\(\int \frac {x^2}{a+b \cos ^2(x)+c \sin ^2(x)} \, dx\) [441]

Optimal result

Integrand size = 20, antiderivative size = 365 \[ \int \frac {x^2}{a+b \cos ^2(x)+c \sin ^2(x)} \, dx=-\frac {i x^2 \log \left (1+\frac {(b-c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )}{2 \sqrt {a+b} \sqrt {a+c}}+\frac {i x^2 \log \left (1+\frac {(b-c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )}{2 \sqrt {a+b} \sqrt {a+c}}-\frac {x \operatorname {PolyLog}\left (2,-\frac {(b-c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )}{2 \sqrt {a+b} \sqrt {a+c}}+\frac {x \operatorname {PolyLog}\left (2,-\frac {(b-c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )}{2 \sqrt {a+b} \sqrt {a+c}}-\frac {i \operatorname {PolyLog}\left (3,-\frac {(b-c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )}{4 \sqrt {a+b} \sqrt {a+c}}+\frac {i \operatorname {PolyLog}\left (3,-\frac {(b-c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )}{4 \sqrt {a+b} \sqrt {a+c}} \] Output:

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

Mathematica [A] (verified)

Time = 4.34 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.76 \[ \int \frac {x^2}{a+b \cos ^2(x)+c \sin ^2(x)} \, dx=-\frac {i \left (2 x^2 \log \left (1+\frac {(b-c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )-2 x^2 \log \left (1+\frac {(b-c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )-2 i x \operatorname {PolyLog}\left (2,\frac {(-b+c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )+2 i x \operatorname {PolyLog}\left (2,\frac {(-b+c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )+\operatorname {PolyLog}\left (3,\frac {(-b+c) e^{2 i x}}{2 a+b+c-2 \sqrt {a+b} \sqrt {a+c}}\right )-\operatorname {PolyLog}\left (3,\frac {(-b+c) e^{2 i x}}{2 a+b+c+2 \sqrt {a+b} \sqrt {a+c}}\right )\right )}{4 \sqrt {a+b} \sqrt {a+c}} \] Input:

Integrate[x^2/(a + b*Cos[x]^2 + c*Sin[x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5098, 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.

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

\(\Big \downarrow \) 5098

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3802

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[x^2/(a + b*Cos[x]^2 + c*Sin[x]^2),x]
 

Output:

4*(((b - c)*(((-1/2*I)*x^2*Log[1 + ((b - c)*E^((2*I)*x))/(2*a + b + c - 2* 
Sqrt[a + b]*Sqrt[a + c])])/(b - c) + (I*((I/2)*x*PolyLog[2, -(((b - c)*E^( 
(2*I)*x))/(2*a + b + c - 2*Sqrt[a + b]*Sqrt[a + c]))] - PolyLog[3, -(((b - 
 c)*E^((2*I)*x))/(2*a + b + c - 2*Sqrt[a + b]*Sqrt[a + c]))]/4))/(b - c))) 
/(4*Sqrt[a + b]*Sqrt[a + c]) - ((b - c)*(((-1/2*I)*x^2*Log[1 + ((b - c)*E^ 
((2*I)*x))/(2*a + b + c + 2*Sqrt[a + b]*Sqrt[a + c])])/(b - c) + (I*((I/2) 
*x*PolyLog[2, -(((b - c)*E^((2*I)*x))/(2*a + b + c + 2*Sqrt[a + b]*Sqrt[a 
+ c]))] - PolyLog[3, -(((b - c)*E^((2*I)*x))/(2*a + b + c + 2*Sqrt[a + b]* 
Sqrt[a + c]))]/4))/(b - c)))/(4*Sqrt[a + b]*Sqrt[a + 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 5098

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + Cos[(d_.) + (e_.)*(x_)]^2*(b_.) + ( 
c_.)*Sin[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> Simp[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. 1160 vs. \(2 (289 ) = 578\).

Time = 0.29 (sec) , antiderivative size = 1161, normalized size of antiderivative = 3.18

method	result	size
risch	\(\text {Expression too large to display}\)	\(1161\)

Input:

int(x^2/(a+cos(x)^2*b+c*sin(x)^2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*b*x*polylog(2,(b 
-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/2*I/((a+b)*(a+c))^(1/2) 
/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*c*x^2*ln(1-(b-c)*exp(2*I*x)/(-2*((a+b)*( 
a+c))^(1/2)-2*a-b-c))-2/3/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a- 
b-c)*a*x^3-1/2*I/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*a*po 
lylog(3,(b-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/2/((a+b)*(a+c 
))^(1/2)*x*polylog(2,(b-c)*exp(2*I*x)/(2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/2 
*I/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*b*x^2*ln(1-(b-c)*e 
xp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-2/3/(-2*((a+b)*(a+c))^(1/2)-2* 
a-b-c)*x^3-1/4*I/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*b*po 
lylog(3,(b-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/(-2*((a+b)*(a 
+c))^(1/2)-2*a-b-c)*x*polylog(2,(b-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2 
*a-b-c))-1/4*I/((a+b)*(a+c))^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*c*poly 
log(3,(b-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/2/((a+b)*(a+c)) 
^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*c*x*polylog(2,(b-c)*exp(2*I*x)/(-2 
*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/2*I/((a+b)*(a+c))^(1/2)*x^2*ln(1-(b-c)*ex 
p(2*I*x)/(2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/3/((a+b)*(a+c))^(1/2)/(-2*((a+ 
b)*(a+c))^(1/2)-2*a-b-c)*c*x^3-1/2*I/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*poly 
log(3,(b-c)*exp(2*I*x)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c))-1/3/((a+b)*(a+c)) 
^(1/2)/(-2*((a+b)*(a+c))^(1/2)-2*a-b-c)*b*x^3-I/((a+b)*(a+c))^(1/2)/(-2...
 

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. 4289 vs. \(2 (288) = 576\).

Time = 3.04 (sec) , antiderivative size = 4289, normalized size of antiderivative = 11.75 \[ \int \frac {x^2}{a+b \cos ^2(x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(x^2/(a+b*cos(x)^2+c*sin(x)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(x**2/(a+b*cos(x)**2+c*sin(x)**2),x)
 

Output:

Integral(x**2/(a + b*cos(x)**2 + c*sin(x)**2), x)
 

Maxima [F]

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

integrate(x^2/(a+b*cos(x)^2+c*sin(x)^2),x, algorithm="maxima")
 

Output:

integrate(x^2/(b*cos(x)^2 + c*sin(x)^2 + a), x)
 

Giac [F]

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

integrate(x^2/(a+b*cos(x)^2+c*sin(x)^2),x, algorithm="giac")
 

Output:

integrate(x^2/(b*cos(x)^2 + c*sin(x)^2 + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**2/(cos(x)**2*b + sin(x)**2*c + a),x)