\(\int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) [2]

Optimal result

Integrand size = 19, antiderivative size = 298 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {b x}{c^2}+\frac {\sqrt {2} b \left (b-\frac {a c}{b}-\frac {b^2}{\sqrt {b^2-4 a c}}+\frac {3 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{c^2 \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} b \left (b-\frac {a c}{b}+\frac {b^2}{\sqrt {b^2-4 a c}}-\frac {3 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{c^2 \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}-\frac {\cos (x)}{c} \] Output:

-b*x/c^2+2^(1/2)*b*(b-a*c/b-b^2/(-4*a*c+b^2)^(1/2)+3*a*c/(-4*a*c+b^2)^(1/2 
))*arctan(1/2*(2*c+(b-(-4*a*c+b^2)^(1/2))*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+ 
c)-b*(-4*a*c+b^2)^(1/2))^(1/2))/c^2/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^( 
1/2)+2^(1/2)*b*(b-a*c/b+b^2/(-4*a*c+b^2)^(1/2)-3*a*c/(-4*a*c+b^2)^(1/2))*a 
rctan(1/2*(2*c+(b+(-4*a*c+b^2)^(1/2))*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+c)+b 
*(-4*a*c+b^2)^(1/2))^(1/2))/c^2/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2) 
-cos(x)/c
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.20 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {-b x+\frac {\left (i b^3-3 i a b c+b^2 \sqrt {-b^2+4 a c}-a c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {\left (-i b^3+3 i a b c+b^2 \sqrt {-b^2+4 a c}-a c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}-c \cos (x)}{c^2} \] Input:

Integrate[Sin[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

(-(b*x) + ((I*b^3 - (3*I)*a*b*c + b^2*Sqrt[-b^2 + 4*a*c] - a*c*Sqrt[-b^2 + 
 4*a*c])*ArcTan[(2*c + (b - I*Sqrt[-b^2 + 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[ 
b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])])/(Sqrt[-1/2*b^2 + 2*a*c]*Sqr 
t[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]]) + (((-I)*b^3 + (3*I)*a*b*c 
+ b^2*Sqrt[-b^2 + 4*a*c] - a*c*Sqrt[-b^2 + 4*a*c])*ArcTan[(2*c + (b + I*Sq 
rt[-b^2 + 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^ 
2 + 4*a*c]])])/(Sqrt[-1/2*b^2 + 2*a*c]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[- 
b^2 + 4*a*c]]) - c*Cos[x])/c^2
 

Rubi [A] (verified)

Time = 2.27 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3737, 2009}

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.

\(\displaystyle \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^3}{a+b \sin (x)+c \sin (x)^2}dx\)

\(\Big \downarrow \) 3737

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {2} b \left (-\frac {b^2}{\sqrt {b^2-4 a c}}+\frac {3 a c}{\sqrt {b^2-4 a c}}-\frac {a c}{b}+b\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c^2 \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\sqrt {2} b \left (\frac {b^2}{\sqrt {b^2-4 a c}}-\frac {3 a c}{\sqrt {b^2-4 a c}}-\frac {a c}{b}+b\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c^2 \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {b x}{c^2}-\frac {\cos (x)}{c}\)

Input:

Int[Sin[x]^3/(a + b*Sin[x] + c*Sin[x]^2),x]
 

Output:

-((b*x)/c^2) + (Sqrt[2]*b*(b - (a*c)/b - b^2/Sqrt[b^2 - 4*a*c] + (3*a*c)/S 
qrt[b^2 - 4*a*c])*ArcTan[(2*c + (b - Sqrt[b^2 - 4*a*c])*Tan[x/2])/(Sqrt[2] 
*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/(c^2*Sqrt[b^2 - 2*c*(a + 
 c) - b*Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*b*(b - (a*c)/b + b^2/Sqrt[b^2 - 4*a 
*c] - (3*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(2*c + (b + Sqrt[b^2 - 4*a*c])*Tan 
[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])])/(c^2*Sqrt 
[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) - Cos[x]/c
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3737

Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n 
_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_), x_Symbol] :> Int[ExpandTr 
ig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x], x] / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ 
ersQ[m, n, p]
 

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {2 \left (\frac {c}{1+\tan \left (\frac {x}{2}\right )^{2}}+b \arctan \left (\tan \left (\frac {x}{2}\right )\right )\right )}{c^{2}}+\frac {2 a \left (\frac {2 \left (-2 \sqrt {-4 a c +b^{2}}\, a c +\sqrt {-4 a c +b^{2}}\, b^{2}+4 a b c -b^{3}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}-\frac {2 \left (2 \sqrt {-4 a c +b^{2}}\, a c -\sqrt {-4 a c +b^{2}}\, b^{2}+4 a b c -b^{3}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{c^{2}}\)	\(305\)
risch	\(\text {Expression too large to display}\)	\(2188\)

Input:

int(sin(x)^3/(a+b*sin(x)+c*sin(x)^2),x,method=_RETURNVERBOSE)
 

Output:

-2/c^2*(c/(1+tan(1/2*x)^2)+b*arctan(tan(1/2*x)))+2/c^2*a*(2*(-2*(-4*a*c+b^ 
2)^(1/2)*a*c+(-4*a*c+b^2)^(1/2)*b^2+4*a*b*c-b^3)/(8*a*c-2*b^2)/(4*a*c-2*b^ 
2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(-4*a*c+b^2 
)^(1/2))/(4*a*c-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))-2*(2*(-4*a*c+b^ 
2)^(1/2)*a*c-(-4*a*c+b^2)^(1/2)*b^2+4*a*b*c-b^3)/(8*a*c-2*b^2)/(4*a*c-2*b^ 
2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2) 
^(1/2)-b)/(4*a*c-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6531 vs. \(2 (261) = 522\).

Time = 1.95 (sec) , antiderivative size = 6531, normalized size of antiderivative = 21.92 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(sin(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(sin(x)**3/(a+b*sin(x)+c*sin(x)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\sin \left (x\right )^{3}}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \] Input:

integrate(sin(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")
 

Output:

-(c^2*integrate(-2*(2*(b^3 - a*b*c)*cos(3*x)^2 + 4*(2*a^2*b + a*b*c)*cos(2 
*x)^2 + 2*(b^3 - a*b*c)*cos(x)^2 + 2*(b^3 - a*b*c)*sin(3*x)^2 + 2*(4*a*b^2 
 - a*c^2 - (2*a^2 - b^2)*c)*cos(x)*sin(2*x) + 4*(2*a^2*b + a*b*c)*sin(2*x) 
^2 + 2*(b^3 - a*b*c)*sin(x)^2 - (2*a*b*c*cos(2*x) + (b^2*c - a*c^2)*sin(3* 
x) - (b^2*c - a*c^2)*sin(x))*cos(4*x) - 2*(2*(b^3 - a*b*c)*cos(x) + (4*a*b 
^2 - a*c^2 - (2*a^2 - b^2)*c)*sin(2*x))*cos(3*x) - 2*(a*b*c + (4*a*b^2 - a 
*c^2 - (2*a^2 - b^2)*c)*sin(x))*cos(2*x) - (2*a*b*c*sin(2*x) - (b^2*c - a* 
c^2)*cos(3*x) + (b^2*c - a*c^2)*cos(x))*sin(4*x) - (b^2*c - a*c^2 - 2*(4*a 
*b^2 - a*c^2 - (2*a^2 - b^2)*c)*cos(2*x) + 4*(b^3 - a*b*c)*sin(x))*sin(3*x 
) + (b^2*c - a*c^2)*sin(x))/(c^4*cos(4*x)^2 + 4*b^2*c^2*cos(3*x)^2 + 4*b^2 
*c^2*cos(x)^2 + c^4*sin(4*x)^2 + 4*b^2*c^2*sin(3*x)^2 + 4*b^2*c^2*sin(x)^2 
 + 4*b*c^3*sin(x) + c^4 + 4*(4*a^2*c^2 + 4*a*c^3 + c^4)*cos(2*x)^2 + 8*(2* 
a*b*c^2 + b*c^3)*cos(x)*sin(2*x) + 4*(4*a^2*c^2 + 4*a*c^3 + c^4)*sin(2*x)^ 
2 - 2*(2*b*c^3*sin(3*x) - 2*b*c^3*sin(x) - c^4 + 2*(2*a*c^3 + c^4)*cos(2*x 
))*cos(4*x) - 8*(b^2*c^2*cos(x) + (2*a*b*c^2 + b*c^3)*sin(2*x))*cos(3*x) - 
 4*(2*a*c^3 + c^4 + 2*(2*a*b*c^2 + b*c^3)*sin(x))*cos(2*x) + 4*(b*c^3*cos( 
3*x) - b*c^3*cos(x) - (2*a*c^3 + c^4)*sin(2*x))*sin(4*x) - 4*(2*b^2*c^2*si 
n(x) + b*c^3 - 2*(2*a*b*c^2 + b*c^3)*cos(2*x))*sin(3*x)), x) + b*x + c*cos 
(x))/c^2
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(sin(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 45.16 (sec) , antiderivative size = 21407, normalized size of antiderivative = 71.84 \[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

int(sin(x)^3/(a + c*sin(x)^2 + b*sin(x)),x)
 

Output:

- 2/(c*(tan(x/2)^2 + 1)) - atan((((8192*(4*a^2*b^7 - 3*a^4*b^5 - 20*a^3*b^ 
5*c + 9*a^5*b^3*c + 20*a^4*b^3*c^2))/c^4 + ((8192*(4*a*b^7*c^2 - 2*a^2*b^7 
*c + 2*a^4*b^5*c + 12*a^5*b*c^4 + 8*a^6*b*c^3 - 24*a^2*b^5*c^3 + 32*a^3*b^ 
3*c^4 + 10*a^3*b^5*c^2 - 10*a^4*b^3*c^3 - 10*a^5*b^3*c^2))/c^4 + ((b^8 - a 
^2*b^6 + 8*a^4*c^4 + 8*a^5*c^3 + b^5*(-(4*a*c - b^2)^3)^(1/2) + 8*a^3*b^4* 
c - a^2*b^3*(-(4*a*c - b^2)^3)^(1/2) + 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - 1 
8*a^4*b^2*c^2 - 10*a*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b^ 
3*c*(-(4*a*c - b^2)^3)^(1/2) + 2*a^3*b*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(16* 
a^2*c^8 + 32*a^3*c^7 + 16*a^4*c^6 + b^4*c^6 - b^6*c^4 - 8*a*b^2*c^7 + 10*a 
*b^4*c^5 - 32*a^2*b^2*c^6 + a^2*b^4*c^4 - 8*a^3*b^2*c^5)))^(1/2)*((8192*(3 
*a*b^7*c^3 - 4*a*b^5*c^5 + 20*a^4*b*c^6 + 9*a^5*b*c^5 + 16*a^2*b^3*c^6 - 1 
3*a^2*b^5*c^4 - 3*a^3*b^5*c^3 + 9*a^4*b^3*c^4))/c^4 + ((b^8 - a^2*b^6 + 8* 
a^4*c^4 + 8*a^5*c^3 + b^5*(-(4*a*c - b^2)^3)^(1/2) + 8*a^3*b^4*c - a^2*b^3 
*(-(4*a*c - b^2)^3)^(1/2) + 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - 18*a^4*b^2*c 
^2 - 10*a*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b^3*c*(-(4*a* 
c - b^2)^3)^(1/2) + 2*a^3*b*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(16*a^2*c^8 + 3 
2*a^3*c^7 + 16*a^4*c^6 + b^4*c^6 - b^6*c^4 - 8*a*b^2*c^7 + 10*a*b^4*c^5 - 
32*a^2*b^2*c^6 + a^2*b^4*c^4 - 8*a^3*b^2*c^5)))^(1/2)*((8192*(3*a*b^5*c^6 
+ 16*a^3*b*c^8 - 4*a^4*b*c^7 - 8*a^5*b*c^6 - 16*a^2*b^3*c^7 - 2*a^2*b^5*c^ 
5 + 9*a^3*b^3*c^6 + 2*a^4*b^3*c^5))/c^4 + ((8192*(3*a*b^5*c^7 - 4*a*b^3...
 

Reduce [F]

\[ \int \frac {\sin ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {-2 \cos \left (x \right ) a c +\cos \left (x \right ) b^{2}-6 \left (\int \frac {\sin \left (x \right )^{2}}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a^{2} b \,c^{2}+5 \left (\int \frac {\sin \left (x \right )^{2}}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a \,b^{3} c -\left (\int \frac {\sin \left (x \right )^{2}}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) b^{5}-4 \left (\int \frac {\sin \left (x \right )}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a^{3} c^{2}+2 \left (\int \frac {\sin \left (x \right )}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a^{2} b^{2} c -2 \left (\int \frac {1}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a^{3} b c +\left (\int \frac {1}{2 \sin \left (x \right )^{2} a \,c^{2}-\sin \left (x \right )^{2} b^{2} c +2 \sin \left (x \right ) a b c -\sin \left (x \right ) b^{3}+2 a^{2} c -a \,b^{2}}d x \right ) a^{2} b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) a^{2}-\mathrm {log}\left (\sin \left (x \right )^{2} c +\sin \left (x \right ) b +a \right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{4} a +2 \tan \left (\frac {x}{2}\right )^{3} b +2 \tan \left (\frac {x}{2}\right )^{2} a +4 \tan \left (\frac {x}{2}\right )^{2} c +2 \tan \left (\frac {x}{2}\right ) b +a \right ) a^{2}+a b x}{c \left (2 a c -b^{2}\right )} \] Input:

int(sin(x)^3/(a+b*sin(x)+c*sin(x)^2),x)
 

Output:

( - 2*cos(x)*a*c + cos(x)*b**2 - 6*int(sin(x)**2/(2*sin(x)**2*a*c**2 - sin 
(x)**2*b**2*c + 2*sin(x)*a*b*c - sin(x)*b**3 + 2*a**2*c - a*b**2),x)*a**2* 
b*c**2 + 5*int(sin(x)**2/(2*sin(x)**2*a*c**2 - sin(x)**2*b**2*c + 2*sin(x) 
*a*b*c - sin(x)*b**3 + 2*a**2*c - a*b**2),x)*a*b**3*c - int(sin(x)**2/(2*s 
in(x)**2*a*c**2 - sin(x)**2*b**2*c + 2*sin(x)*a*b*c - sin(x)*b**3 + 2*a**2 
*c - a*b**2),x)*b**5 - 4*int(sin(x)/(2*sin(x)**2*a*c**2 - sin(x)**2*b**2*c 
 + 2*sin(x)*a*b*c - sin(x)*b**3 + 2*a**2*c - a*b**2),x)*a**3*c**2 + 2*int( 
sin(x)/(2*sin(x)**2*a*c**2 - sin(x)**2*b**2*c + 2*sin(x)*a*b*c - sin(x)*b* 
*3 + 2*a**2*c - a*b**2),x)*a**2*b**2*c - 2*int(1/(2*sin(x)**2*a*c**2 - sin 
(x)**2*b**2*c + 2*sin(x)*a*b*c - sin(x)*b**3 + 2*a**2*c - a*b**2),x)*a**3* 
b*c + int(1/(2*sin(x)**2*a*c**2 - sin(x)**2*b**2*c + 2*sin(x)*a*b*c - sin( 
x)*b**3 + 2*a**2*c - a*b**2),x)*a**2*b**3 - 2*log(tan(x/2)**2 + 1)*a**2 - 
log(sin(x)**2*c + sin(x)*b + a)*a**2 + log(tan(x/2)**4*a + 2*tan(x/2)**3*b 
 + 2*tan(x/2)**2*a + 4*tan(x/2)**2*c + 2*tan(x/2)*b + a)*a**2 + a*b*x)/(c* 
(2*a*c - b**2))