\(\int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx\) [523]

Optimal result

Integrand size = 14, antiderivative size = 340 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=-\frac {i x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {i x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}+\frac {x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{\sqrt {4 a^2-b^2}}-\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a-\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}}+\frac {i \operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )}{2 \sqrt {4 a^2-b^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=-\frac {i \left (2 x^2 \log \left (1+\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )-2 x^2 \log \left (1-\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )-2 i x \operatorname {PolyLog}\left (2,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )+2 i x \operatorname {PolyLog}\left (2,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )+\operatorname {PolyLog}\left (3,-\frac {i b e^{2 i x}}{-2 a+\sqrt {4 a^2-b^2}}\right )-\operatorname {PolyLog}\left (3,\frac {i b e^{2 i x}}{2 a+\sqrt {4 a^2-b^2}}\right )\right )}{2 \sqrt {4 a^2-b^2}} \] Input:

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

Output:

((-1/2*I)*(2*x^2*Log[1 + (I*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] - 2 
*x^2*Log[1 - (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] - (2*I)*x*PolyLo 
g[2, ((-I)*b*E^((2*I)*x))/(-2*a + Sqrt[4*a^2 - b^2])] + (2*I)*x*PolyLog[2, 
 (I*b*E^((2*I)*x))/(2*a + Sqrt[4*a^2 - b^2])] + PolyLog[3, ((-I)*b*E^((2*I 
)*x))/(-2*a + Sqrt[4*a^2 - b^2])] - PolyLog[3, (I*b*E^((2*I)*x))/(2*a + Sq 
rt[4*a^2 - b^2])]))/Sqrt[4*a^2 - b^2]
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.94, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5095, 3042, 3804, 27, 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.

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

\(\Big \downarrow \) 5095

\(\displaystyle \int \frac {x^2}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^2}{a+\frac {1}{2} b \sin (2 x)}dx\)

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 3804

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

rule 5095

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + Cos[(c_.) + (d_.)*(x_)]*(b_.)*Sin[(c 
_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[(e + f*x)^m*(a + b*(Sin[2*c + 2* 
d*x]/2))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

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. 1781 vs. \(2 (290 ) = 580\).

Time = 0.16 (sec) , antiderivative size = 1782, normalized size of antiderivative = 5.24

method	result	size
risch	\(\text {Expression too large to display}\)	\(1782\)

Input:

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

Output:

8/3/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/ 
2)*a*x^3+4*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*(-(2*a-b)*(2* 
a+b))^(1/2)*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a*x^2-8*I 
/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(2,b*exp(2*I*x)/(- 
2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a^2*x+2*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b) 
*(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*polylog(3,b*exp(2*I*x)/(-2*I*a+( 
-(2*a-b)*(2*a+b))^(1/2)))*a+8/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/ 
2))*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a^2*x^2-2/(8*a^2- 
2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*ln(1-b*exp(2*I*x)/(-2*I*a+(-(2*a- 
b)*(2*a+b))^(1/2)))*b^2*x^2+4/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/ 
2))*(-(2*a-b)*(2*a+b))^(1/2)*polylog(2,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a 
+b))^(1/2)))*a*x-16/3*I/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*a^ 
2*x^3-8*I/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2))*polylog(2,b*exp( 
2*I*x)/(-2*I*a-(-(2*a-b)*(2*a+b))^(1/2)))*a^2*x+4/3*I/(8*a^2-2*b^2)/(-2*I* 
a+(-(2*a-b)*(2*a+b))^(1/2))*b^2*x^3+4/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a 
+b))^(1/2))*polylog(3,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*a^2- 
1/(8*a^2-2*b^2)/(-2*I*a+(-(2*a-b)*(2*a+b))^(1/2))*polylog(3,b*exp(2*I*x)/( 
-2*I*a+(-(2*a-b)*(2*a+b))^(1/2)))*b^2-8/3/(8*a^2-2*b^2)/(-2*I*a-(-(2*a-b)* 
(2*a+b))^(1/2))*(-(2*a-b)*(2*a+b))^(1/2)*a*x^3+2*I/(8*a^2-2*b^2)/(-2*I*a+( 
-(2*a-b)*(2*a+b))^(1/2))*polylog(2,b*exp(2*I*x)/(-2*I*a+(-(2*a-b)*(2*a+...
 

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. 2492 vs. \(2 (272) = 544\).

Time = 0.72 (sec) , antiderivative size = 2492, normalized size of antiderivative = 7.33 \[ \int \frac {x^2}{a+b \cos (x) \sin (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) - (b 
*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2 
)/b^2) + 2*I*a)/b) - b)/b) + b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a* 
cos(x) - 2*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sq 
rt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) - b*x^2*sqrt(-(4*a^2 - 
b^2)/b^2)*log(-((2*I*a*cos(x) - 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt( 
-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b) - b)/b) 
 - b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos(x) + 2*a*sin(x) + (b*c 
os(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2) 
/b^2) + 2*I*a)/b) - b)/b) + b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*co 
s(x) - 2*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt 
((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) + b*x^2*sqrt(-(4*a^2 - b^ 
2)/b^2)*log(-((-2*I*a*cos(x) + 2*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt(- 
(4*a^2 - b^2)/b^2))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) - 
 b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((2*I*a*cos(x) + 2*a*sin(x) + (b*cos( 
x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(-(b*sqrt(-(4*a^2 - b^2)/b^ 
2) - 2*I*a)/b) - b)/b) - b*x^2*sqrt(-(4*a^2 - b^2)/b^2)*log(-((-2*I*a*cos( 
x) - 2*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(- 
(b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b) - b)/b) + 2*I*b*x*sqrt(-(4*a^2 - b 
^2)/b^2)*dilog(((2*I*a*cos(x) + 2*a*sin(x) - (b*cos(x) - I*b*sin(x))*sq...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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