\(\int \frac {x^2 \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx\) [85]

Optimal result

Integrand size = 36, antiderivative size = 407 \[ \int \frac {x^2 \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=-\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{2 \sqrt {a+c} \sqrt {b+c} d}+\frac {i x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{2 \sqrt {a+c} \sqrt {b+c} d}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{2 \sqrt {a+c} \sqrt {b+c} d^2}+\frac {x \operatorname {PolyLog}\left (2,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{2 \sqrt {a+c} \sqrt {b+c} d^2}-\frac {i \operatorname {PolyLog}\left (3,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )}{4 \sqrt {a+c} \sqrt {b+c} d^3}+\frac {i \operatorname {PolyLog}\left (3,-\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 \left (c+\sqrt {a+c} \sqrt {b+c}\right )}\right )}{4 \sqrt {a+c} \sqrt {b+c} d^3} \] Output:

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

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=-\frac {i \left (2 d^2 x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )-2 d^2 x^2 \log \left (1+\frac {(a-b) e^{2 i (c+d x)}}{a+b+2 c+2 \sqrt {a+c} \sqrt {b+c}}\right )-2 i d x \operatorname {PolyLog}\left (2,\frac {(-a+b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )+2 i d x \operatorname {PolyLog}\left (2,\frac {(-a+b) e^{2 i (c+d x)}}{a+b+2 c+2 \sqrt {a+c} \sqrt {b+c}}\right )+\operatorname {PolyLog}\left (3,\frac {(-a+b) e^{2 i (c+d x)}}{a+b+2 c-2 \sqrt {a+c} \sqrt {b+c}}\right )-\operatorname {PolyLog}\left (3,\frac {(-a+b) e^{2 i (c+d x)}}{a+b+2 c+2 \sqrt {a+c} \sqrt {b+c}}\right )\right )}{4 \sqrt {a+c} \sqrt {b+c} d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5100, 3042, 3802, 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.

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

\(\Big \downarrow \) 5100

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3802

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 3802

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

rule 5100

Int[(((f_.) + (g_.)*(x_))^(m_.)*Sec[(d_.) + (e_.)*(x_)]^2)/((b_.) + (a_.)*S 
ec[(d_.) + (e_.)*(x_)]^2 + (c_.)*Tan[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> S 
imp[2   Int[(f + g*x)^m/(2*a + b + c + (b - c)*Cos[2*d + 2*e*x]), x], x] /; 
 FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && NeQ[a + b, 0] && NeQ[a + 
c, 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)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2060 vs. \(2 (331 ) = 662\).

Time = 0.90 (sec) , antiderivative size = 2061, normalized size of antiderivative = 5.06

method	result	size
risch	\(\text {Expression too large to display}\)	\(2061\)

Input:

int(x^2*sec(d*x+c)^2/(a+c*sec(d*x+c)^2+tan(d*x+c)^2*b),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5696 vs. \(2 (325) = 650\).

Time = 3.00 (sec) , antiderivative size = 5696, normalized size of antiderivative = 14.00 \[ \int \frac {x^2 \sec ^2(c+d x)}{a+c \sec ^2(c+d x)+b \tan ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

integrate(x**2*sec(d*x+c)**2/(a+c*sec(d*x+c)**2+b*tan(d*x+c)**2),x)
 

Output:

Integral(x**2*sec(c + d*x)**2/(a + b*tan(c + d*x)**2 + c*sec(c + d*x)**2), 
 x)
 

Maxima [F]

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

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

Output:

integrate(x^2*sec(d*x + c)^2/(c*sec(d*x + c)^2 + b*tan(d*x + c)^2 + a), x)
 

Giac [F]

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

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

Output:

integrate(x^2*sec(d*x + c)^2/(c*sec(d*x + c)^2 + b*tan(d*x + c)^2 + a), x)
 

Mupad [F(-1)]

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

int(x^2/(cos(c + d*x)^2*(a + c/cos(c + d*x)^2 + b*tan(c + d*x)^2)),x)
 

Output:

int(x^2/(cos(c + d*x)^2*(a + c/cos(c + d*x)^2 + b*tan(c + d*x)^2)), x)
 

Reduce [F]

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

int(x^2*sec(d*x+c)^2/(a+c*sec(d*x+c)^2+b*tan(d*x+c)^2),x)
 

Output:

int((sec(c + d*x)**2*x**2)/(sec(c + d*x)**2*c + tan(c + d*x)**2*b + a),x)