Integrand size = 19, antiderivative size = 323 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {x}{2 c}+\frac {\left (b^2-a c\right ) x}{c^3}-\frac {\sqrt {2} \left (b^3-2 a b c-\frac {b^4-4 a b^2 c+2 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \left (b^3-2 a b c+\frac {b^4-4 a b^2 c+2 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{c^3 \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}+\frac {b \cos (x)}{c^2}-\frac {\cos (x) \sin (x)}{2 c} \]
1/2*x/c+(-a*c+b^2)*x/c^3+b*cos(x)/c^2-1/2*cos(x)*sin(x)/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^3-2*a*b*c+(-2*a^2*c^2+4*a*b^2*c-b^4)/(-4*a*c+b^2)^ (1/2))/c^3/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(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))*2^(1/2)*(b^3-2*a*b*c+(2*a^2*c^2-4*a*b^2*c+b^4)/(-4*a*c+b^2)^(1/2) )/c^3/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 1.22 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {4 b^2 x+2 c (-2 a+c) x-\frac {4 \left (i b^4-4 i a b^2 c+2 i a^2 c^2+b^3 \sqrt {-b^2+4 a c}-2 a b c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}-\frac {4 \left (-i b^4+4 i a b^2 c-2 i a^2 c^2+b^3 \sqrt {-b^2+4 a c}-2 a b c \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}+4 b c \cos (x)-c^2 \sin (2 x)}{4 c^3} \]
(4*b^2*x + 2*c*(-2*a + c)*x - (4*(I*b^4 - (4*I)*a*b^2*c + (2*I)*a^2*c^2 + b^3*Sqrt[-b^2 + 4*a*c] - 2*a*b*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*((-I)*b^4 + (4*I)*a*b^2*c - (2*I)*a^2*c^2 + b^3*Sqrt [-b^2 + 4*a*c] - 2*a*b*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*b*c*Cos[x] - c^2*Sin[2*x])/(4*c^3)
Time = 2.24 (sec) , antiderivative size = 323, 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.
| \(\displaystyle \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)^4}{a+b \sin (x)+c \sin (x)^2}dx\) |
| \(\Big \downarrow \) 3737 |
| \(\displaystyle \int \left (\frac {b^2-a c}{c^3}+\frac {-a b^2 \left (1-\frac {a c}{b^2}\right )-\left (b^3 \sin (x) \left (1-\frac {2 a c}{b^2}\right )\right )}{c^3 \left (a+b \sin (x)+c \sin ^2(x)\right )}-\frac {b \sin (x)}{c^2}+\frac {\sin ^2(x)}{c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {2} \left (-\frac {2 a^2 c^2-4 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-2 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c^3 \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {\sqrt {2} \left (\frac {2 a^2 c^2-4 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-2 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{c^3 \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {x \left (b^2-a c\right )}{c^3}+\frac {b \cos (x)}{c^2}+\frac {x}{2 c}-\frac {\sin (x) \cos (x)}{2 c}\) |
x/(2*c) + ((b^2 - a*c)*x)/c^3 - (Sqrt[2]*(b^3 - 2*a*b*c - (b^4 - 4*a*b^2*c + 2*a^2*c^2)/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^3*Sqrt [b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]]) - (Sqrt[2]*(b^3 - 2*a*b*c + (b^ 4 - 4*a*b^2*c + 2*a^2*c^2)/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^3*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + (b*Cos[x])/c^2 - (Cos[x]*Sin[x])/(2*c)
3.1.1.3.1 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 = 3.04 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {2 a \left (-\frac {2 \left (-3 \sqrt {-4 a c +b^{2}}\, a b c +\sqrt {-4 a c +b^{2}}\, b^{3}+4 a^{2} c^{2}-5 a \,b^{2} c +b^{4}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {2 \left (3 \sqrt {-4 a c +b^{2}}\, a b c -\sqrt {-4 a c +b^{2}}\, b^{3}+4 a^{2} c^{2}-5 a \,b^{2} c +b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{c^{3}}-\frac {2 \left (\frac {-\frac {c^{2} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-b c \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+\frac {c^{2} \tan \left (\frac {x}{2}\right )}{2}-b c}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}+\frac {\left (2 a c -2 b^{2}-c^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{c^{3}}\) | \(372\) |
| risch | \(\text {Expression too large to display}\) | \(3919\) |
2/c^3*a*(-2*(-3*(-4*a*c+b^2)^(1/2)*a*b*c+(-4*a*c+b^2)^(1/2)*b^3+4*a^2*c^2- 5*a*b^2*c+b^4)/(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))+2*(3*(-4*a*c+b^2)^(1/2)*a*b*c-(-4*a*c+b^2)^(1/2) *b^3+4*a^2*c^2-5*a*b^2*c+b^4)/(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/c^3*((-1/2*c^2*tan(1/2*x)^3-b*c *tan(1/2*x)^2+1/2*c^2*tan(1/2*x)-b*c)/(1+tan(1/2*x)^2)^2+1/2*(2*a*c-2*b^2- c^2)*arctan(tan(1/2*x)))
Leaf count of result is larger than twice the leaf count of optimal. 8169 vs. \(2 (285) = 570\).
Time = 4.77 (sec) , antiderivative size = 8169, normalized size of antiderivative = 25.29 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\sin \left (x\right )^{4}}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \]
1/4*(4*c^3*integrate(-2*(2*(b^4 - 2*a*b^2*c)*cos(3*x)^2 + 4*(2*a^2*b^2 - a ^2*c^2 - (2*a^3 - a*b^2)*c)*cos(2*x)^2 + 2*(b^4 - 2*a*b^2*c)*cos(x)^2 + 2* (b^4 - 2*a*b^2*c)*sin(3*x)^2 + 2*(4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c) *cos(x)*sin(2*x) + 4*(2*a^2*b^2 - a^2*c^2 - (2*a^3 - a*b^2)*c)*sin(2*x)^2 + 2*(b^4 - 2*a*b^2*c)*sin(x)^2 - (2*(a*b^2*c - a^2*c^2)*cos(2*x) + (b^3*c - 2*a*b*c^2)*sin(3*x) - (b^3*c - 2*a*b*c^2)*sin(x))*cos(4*x) - 2*(2*(b^4 - 2*a*b^2*c)*cos(x) + (4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c)*sin(2*x))*c os(3*x) - 2*(a*b^2*c - a^2*c^2 + (4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c) *sin(x))*cos(2*x) + ((b^3*c - 2*a*b*c^2)*cos(3*x) - (b^3*c - 2*a*b*c^2)*co s(x) - 2*(a*b^2*c - a^2*c^2)*sin(2*x))*sin(4*x) - (b^3*c - 2*a*b*c^2 - 2*( 4*a*b^3 - 2*a*b*c^2 - (6*a^2*b - b^3)*c)*cos(2*x) + 4*(b^4 - 2*a*b^2*c)*si n(x))*sin(3*x) + (b^3*c - 2*a*b*c^2)*sin(x))/(c^5*cos(4*x)^2 + 4*b^2*c^3*c os(3*x)^2 + 4*b^2*c^3*cos(x)^2 + c^5*sin(4*x)^2 + 4*b^2*c^3*sin(3*x)^2 + 4 *b^2*c^3*sin(x)^2 + 4*b*c^4*sin(x) + c^5 + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*c os(2*x)^2 + 8*(2*a*b*c^3 + b*c^4)*cos(x)*sin(2*x) + 4*(4*a^2*c^3 + 4*a*c^4 + c^5)*sin(2*x)^2 - 2*(2*b*c^4*sin(3*x) - 2*b*c^4*sin(x) - c^5 + 2*(2*a*c ^4 + c^5)*cos(2*x))*cos(4*x) - 8*(b^2*c^3*cos(x) + (2*a*b*c^3 + b*c^4)*sin (2*x))*cos(3*x) - 4*(2*a*c^4 + c^5 + 2*(2*a*b*c^3 + b*c^4)*sin(x))*cos(2*x ) + 4*(b*c^4*cos(3*x) - b*c^4*cos(x) - (2*a*c^4 + c^5)*sin(2*x))*sin(4*x) - 4*(2*b^2*c^3*sin(x) + b*c^4 - 2*(2*a*b*c^3 + b*c^4)*cos(2*x))*sin(3*x...
Timed out. \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 25.96 (sec) , antiderivative size = 39682, normalized size of antiderivative = 122.85 \[ \int \frac {\sin ^4(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
((2*b)/c^2 - tan(x/2)/c + tan(x/2)^3/c + (2*b*tan(x/2)^2)/c^2)/(2*tan(x/2) ^2 + tan(x/2)^4 + 1) - atan(((((2048*(44*a^5*c^9 - 16*a^4*c^10 - 4*a^6*c^8 - 64*a^7*c^7 + 12*a^8*c^6 + 4*a*b^6*c^7 + 15*a*b^8*c^5 + 14*a*b^10*c^3 - 28*a^2*b^4*c^8 - 119*a^2*b^6*c^6 - 128*a^2*b^8*c^4 - 8*a^2*b^10*c^2 + 52*a ^3*b^2*c^9 + 290*a^3*b^4*c^7 + 397*a^3*b^6*c^5 + 62*a^3*b^8*c^3 - 227*a^4* b^2*c^8 - 491*a^4*b^4*c^6 - 148*a^4*b^6*c^4 + 8*a^4*b^8*c^2 + 221*a^5*b^2* c^7 + 102*a^5*b^4*c^5 - 60*a^5*b^6*c^3 + 68*a^6*b^2*c^6 + 136*a^6*b^4*c^4 - 100*a^7*b^2*c^5))/c^8 - (-(a^2*b^8 - b^10 + 8*a^5*c^5 + 8*a^6*c^4 - b^7* (-(4*a*c - b^2)^3)^(1/2) - 10*a^3*b^6*c + a^2*b^5*(-(4*a*c - b^2)^3)^(1/2) - 52*a^2*b^6*c^2 + 96*a^3*b^4*c^3 - 66*a^4*b^2*c^4 + 33*a^4*b^4*c^2 - 38* a^5*b^2*c^3 + 12*a*b^8*c + 4*a^3*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 4*a^3*b^ 3*c*(-(4*a*c - b^2)^3)^(1/2) + 3*a^4*b*c^2*(-(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*(16*a^2*c^10 + 32*a^3*c^9 + 16*a^4*c^8 + b^4*c^8 - b^6*c^6 - 8*a*b^2*c^ 9 + 10*a*b^4*c^7 - 32*a^2*b^2*c^8 + a^2*b^4*c^6 - 8*a^3*b^2*c^7)))^(1/2)*( (2048*(4*a*b^3*c^11 + 13*a*b^5*c^9 + 4*a*b^7*c^7 - 12*a*b^9*c^5 - 16*a^2*b *c^12 + 44*a^3*b*c^11 + 4*a^4*b*c^10 + 80*a^5*b*c^9 + 12*a^6*b*c^8 - 63*a^ 2*b^3*c^10 - 16*a^2*b^5*c^8 + 76*a^2*b^7*c^6 - a^3*b^3*c^9 - 104*a^3*b^5*c ^7 + 12*a^3*b^7*c^5 - 56*a^4*b^3*c^8 - 60*a^4*b^5*c^6 + 48*a^5*b^3*c^7))/c ^8 - (((2048*(12*a*b^5*c^11 - 16*a*b^3*c^13 + 64*a^2*b*c^14 + 80*a^3*b*...