\(\int \frac {x \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [145]

Optimal result

Integrand size = 24, antiderivative size = 231 \[ \int \frac {x \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}+\frac {\operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2}-\frac {\operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^2} \] Output:

1/2*x*ln(1+(a+b)*exp(2*d*x+2*c)/(a-2*(-a)^(1/2)*b^(1/2)-b))/(-a)^(1/2)/b^( 
1/2)/d-1/2*x*ln(1+(a+b)*exp(2*d*x+2*c)/(a+2*(-a)^(1/2)*b^(1/2)-b))/(-a)^(1 
/2)/b^(1/2)/d+1/4*polylog(2,-(a+b)*exp(2*d*x+2*c)/(a-2*(-a)^(1/2)*b^(1/2)- 
b))/(-a)^(1/2)/b^(1/2)/d^2-1/4*polylog(2,-(a+b)*exp(2*d*x+2*c)/(a+2*(-a)^( 
1/2)*b^(1/2)-b))/(-a)^(1/2)/b^(1/2)/d^2
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.08 \[ \int \frac {x \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {-4 \sqrt {-a} c \arctan \left (\frac {a-b+(a+b) e^{2 (c+d x)}}{2 \sqrt {a} \sqrt {b}}\right )+2 \sqrt {a} (c+d x) \log \left (1+\frac {(a+b) e^{2 (c+d x)}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )-2 \sqrt {a} (c+d x) \log \left (1+\frac {(a+b) e^{2 (c+d x)}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )+\sqrt {a} \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 (c+d x)}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )-\sqrt {a} \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 (c+d x)}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a^2} \sqrt {b} d^2} \] Input:

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

Output:

(-4*Sqrt[-a]*c*ArcTan[(a - b + (a + b)*E^(2*(c + d*x)))/(2*Sqrt[a]*Sqrt[b] 
)] + 2*Sqrt[a]*(c + d*x)*Log[1 + ((a + b)*E^(2*(c + d*x)))/(a - 2*Sqrt[-a] 
*Sqrt[b] - b)] - 2*Sqrt[a]*(c + d*x)*Log[1 + ((a + b)*E^(2*(c + d*x)))/(a 
+ 2*Sqrt[-a]*Sqrt[b] - b)] + Sqrt[a]*PolyLog[2, -(((a + b)*E^(2*(c + d*x)) 
)/(a - 2*Sqrt[-a]*Sqrt[b] - b))] - Sqrt[a]*PolyLog[2, -(((a + b)*E^(2*(c + 
 d*x)))/(a + 2*Sqrt[-a]*Sqrt[b] - b))])/(4*Sqrt[-a^2]*Sqrt[b]*d^2)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6166, 3042, 3801, 2694, 27, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 6166

\(\displaystyle 2 \int \frac {x}{a-b+(a+b) \cosh (2 c+2 d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x}{a-b+(a+b) \sin \left (2 i c+2 i d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3801

\(\displaystyle 4 \int \frac {e^{2 c+2 d x} x}{a+(a+b) e^{4 (c+d x)}+2 (a-b) e^{2 c+2 d x}+b}dx\)

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle 4 \left (\frac {(a+b) \left (\frac {\operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-b-2 \sqrt {-a} \sqrt {b}}\right )}{4 d^2 (a+b)}+\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{-2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 d (a+b)}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {(a+b) \left (\frac {\operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-b+2 \sqrt {-a} \sqrt {b}}\right )}{4 d^2 (a+b)}+\frac {x \log \left (\frac {(a+b) e^{2 c+2 d x}}{2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 d (a+b)}\right )}{4 \sqrt {-a} \sqrt {b}}\right )\)

Input:

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

Output:

4*(((a + b)*((x*Log[1 + ((a + b)*E^(2*c + 2*d*x))/(a - 2*Sqrt[-a]*Sqrt[b] 
- b)])/(2*(a + b)*d) + PolyLog[2, -(((a + b)*E^(2*c + 2*d*x))/(a - 2*Sqrt[ 
-a]*Sqrt[b] - b))]/(4*(a + b)*d^2)))/(4*Sqrt[-a]*Sqrt[b]) - ((a + b)*((x*L 
og[1 + ((a + b)*E^(2*c + 2*d*x))/(a + 2*Sqrt[-a]*Sqrt[b] - b)])/(2*(a + b) 
*d) + PolyLog[2, -(((a + b)*E^(2*c + 2*d*x))/(a + 2*Sqrt[-a]*Sqrt[b] - b)) 
]/(4*(a + b)*d^2)))/(4*Sqrt[-a]*Sqrt[b]))
 

Defintions of rubi rules used

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 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 3801

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (Comple 
x[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Simp[2   Int[((c + d*x)^m*(E^((-I)*e 
+ f*fz*x)/(b + (2*a*E^((-I)*e + f*fz*x))/E^(I*Pi*(k - 1/2)) - (b*E^(2*((-I) 
*e + f*fz*x)))/E^(2*I*k*Pi))))/E^(I*Pi*(k - 1/2)), x], x] /; FreeQ[{a, b, c 
, d, e, f, fz}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 6166

Int[(((f_.) + (g_.)*(x_))^(m_.)*Sech[(d_.) + (e_.)*(x_)]^2)/((b_) + (c_.)*T 
anh[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> Simp[2   Int[(f + g*x)^m/(b - c + 
(b + c)*Cosh[2*d + 2*e*x]), x], x] /; FreeQ[{b, c, d, e, f, g}, x] && IGtQ[ 
m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(952\) vs. \(2(187)=374\).

Time = 6.42 (sec) , antiderivative size = 953, normalized size of antiderivative = 4.13

method	result	size
risch	\(-\frac {a c x}{d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b c x}{d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a c}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b c}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {x^{2}}{-2 \sqrt {-a b}-a +b}-\frac {x^{2}}{2 \sqrt {-a b}}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a x}{2 d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b x}{2 d \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {a \,c^{2}}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b \,c^{2}}{2 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) a}{4 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {\operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) b}{4 d^{2} \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}-\frac {c^{2}}{2 d^{2} \sqrt {-a b}}+\frac {\operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right )}{4 d^{2} \sqrt {-a b}}-\frac {c^{2}}{d^{2} \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right )}{2 d^{2} \left (-2 \sqrt {-a b}-a +b \right )}-\frac {a \,x^{2}}{2 \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {b \,x^{2}}{2 \sqrt {-a b}\, \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) c}{d^{2} \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{-2 \sqrt {-a b}-a +b}\right ) x}{d \left (-2 \sqrt {-a b}-a +b \right )}-\frac {2 c x}{d \left (-2 \sqrt {-a b}-a +b \right )}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right ) c}{2 d^{2} \sqrt {-a b}}-\frac {c \arctan \left (\frac {2 \left (a +b \right ) {\mathrm e}^{2 d x +2 c}+2 a -2 b}{4 \sqrt {a b}}\right )}{d^{2} \sqrt {a b}}+\frac {\ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 d x +2 c}}{2 \sqrt {-a b}-a +b}\right ) x}{2 d \sqrt {-a b}}-\frac {c x}{d \sqrt {-a b}}\)	\(953\)

Input:

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

Output:

-1/d/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*c*x+1/d/(-a*b)^(1/2)/(-2*(-a*b)^ 
(1/2)-a+b)*b*c*x+1/2/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*ln(1-(a+b)*exp 
(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*a*c-1/2/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1/ 
2)-a+b)*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*b*c-1/(-2*(-a*b)^ 
(1/2)-a+b)*x^2-1/2/(-a*b)^(1/2)*x^2+1/2/d/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+ 
b)*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*a*x-1/2/d/(-a*b)^(1/2) 
/(-2*(-a*b)^(1/2)-a+b)*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*b* 
x-1/2/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*c^2+1/2/d^2/(-a*b)^(1/2)/(- 
2*(-a*b)^(1/2)-a+b)*b*c^2+1/4/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*polyl 
og(2,(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*a-1/4/d^2/(-a*b)^(1/2)/(- 
2*(-a*b)^(1/2)-a+b)*polylog(2,(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))* 
b-1/2/d^2/(-a*b)^(1/2)*c^2+1/4/d^2/(-a*b)^(1/2)*polylog(2,(a+b)*exp(2*d*x+ 
2*c)/(2*(-a*b)^(1/2)-a+b))-1/d^2/(-2*(-a*b)^(1/2)-a+b)*c^2+1/2/d^2/(-2*(-a 
*b)^(1/2)-a+b)*polylog(2,(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))-1/2/( 
-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*x^2+1/2/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)- 
a+b)*b*x^2+1/d^2/(-2*(-a*b)^(1/2)-a+b)*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b 
)^(1/2)-a+b))*c+1/d/(-2*(-a*b)^(1/2)-a+b)*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(- 
a*b)^(1/2)-a+b))*x-2/d/(-2*(-a*b)^(1/2)-a+b)*c*x+1/2/d^2/(-a*b)^(1/2)*ln(1 
-(a+b)*exp(2*d*x+2*c)/(2*(-a*b)^(1/2)-a+b))*c-1/d^2*c/(a*b)^(1/2)*arctan(1 
/4*(2*(a+b)*exp(2*d*x+2*c)+2*a-2*b)/(a*b)^(1/2))+1/2/d/(-a*b)^(1/2)*ln(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1516 vs. \(2 (185) = 370\).

Time = 0.20 (sec) , antiderivative size = 1516, normalized size of antiderivative = 6.56 \[ \int \frac {x \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*c*log(2*sqrt(-(2*(a + b)*sqrt 
(-a*b/(a^2 + 2*a*b + b^2)) + a - b)/(a + b)) + 2*cosh(d*x + c) + 2*sinh(d* 
x + c)) + (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*c*log(-2*sqrt(-(2*(a + b) 
*sqrt(-a*b/(a^2 + 2*a*b + b^2)) + a - b)/(a + b)) + 2*cosh(d*x + c) + 2*si 
nh(d*x + c)) - (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*c*log(2*sqrt((2*(a + 
 b)*sqrt(-a*b/(a^2 + 2*a*b + b^2)) - a + b)/(a + b)) + 2*cosh(d*x + c) + 2 
*sinh(d*x + c)) - (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*c*log(-2*sqrt((2* 
(a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2)) - a + b)/(a + b)) + 2*cosh(d*x + c) 
 + 2*sinh(d*x + c)) - (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*dilog(-(((a - 
 b)*cosh(d*x + c) + (a - b)*sinh(d*x + c) - 2*((a + b)*cosh(d*x + c) + (a 
+ b)*sinh(d*x + c))*sqrt(-a*b/(a^2 + 2*a*b + b^2)))*sqrt(-(2*(a + b)*sqrt( 
-a*b/(a^2 + 2*a*b + b^2)) + a - b)/(a + b)) + a + b)/(a + b) + 1) - (a + b 
)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*dilog((((a - b)*cosh(d*x + c) + (a - b)*s 
inh(d*x + c) - 2*((a + b)*cosh(d*x + c) + (a + b)*sinh(d*x + c))*sqrt(-a*b 
/(a^2 + 2*a*b + b^2)))*sqrt(-(2*(a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2)) + a 
 - b)/(a + b)) - a - b)/(a + b) + 1) + (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^ 
2))*dilog(-(((a - b)*cosh(d*x + c) + (a - b)*sinh(d*x + c) + 2*((a + b)*co 
sh(d*x + c) + (a + b)*sinh(d*x + c))*sqrt(-a*b/(a^2 + 2*a*b + b^2)))*sqrt( 
(2*(a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2)) - a + b)/(a + b)) + a + b)/(a + 
b) + 1) + (a + b)*sqrt(-a*b/(a^2 + 2*a*b + b^2))*dilog((((a - b)*cosh(d...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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