\(\int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) [82]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b} d}-\frac {(a-b) \tanh (c+d x)}{b^2 d}-\frac {\tanh ^3(c+d x)}{3 b d} \] Output:

a^2*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/b^(5/2)/(a+b)^(1/2)/d-(a-b)*t 
anh(d*x+c)/b^2/d-1/3*tanh(d*x+c)^3/b/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(214\) vs. \(2(77)=154\).

Time = 2.73 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.78 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (3 a^2 \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (\cosh (2 c)-\sinh (2 c))+\sqrt {a+b} \text {sech}(c+d x) \sqrt {b (\cosh (c)-\sinh (c))^4} \left (\text {sech}(c) \left (-3 a+2 b+b \text {sech}^2(c+d x)\right ) \sinh (d x)+b \text {sech}(c+d x) \tanh (c)\right )\right )}{6 b^2 \sqrt {a+b} d \left (a+b \text {sech}^2(c+d x)\right ) \sqrt {b (\cosh (c)-\sinh (c))^4}} \] Input:

Integrate[Sech[c + d*x]^6/(a + b*Sech[c + d*x]^2),x]
 

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*a^2*ArcTanh[(Sech[d*x] 
*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqr 
t[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]) + Sqrt[a 
+ b]*Sech[c + d*x]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(Sech[c]*(-3*a + 2*b + b* 
Sech[c + d*x]^2)*Sinh[d*x] + b*Sech[c + d*x]*Tanh[c])))/(6*b^2*Sqrt[a + b] 
*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 300, 2009}

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.

\(\displaystyle \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)^6}{a+b \sec (i c+i d x)^2}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b}}-\frac {(a-b) \tanh (c+d x)}{b^2}-\frac {\tanh ^3(c+d x)}{3 b}}{d}\)

Input:

Int[Sech[c + d*x]^6/(a + b*Sech[c + d*x]^2),x]
 

Output:

((a^2*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(b^(5/2)*Sqrt[a + b]) 
- ((a - b)*Tanh[c + d*x])/b^2 - Tanh[c + d*x]^3/(3*b))/d
 

Defintions of rubi rules used

rule 300
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4634
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ 
[m/2] && IntegerQ[n/2]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(67)=134\).

Time = 1.46 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.38

method result size
derivativedivides \(\frac {-\frac {2 a^{2} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{b^{2}}+\frac {2 \left (-a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 \left (-2 a +\frac {2 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}}{d}\) \(183\)
default \(\frac {-\frac {2 a^{2} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{b^{2}}+\frac {2 \left (-a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 \left (-2 a +\frac {2 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}}{d}\) \(183\)
risch \(\frac {2 \,{\mathrm e}^{4 d x +4 c} a +4 a \,{\mathrm e}^{2 d x +2 c}-4 b \,{\mathrm e}^{2 d x +2 c}+2 a -\frac {4 b}{3}}{d \,b^{2} \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{2 \sqrt {a b +b^{2}}\, d \,b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{2 \sqrt {a b +b^{2}}\, d \,b^{2}}\) \(220\)

Input:

int(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 832 vs. \(2 (67) = 134\).

Time = 0.39 (sec) , antiderivative size = 1905, normalized size of antiderivative = 24.74 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="fricas")
 

Output:

[1/6*(12*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 48*(a^2*b + a*b^2)*cosh(d*x + c 
)*sinh(d*x + c)^3 + 12*(a^2*b + a*b^2)*sinh(d*x + c)^4 + 12*a^2*b + 4*a*b^ 
2 - 8*b^3 + 24*(a^2*b - b^3)*cosh(d*x + c)^2 + 24*(a^2*b - b^3 + 3*(a^2*b 
+ a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*(a^2*cosh(d*x + c)^6 + 6*a^2 
*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*sinh(d*x + c)^6 + 3*a^2*cosh(d*x + c) 
^4 + 3*(5*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^4 + 3*a^2*cosh(d*x + c) 
^2 + 4*(5*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*( 
5*a^2*cosh(d*x + c)^4 + 6*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^2 + a^2 
 + 6*(a^2*cosh(d*x + c)^5 + 2*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sin 
h(d*x + c))*sqrt(a*b + b^2)*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*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c 
)^2 + a + 2*b)*sqrt(a*b + b^2))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sin 
h(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*co 
sh(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)) + 48*((a^2*b + a*b^2)*cosh(d*x + c)^3 
 + (a^2*b - b^3)*cosh(d*x + c))*sinh(d*x + c))/((a*b^3 + b^4)*d*cosh(d*x + 
 c)^6 + 6*(a*b^3 + b^4)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a*b^3 + b^4)...
 

Sympy [F]

\[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\int \frac {\operatorname {sech}^{6}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(sech(d*x+c)**6/(a+b*sech(d*x+c)**2),x)
 

Output:

Integral(sech(c + d*x)**6/(a + b*sech(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (67) = 134\).

Time = 0.15 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.08 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {a^{2} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} b^{2} d} - \frac {2 \, {\left (6 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, a - 2 \, b\right )}}{3 \, {\left (3 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )} + b^{2} e^{\left (-6 \, d x - 6 \, c\right )} + b^{2}\right )} d} \] Input:

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.53 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {\frac {3 \, a^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} b^{2}} + \frac {2 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a - 2 \, b\right )}}{b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \] Input:

integrate(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x, algorithm="giac")
 

Output:

1/3*(3*a^2*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/(sqr 
t(-a*b - b^2)*b^2) + 2*(3*a*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) - 6*b*e^ 
(2*d*x + 2*c) + 3*a - 2*b)/(b^2*(e^(2*d*x + 2*c) + 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 2.88 (sec) , antiderivative size = 334, normalized size of antiderivative = 4.34 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {8}{3\,b\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {4}{b\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {2\,a}{b^2\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^2\,\ln \left (\frac {4\,a^2\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^5\,\left (a+b\right )}-\frac {8\,a^2\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^{9/2}\,\sqrt {a+b}}\right )}{2\,b^{5/2}\,d\,\sqrt {a+b}}+\frac {a^2\,\ln \left (\frac {8\,a^2\,\left (a+2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}+4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^{9/2}\,\sqrt {a+b}}+\frac {4\,a^2\,\left (2\,a\,b+a^2+a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^5\,\left (a+b\right )}\right )}{2\,b^{5/2}\,d\,\sqrt {a+b}} \] Input:

int(1/(cosh(c + d*x)^6*(a + b/cosh(c + d*x)^2)),x)
 

Output:

8/(3*b*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) 
 - 4/(b*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + (2*a)/(b^2*d*(exp 
(2*c + 2*d*x) + 1)) - (a^2*log((4*a^2*(2*a*b + a^2 + a^2*exp(2*c + 2*d*x) 
+ 8*b^2*exp(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/(b^5*(a + b)) - (8*a^2 
*(a + 2*a*exp(2*c + 2*d*x) + 4*b*exp(2*c + 2*d*x)))/(b^(9/2)*(a + b)^(1/2) 
)))/(2*b^(5/2)*d*(a + b)^(1/2)) + (a^2*log((8*a^2*(a + 2*a*exp(2*c + 2*d*x 
) + 4*b*exp(2*c + 2*d*x)))/(b^(9/2)*(a + b)^(1/2)) + (4*a^2*(2*a*b + a^2 + 
 a^2*exp(2*c + 2*d*x) + 8*b^2*exp(2*c + 2*d*x) + 8*a*b*exp(2*c + 2*d*x)))/ 
(b^5*(a + b))))/(2*b^(5/2)*d*(a + b)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 738, normalized size of antiderivative = 9.58 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {3 e^{6 d x +6 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}+3 e^{6 d x +6 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}-3 e^{6 d x +6 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) a^{2}+9 e^{4 d x +4 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}+9 e^{4 d x +4 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}-9 e^{4 d x +4 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) a^{2}+9 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}+9 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}-9 e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) a^{2}+3 \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}+3 \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) a^{2}-3 \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) a^{2}-4 e^{6 d x +6 c} a^{2} b -4 e^{6 d x +6 c} a \,b^{2}+12 e^{2 d x +2 c} a^{2} b -12 e^{2 d x +2 c} a \,b^{2}-24 e^{2 d x +2 c} b^{3}+8 a^{2} b -8 b^{3}}{6 b^{3} d \left (e^{6 d x +6 c} a +e^{6 d x +6 c} b +3 e^{4 d x +4 c} a +3 e^{4 d x +4 c} b +3 e^{2 d x +2 c} a +3 e^{2 d x +2 c} b +a +b \right )} \] Input:

int(sech(d*x+c)^6/(a+b*sech(d*x+c)^2),x)
 

Output:

(3*e**(6*c + 6*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 + 3*e**(6*c + 6*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 - 3*e**(6*c + 6*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 + 9*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 
 + 9*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 - 9*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 + 9*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 + 9*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 
- 9*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 + 3*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b 
)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 + 3*sqrt(b)*sqrt(a + 
 b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 
 - 3*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + 
a + 2*b)*a**2 - 4*e**(6*c + 6*d*x)*a**2*b - 4*e**(6*c + 6*d*x)*a*b**2 + 12 
*e**(2*c + 2*d*x)*a**2*b - 12*e**(2*c + 2*d*x)*a*b**2 - 24*e**(2*c + 2*d*x 
)*b**3 + 8*a**2*b - 8*b**3)/(6*b**3*d*(e**(6*c + 6*d*x)*a + e**(6*c + 6...