\(\int \frac {x^2 \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [146]

Optimal result

Integrand size = 26, antiderivative size = 351 \[ \int \frac {x^2 \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {x^2 \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}-\frac {x^2 \log \left (1+\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d}+\frac {x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{2 \sqrt {-a} \sqrt {b} d^2}-\frac {\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 c+2 d x}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^3}+\frac {\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 c+2 d x}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^3} \] Output:

1/2*x^2*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^2*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*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^2-1/2*x*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(3,-(a+b)*exp(2 
*d*x+2*c)/(a-2*(-a)^(1/2)*b^(1/2)-b))/(-a)^(1/2)/b^(1/2)/d^3+1/4*polylog(3 
,-(a+b)*exp(2*d*x+2*c)/(a+2*(-a)^(1/2)*b^(1/2)-b))/(-a)^(1/2)/b^(1/2)/d^3
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {2 d^2 x^2 \log \left (1+\frac {(a+b) e^{2 (c+d x)}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )-2 d^2 x^2 \log \left (1+\frac {(a+b) e^{2 (c+d x)}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )+2 d x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 (c+d x)}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )-2 d x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 (c+d x)}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )-\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 (c+d x)}}{a-2 \sqrt {-a} \sqrt {b}-b}\right )+\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 (c+d x)}}{a+2 \sqrt {-a} \sqrt {b}-b}\right )}{4 \sqrt {-a} \sqrt {b} d^3} \] Input:

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

Output:

(2*d^2*x^2*Log[1 + ((a + b)*E^(2*(c + d*x)))/(a - 2*Sqrt[-a]*Sqrt[b] - b)] 
 - 2*d^2*x^2*Log[1 + ((a + b)*E^(2*(c + d*x)))/(a + 2*Sqrt[-a]*Sqrt[b] - b 
)] + 2*d*x*PolyLog[2, -(((a + b)*E^(2*(c + d*x)))/(a - 2*Sqrt[-a]*Sqrt[b] 
- b))] - 2*d*x*PolyLog[2, -(((a + b)*E^(2*(c + d*x)))/(a + 2*Sqrt[-a]*Sqrt 
[b] - b))] - PolyLog[3, -(((a + b)*E^(2*(c + d*x)))/(a - 2*Sqrt[-a]*Sqrt[b 
] - b))] + PolyLog[3, -(((a + b)*E^(2*(c + d*x)))/(a + 2*Sqrt[-a]*Sqrt[b] 
- b))])/(4*Sqrt[-a]*Sqrt[b]*d^3)
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {6166, 3042, 3801, 2694, 27, 2620, 3011, 2720, 7143}

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.

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

\(\Big \downarrow \) 6166

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3801

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

\(\displaystyle 4 \left (\frac {(a+b) \left (\frac {x^2 \log \left (\frac {(a+b) e^{2 c+2 d x}}{-2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 d (a+b)}-\frac {\frac {\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 c+2 d x}}{a-b-2 \sqrt {-a} \sqrt {b}}\right )}{4 d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-b-2 \sqrt {-a} \sqrt {b}}\right )}{2 d}}{d (a+b)}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {(a+b) \left (\frac {x^2 \log \left (\frac {(a+b) e^{2 c+2 d x}}{2 \sqrt {-a} \sqrt {b}+a-b}+1\right )}{2 d (a+b)}-\frac {\frac {\operatorname {PolyLog}\left (3,-\frac {(a+b) e^{2 c+2 d x}}{a-b+2 \sqrt {-a} \sqrt {b}}\right )}{4 d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(a+b) e^{2 c+2 d x}}{a-b+2 \sqrt {-a} \sqrt {b}}\right )}{2 d}}{d (a+b)}\right )}{4 \sqrt {-a} \sqrt {b}}\right )\)

Input:

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

Output:

4*(((a + b)*((x^2*Log[1 + ((a + b)*E^(2*c + 2*d*x))/(a - 2*Sqrt[-a]*Sqrt[b 
] - b)])/(2*(a + b)*d) - (-1/2*(x*PolyLog[2, -(((a + b)*E^(2*c + 2*d*x))/( 
a - 2*Sqrt[-a]*Sqrt[b] - b))])/d + PolyLog[3, -(((a + b)*E^(2*c + 2*d*x))/ 
(a - 2*Sqrt[-a]*Sqrt[b] - b))]/(4*d^2))/((a + b)*d)))/(4*Sqrt[-a]*Sqrt[b]) 
 - ((a + b)*((x^2*Log[1 + ((a + b)*E^(2*c + 2*d*x))/(a + 2*Sqrt[-a]*Sqrt[b 
] - b)])/(2*(a + b)*d) - (-1/2*(x*PolyLog[2, -(((a + b)*E^(2*c + 2*d*x))/( 
a + 2*Sqrt[-a]*Sqrt[b] - b))])/d + PolyLog[3, -(((a + b)*E^(2*c + 2*d*x))/ 
(a + 2*Sqrt[-a]*Sqrt[b] - b))]/(4*d^2))/((a + b)*d)))/(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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

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]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1185\) vs. \(2(285)=570\).

Time = 6.41 (sec) , antiderivative size = 1186, normalized size of antiderivative = 3.38

method	result	size
risch	\(\text {Expression too large to display}\)	\(1186\)

Input:

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

Output:

2/3/d^3/(-a*b)^(1/2)*c^3-1/4/d^3/(-a*b)^(1/2)*polylog(3,(a+b)*exp(2*d*x+2* 
c)/(2*(-a*b)^(1/2)-a+b))-1/2/d^3/(-2*(-a*b)^(1/2)-a+b)*polylog(3,(a+b)*exp 
(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))+4/3/d^3/(-2*(-a*b)^(1/2)-a+b)*c^3+1/2/d 
/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b) 
^(1/2)-a+b))*x^2+1/2/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*polylog(2,(a 
+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*x-1/2/d/(-a*b)^(1/2)/(-2*(-a*b)^ 
(1/2)-a+b)*b*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*x^2-1/2/d^2/ 
(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*b*polylog(2,(a+b)*exp(2*d*x+2*c)/(-2*(- 
a*b)^(1/2)-a+b))*x-1/2/d^3/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*ln(1-(a+b) 
*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*c^2+1/2/d^3/(-a*b)^(1/2)/(-2*(-a*b) 
^(1/2)-a+b)*b*ln(1-(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))*c^2-1/d^2/( 
-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*b*c^2*x+1/d^2/(-a*b)^(1/2)/(-2*(-a*b)^(1 
/2)-a+b)*a*c^2*x-2/3/d^3/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*b*c^3+2/3/d^3/ 
(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*a*c^3-1/4/d^3/(-a*b)^(1/2)/(-2*(-a*b)^( 
1/2)-a+b)*a*polylog(3,(a+b)*exp(2*d*x+2*c)/(-2*(-a*b)^(1/2)-a+b))+1/4/d^3/ 
(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*b*polylog(3,(a+b)*exp(2*d*x+2*c)/(-2*(- 
a*b)^(1/2)-a+b))+1/3/(-a*b)^(1/2)/(-2*(-a*b)^(1/2)-a+b)*b*x^3-1/3/(-a*b)^( 
1/2)/(-2*(-a*b)^(1/2)-a+b)*a*x^3+1/d^3*c^2/(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/d^3/(-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^2+1/d^2/(-a*b)^(1/2)*c^2*x+1...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2110 vs. \(2 (283) = 566\).

Time = 0.17 (sec) , antiderivative size = 2110, normalized size of antiderivative = 6.01 \[ \int \frac {x^2 \text {sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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