 

Output:

(((2*b*ArcTanh[Tanh[c + d*x]])/a + ((a - 2*b)*Sqrt[a + b]*ArcTanh[(Sqrt[b] 
*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[b]))/(2*a*b) - ((a + b)*Tanh[c + d*x 
])/(2*a*b*(a + b - b*Tanh[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(258\) vs. \(2(79)=158\).

Time = 21.77 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.85

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (-\frac {\left (a +b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 b}-\frac {\left (a +b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 b}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (a^{2}-a b -2 b^{2}\right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{b}}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(259\)
default	\(\frac {\frac {\frac {2 \left (-\frac {\left (a +b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 b}-\frac {\left (a +b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 b}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (a^{2}-a b -2 b^{2}\right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{b}}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\)	\(259\)
risch	\(\frac {x}{a^{2}}+\frac {a^{2} {\mathrm e}^{2 d x +2 c}+3 a b \,{\mathrm e}^{2 d x +2 c}+2 b^{2} {\mathrm e}^{2 d x +2 c}+a^{2}+a b}{b \,a^{2} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{4 b^{2} d a}-\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{2 b d \,a^{2}}-\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{4 b^{2} d a}+\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{2 b d \,a^{2}}\)	\(297\)

Input:

int(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 611 vs. \(2 (82) = 164\).

Time = 0.25 (sec) , antiderivative size = 1479, normalized size of antiderivative = 16.25 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

[1/4*(4*a*b*d*x*cosh(d*x + c)^4 + 16*a*b*d*x*cosh(d*x + c)*sinh(d*x + c)^3 
 + 4*a*b*d*x*sinh(d*x + c)^4 + 4*a*b*d*x + 4*(2*(a*b + 2*b^2)*d*x + a^2 + 
3*a*b + 2*b^2)*cosh(d*x + c)^2 + 4*(6*a*b*d*x*cosh(d*x + c)^2 + 2*(a*b + 2 
*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*sinh(d*x + c)^2 - ((a^2 - 2*a*b)*cosh(d*x 
 + c)^4 + 4*(a^2 - 2*a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 - 2*a*b)*si 
nh(d*x + c)^4 + 2*(a^2 - 4*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - 2*a*b)*cosh( 
d*x + c)^2 + a^2 - 4*b^2)*sinh(d*x + c)^2 + a^2 - 2*a*b + 4*((a^2 - 2*a*b) 
*cosh(d*x + c)^3 + (a^2 - 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a + b 
)/b)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2* 
sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c) 
^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x 
+ c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*b*cosh(d*x + c) 
^2 + 2*a*b*cosh(d*x + c)*sinh(d*x + c) + a*b*sinh(d*x + c)^2 + a*b + 2*b^2 
)*sqrt((a + b)/b))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 
+ a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 
 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + 
c))*sinh(d*x + c) + a)) + 4*a^2 + 4*a*b + 8*(2*a*b*d*x*cosh(d*x + c)^3 + ( 
2*(a*b + 2*b^2)*d*x + a^2 + 3*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))/( 
a^3*b*d*cosh(d*x + c)^4 + 4*a^3*b*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*b* 
d*sinh(d*x + c)^4 + a^3*b*d + 2*(a^3*b + 2*a^2*b^2)*d*cosh(d*x + c)^2 +...
 

Sympy [F]

\[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {\tanh ^{4}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(tanh(d*x+c)**4/(a+b*sech(d*x+c)**2)**2,x)
 

Output:

Integral(tanh(c + d*x)**4/(a + b*sech(c + d*x)**2)**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1053 vs. \(2 (82) = 164\).

Time = 0.32 (sec) , antiderivative size = 1053, normalized size of antiderivative = 11.57 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

1/64*(a^3 - 6*a^2*b - 24*a*b^2 - 16*b^3)*log((a*e^(2*d*x + 2*c) + a + 2*b 
- 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(( 
a^3*b + a^2*b^2)*sqrt((a + b)*b)*d) + 1/16*a*log((a*e^(2*d*x + 2*c) + a + 
2*b - 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)) 
)/(sqrt((a + b)*b)*(a*b + b^2)*d) - 1/64*(a^3 - 6*a^2*b - 24*a*b^2 - 16*b^ 
3)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2 
*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3*b + a^2*b^2)*sqrt((a + b)*b)*d) 
- 3/32*(a + 2*b)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a 
*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*(a*b + 
b^2)*d) - 1/16*a*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a 
*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/(sqrt((a + b)*b)*(a*b + 
b^2)*d) + 1/16*(a^3 + 8*a^2*b + 8*a*b^2 + (a^3 + 18*a^2*b + 48*a*b^2 + 32* 
b^3)*e^(2*d*x + 2*c))/((a^4*b + a^3*b^2 + (a^4*b + a^3*b^2)*e^(4*d*x + 4*c 
) + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*e^(2*d*x + 2*c))*d) - 1/16*(a^3 + 8* 
a^2*b + 8*a*b^2 + (a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*e^(-2*d*x - 2*c))/( 
(a^4*b + a^3*b^2 + 2*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*e^(-2*d*x - 2*c) + (a 
^4*b + a^3*b^2)*e^(-4*d*x - 4*c))*d) + 1/4*(a^2 + 2*a*b + (a^2 + 8*a*b + 8 
*b^2)*e^(2*d*x + 2*c))/((a^3*b + a^2*b^2 + (a^3*b + a^2*b^2)*e^(4*d*x + 4* 
c) + 2*(a^3*b + 3*a^2*b^2 + 2*a*b^3)*e^(2*d*x + 2*c))*d) - 1/4*(a^2 + 2*a* 
b + (a^2 + 8*a*b + 8*b^2)*e^(-2*d*x - 2*c))/((a^3*b + a^2*b^2 + 2*(a^3*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (82) = 164\).

Time = 0.89 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.90 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, {\left (d x + c\right )}}{a^{2}} + \frac {{\left (a^{2} - a b - 2 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2} b} + \frac {2 \, {\left (a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + a b\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} a^{2} b}}{2 \, d} \] Input:

integrate(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

1/2*(2*(d*x + c)/a^2 + (a^2 - a*b - 2*b^2)*arctan(1/2*(a*e^(2*d*x + 2*c) + 
 a + 2*b)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^2*b) + 2*(a^2*e^(2*d*x + 2 
*c) + 3*a*b*e^(2*d*x + 2*c) + 2*b^2*e^(2*d*x + 2*c) + a^2 + a*b)/((a*e^(4* 
d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)*a^2*b))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {{\left ({\mathrm {cosh}\left (c+d\,x\right )}^2-1\right )}^2}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \] Input:

int(tanh(c + d*x)^4/(a + b/cosh(c + d*x)^2)^2,x)
 

Output:

int((cosh(c + d*x)^2 - 1)^2/(b + a*cosh(c + d*x)^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 971, normalized size of antiderivative = 10.67 \[ \int \frac {\tanh ^4(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(tanh(d*x+c)^4/(a+b*sech(d*x+c)^2)^2,x)
 

Output:

(e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - 
a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 - 2*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a 
+ b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))* 
a*b + e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) 
- a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 - 2*e**(4*c + 4*d*x)*sqrt(b)*sqrt( 
a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a 
*b - e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**( 
2*c + 2*d*x)*a + a + 2*b)*a**2 + 2*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*lo 
g(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a*b + 2*e**(2*c + 
2*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + 
e**(c + d*x)*sqrt(a))*a**2 - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log( - 
 sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 + 2*e* 
*(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2* 
b) + e**(c + d*x)*sqrt(a))*a**2 - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*l 
og(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 - 2* 
e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 
 2*d*x)*a + a + 2*b)*a**2 + 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(2*s 
qrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*b**2 + sqrt(b)*sqrt(a + 
 b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a 
**2 - 2*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b...