Integrand size = 19, antiderivative size = 331 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {\sqrt {2} c \left (b^3-3 a b c+\sqrt {b^2-4 a c} \left (b^2-a c\right )\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} c \left (b^3-3 a b c-\sqrt {b^2-4 a c} \left (b^2-a c\right )\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}-\frac {\text {arctanh}(\cos (x))}{2 a}-\frac {\left (b^2-a c\right ) \text {arctanh}(\cos (x))}{a^3}+\frac {b \cot (x)}{a^2}-\frac {\cot (x) \csc (x)}{2 a} \]
-1/2*arctanh(cos(x))/a-(-a*c+b^2)*arctanh(cos(x))/a^3+b*cot(x)/a^2-1/2*cot (x)*csc(x)/a-c*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))*2^(1/2)*(b^3-3*a*b*c+(-a*c+b^2 )*(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(1/2)/(b^2-2*c*(a+c)-b*(-4*a*c+b^2) ^(1/2))^(1/2)+c*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))*2^(1/2)*(b^3-3*a*b*c-(-a*c+b^ 2)*(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(1/2)/(b^2-2*c*(a+c)+b*(-4*a*c+b^2 )^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 2.94 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.45 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\csc ^2(x) (-2 a-c+c \cos (2 x)-2 b \sin (x)) \left (\frac {8 c \left (-i b^3+3 i a b c+b^2 \sqrt {-b^2+4 a c}-a c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {8 c \left (i b^3-3 i a b c+b^2 \sqrt {-b^2+4 a c}-a c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}-4 a b \cot \left (\frac {x}{2}\right )+a^2 \csc ^2\left (\frac {x}{2}\right )+4 \left (a^2+2 b^2-2 a c\right ) \log \left (\cos \left (\frac {x}{2}\right )\right )-4 \left (a^2+2 b^2-2 a c\right ) \log \left (\sin \left (\frac {x}{2}\right )\right )-a^2 \sec ^2\left (\frac {x}{2}\right )+4 a b \tan \left (\frac {x}{2}\right )\right )}{16 a^3 \left (c+b \csc (x)+a \csc ^2(x)\right )} \]
(Csc[x]^2*(-2*a - c + c*Cos[2*x] - 2*b*Sin[x])*((8*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 *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]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqr t[-b^2 + 4*a*c]]) + (8*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*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]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]]) - 4*a*b*Cot[x /2] + a^2*Csc[x/2]^2 + 4*(a^2 + 2*b^2 - 2*a*c)*Log[Cos[x/2]] - 4*(a^2 + 2* b^2 - 2*a*c)*Log[Sin[x/2]] - a^2*Sec[x/2]^2 + 4*a*b*Tan[x/2]))/(16*a^3*(c + b*Csc[x] + a*Csc[x]^2))
Time = 2.37 (sec) , antiderivative size = 331, 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.
| \(\displaystyle \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (x)^3 \left (a+b \sin (x)+c \sin (x)^2\right )}dx\) |
| \(\Big \downarrow \) 3737 |
| \(\displaystyle \int \left (\frac {\csc (x) \left (b^2-a c\right )}{a^3}+\frac {-b^2 c \sin (x) \left (1-\frac {a c}{b^2}\right )-\left (b^3 \left (1-\frac {2 a c}{b^2}\right )\right )}{a^3 \left (a+b \sin (x)+c \sin ^2(x)\right )}-\frac {b \csc ^2(x)}{a^2}+\frac {\csc ^3(x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {2} c \left (\sqrt {b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\sqrt {2} c \left (-\sqrt {b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {\left (b^2-a c\right ) \text {arctanh}(\cos (x))}{a^3}+\frac {b \cot (x)}{a^2}-\frac {\text {arctanh}(\cos (x))}{2 a}-\frac {\cot (x) \csc (x)}{2 a}\) |
-((Sqrt[2]*c*(b^3 - 3*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 - 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*Sqr t[b^2 - 4*a*c]])])/(a^3*Sqrt[b^2 - 4*a*c]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[ b^2 - 4*a*c]])) + (Sqrt[2]*c*(b^3 - 3*a*b*c - Sqrt[b^2 - 4*a*c]*(b^2 - 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]])])/(a^3*Sqrt[b^2 - 4*a*c]*Sqrt[b^2 - 2*c*( a + c) + b*Sqrt[b^2 - 4*a*c]]) - ArcTanh[Cos[x]]/(2*a) - ((b^2 - a*c)*ArcT anh[Cos[x]])/a^3 + (b*Cot[x])/a^2 - (Cot[x]*Csc[x])/(2*a)
3.1.8.3.1 Defintions of rubi rules used
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]
Time = 4.32 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.25
| method | result | size |
| default | \(\frac {\frac {a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-2 b \tan \left (\frac {x}{2}\right )}{4 a^{2}}+\frac {-\frac {2 \left (-2 \sqrt {-4 a c +b^{2}}\, a^{2} c^{2}+4 \sqrt {-4 a c +b^{2}}\, b^{2} c a -\sqrt {-4 a c +b^{2}}\, b^{4}+8 a^{2} b \,c^{2}-6 a \,b^{3} c +b^{5}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {2 \left (2 \sqrt {-4 a c +b^{2}}\, a^{2} c^{2}-4 \sqrt {-4 a c +b^{2}}\, b^{2} c a +\sqrt {-4 a c +b^{2}}\, b^{4}+8 a^{2} b \,c^{2}-6 a \,b^{3} c +b^{5}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}}{a^{2}}-\frac {1}{8 a \tan \left (\frac {x}{2}\right )^{2}}+\frac {\left (2 a^{2}-4 a c +4 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {x}{2}\right )}\) | \(414\) |
| risch | \(\text {Expression too large to display}\) | \(4146\) |
1/4/a^2*(1/2*a*tan(1/2*x)^2-2*b*tan(1/2*x))+2/a^2*(-(-2*(-4*a*c+b^2)^(1/2) *a^2*c^2+4*(-4*a*c+b^2)^(1/2)*b^2*c*a-(-4*a*c+b^2)^(1/2)*b^4+8*a^2*b*c^2-6 *a*b^3*c+b^5)/a/(4*a*c-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))+(2*(-4*a*c+b^2)^(1/2)*a^2*c^2-4*(-4*a*c+b^2)^(1/2 )*b^2*c*a+(-4*a*c+b^2)^(1/2)*b^4+8*a^2*b*c^2-6*a*b^3*c+b^5)/a/(4*a*c-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)))-1/8 /a/tan(1/2*x)^2+1/4/a^3*(2*a^2-4*a*c+4*b^2)*ln(tan(1/2*x))+1/2/a^2*b/tan(1 /2*x)
Timed out. \[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int \frac {\csc ^{3}{\left (x \right )}}{a + b \sin {\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \]
\[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\csc \left (x\right )^{3}}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \]
-1/4*(8*a^2*cos(2*x)*cos(x) + 8*a^2*sin(3*x)*sin(2*x) - 4*a^2*cos(x) - 4*( a^2*cos(3*x) + a^2*cos(x) - 2*a*b*sin(2*x))*cos(4*x) + 4*(2*a^2*cos(2*x) - a^2)*cos(3*x) - 4*(a^3*cos(4*x)^2 + 4*a^3*cos(2*x)^2 + a^3*sin(4*x)^2 - 4 *a^3*sin(4*x)*sin(2*x) + 4*a^3*sin(2*x)^2 - 4*a^3*cos(2*x) + a^3 - 2*(2*a^ 3*cos(2*x) - a^3)*cos(4*x))*integrate(-2*(2*(b^3*c - a*b*c^2)*cos(3*x)^2 + 4*(2*a*b^3 - 2*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(2*x)^2 + 2*(b^3*c - a*b*c ^2)*cos(x)^2 + 2*(b^3*c - a*b*c^2)*sin(3*x)^2 + 2*(2*b^4 - 2*a*b^2*c - a*c ^3 - (2*a^2 - b^2)*c^2)*cos(x)*sin(2*x) + 4*(2*a*b^3 - 2*a*b*c^2 - (4*a^2* b - b^3)*c)*sin(2*x)^2 + 2*(b^3*c - a*b*c^2)*sin(x)^2 - (2*(b^3*c - 2*a*b* c^2)*cos(2*x) + (b^2*c^2 - a*c^3)*sin(3*x) - (b^2*c^2 - a*c^3)*sin(x))*cos (4*x) - 2*(2*(b^3*c - a*b*c^2)*cos(x) + (2*b^4 - 2*a*b^2*c - a*c^3 - (2*a^ 2 - b^2)*c^2)*sin(2*x))*cos(3*x) - 2*(b^3*c - 2*a*b*c^2 + (2*b^4 - 2*a*b^2 *c - a*c^3 - (2*a^2 - b^2)*c^2)*sin(x))*cos(2*x) + ((b^2*c^2 - a*c^3)*cos( 3*x) - (b^2*c^2 - a*c^3)*cos(x) - 2*(b^3*c - 2*a*b*c^2)*sin(2*x))*sin(4*x) - (b^2*c^2 - a*c^3 - 2*(2*b^4 - 2*a*b^2*c - a*c^3 - (2*a^2 - b^2)*c^2)*co s(2*x) + 4*(b^3*c - a*b*c^2)*sin(x))*sin(3*x) + (b^2*c^2 - a*c^3)*sin(x))/ (a^3*c^2*cos(4*x)^2 + 4*a^3*b^2*cos(3*x)^2 + 4*a^3*b^2*cos(x)^2 + a^3*c^2* sin(4*x)^2 + 4*a^3*b^2*sin(3*x)^2 + 4*a^3*b^2*sin(x)^2 + 4*a^3*b*c*sin(x) + a^3*c^2 + 4*(4*a^5 + 4*a^4*c + a^3*c^2)*cos(2*x)^2 + 8*(2*a^4*b + a^3*b* c)*cos(x)*sin(2*x) + 4*(4*a^5 + 4*a^4*c + a^3*c^2)*sin(2*x)^2 - 2*(2*a^...
Timed out. \[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 24.74 (sec) , antiderivative size = 21909, normalized size of antiderivative = 66.19 \[ \int \frac {\csc ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
atan(-(((8*a^4*c^6 - b^10 + 8*a^5*c^5 - b^7*(-(4*a*c - b^2)^3)^(1/2) + b^8 *c^2 - 10*a*b^6*c^3 + 33*a^2*b^4*c^4 - 52*a^2*b^6*c^2 - 38*a^3*b^2*c^5 + 9 6*a^3*b^4*c^3 - 66*a^4*b^2*c^4 + b^5*c^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a*b ^8*c - 4*a*b^3*c^3*(-(4*a*c - b^2)^3)^(1/2) + 3*a^2*b*c^4*(-(4*a*c - b^2)^ 3)^(1/2) + 4*a^3*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 10*a^2*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) + 6*a*b^5*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^8*b^4 - a^6*b^ 6 + 16*a^8*c^4 + 32*a^9*c^3 + 16*a^10*c^2 + 10*a^7*b^4*c - 8*a^9*b^2*c + a ^6*b^4*c^2 - 8*a^7*b^2*c^3 - 32*a^8*b^2*c^2)))^(1/2)*(((8*a^4*c^6 - b^10 + 8*a^5*c^5 - b^7*(-(4*a*c - b^2)^3)^(1/2) + b^8*c^2 - 10*a*b^6*c^3 + 33*a^ 2*b^4*c^4 - 52*a^2*b^6*c^2 - 38*a^3*b^2*c^5 + 96*a^3*b^4*c^3 - 66*a^4*b^2* c^4 + b^5*c^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a*b^8*c - 4*a*b^3*c^3*(-(4*a*c - b^2)^3)^(1/2) + 3*a^2*b*c^4*(-(4*a*c - b^2)^3)^(1/2) + 4*a^3*b*c^3*(-(4 *a*c - b^2)^3)^(1/2) - 10*a^2*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) + 6*a*b^5*c *(-(4*a*c - b^2)^3)^(1/2))/(2*(a^8*b^4 - a^6*b^6 + 16*a^8*c^4 + 32*a^9*c^3 + 16*a^10*c^2 + 10*a^7*b^4*c - 8*a^9*b^2*c + a^6*b^4*c^2 - 8*a^7*b^2*c^3 - 32*a^8*b^2*c^2)))^(1/2)*(((8*a^4*c^6 - b^10 + 8*a^5*c^5 - b^7*(-(4*a*c - b^2)^3)^(1/2) + b^8*c^2 - 10*a*b^6*c^3 + 33*a^2*b^4*c^4 - 52*a^2*b^6*c^2 - 38*a^3*b^2*c^5 + 96*a^3*b^4*c^3 - 66*a^4*b^2*c^4 + b^5*c^2*(-(4*a*c - b^ 2)^3)^(1/2) + 12*a*b^8*c - 4*a*b^3*c^3*(-(4*a*c - b^2)^3)^(1/2) + 3*a^2*b* c^4*(-(4*a*c - b^2)^3)^(1/2) + 4*a^3*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 1...