Integrand size = 19, antiderivative size = 324 \[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {\sqrt {2} b c \left (1+\frac {b^2-2 c (a+c)}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {\sqrt {2} b c \left (1-\frac {b^2-2 c (a+c)}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}+\frac {\cos (x)}{2 (a+b+c) (1-\sin (x))}-\frac {\cos (x)}{2 (a-b+c) (1+\sin (x))} \]
1/2*cos(x)/(a+b+c)/(1-sin(x))-1/2*cos(x)/(a-b+c)/(1+sin(x))-b*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)*(1+(b^2-2*c*(a+c))/b/(-4*a*c+b^2)^(1/2))/(a-b+c )/(a+b+c)/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2)-b*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)*(1+(-b^2+2*c*(a+c))/b/(-4*a*c+b^2)^(1/2))/(a-b+c)/(a+b +c)/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 1.20 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {c \left (-i b^2+2 i c (a+c)+b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \left (a^2-b^2+2 a c+c^2\right ) \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}-\frac {c \left (i b^2-2 i c (a+c)+b \sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c} \left (a^2-b^2+2 a c+c^2\right ) \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}+\frac {\sin \left (\frac {x}{2}\right )}{(a+b+c) \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )}+\frac {\sin \left (\frac {x}{2}\right )}{(a-b+c) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )} \]
-((c*((-I)*b^2 + (2*I)*c*(a + c) + b*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*Sq rt[-b^2 + 4*a*c]])])/(Sqrt[-1/2*b^2 + 2*a*c]*(a^2 - b^2 + 2*a*c + c^2)*Sqr t[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])) - (c*(I*b^2 - (2*I)*c*(a + c) + b*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]*(a^2 - b^2 + 2*a*c + c^2)*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqr t[-b^2 + 4*a*c]]) + Sin[x/2]/((a + b + c)*(Cos[x/2] - Sin[x/2])) + Sin[x/2 ]/((a - b + c)*(Cos[x/2] + Sin[x/2]))
Time = 1.58 (sec) , antiderivative size = 324, 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, 3747, 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.
| \(\displaystyle \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (x)^2 \left (a+b \sin (x)+c \sin (x)^2\right )}dx\) |
| \(\Big \downarrow \) 3747 |
| \(\displaystyle \int \left (\frac {-\left (b^2 \left (1-\frac {c (a+c)}{b^2}\right )\right )-b c \sin (x)}{(a-b+c) (a+b+c) \left (a+b \sin (x)+c \sin ^2(x)\right )}-\frac {1}{2 (\sin (x)-1) (a+b+c)}+\frac {1}{2 (\sin (x)+1) (a-b+c)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {2} b c \left (\frac {b^2-2 c (a+c)}{b \sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {\sqrt {2} b c \left (1-\frac {b^2-2 c (a+c)}{b \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{(a-b+c) (a+b+c) \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\cos (x)}{2 (1-\sin (x)) (a+b+c)}-\frac {\cos (x)}{2 (\sin (x)+1) (a-b+c)}\) |
-((Sqrt[2]*b*c*(1 + (b^2 - 2*c*(a + c))/(b*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*S qrt[b^2 - 4*a*c]])])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 2*c*(a + c) - b*S qrt[b^2 - 4*a*c]])) - (Sqrt[2]*b*c*(1 - (b^2 - 2*c*(a + c))/(b*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]])])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + Cos[x]/(2*(a + b + c)*(1 - Sin[x] )) - Cos[x]/(2*(a - b + c)*(1 + Sin[x]))
3.1.13.3.1 Defintions of rubi rules used
Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n _.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_.), x_Symbol] :> Int[ExpandT rig[(1 - sin[d + e*x]^2)^(m/2)*(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] && IntegerQ[m/2] & & NeQ[b^2 - 4*a*c, 0] && IntegersQ[n, p]
Time = 7.06 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {2 a \left (\frac {\left (-3 \sqrt {-4 a c +b^{2}}\, a b c +\sqrt {-4 a c +b^{2}}\, b^{3}-\sqrt {-4 a c +b^{2}}\, b \,c^{2}+4 a^{2} c^{2}-5 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}-\frac {\left (3 \sqrt {-4 a c +b^{2}}\, a b c -\sqrt {-4 a c +b^{2}}\, b^{3}+\sqrt {-4 a c +b^{2}}\, b \,c^{2}+4 a^{2} c^{2}-5 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (a -b +c \right ) \left (a +b +c \right )}-\frac {2}{\left (2 a +2 b +2 c \right ) \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {2}{\left (2 a -2 b +2 c \right ) \left (\tan \left (\frac {x}{2}\right )+1\right )}\) | \(417\) |
| risch | \(\text {Expression too large to display}\) | \(6071\) |
2/(a-b+c)/(a+b+c)*a*((-3*(-4*a*c+b^2)^(1/2)*a*b*c+(-4*a*c+b^2)^(1/2)*b^3-( -4*a*c+b^2)^(1/2)*b*c^2+4*a^2*c^2-5*a*b^2*c+4*a*c^3+b^4-b^2*c^2)/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)) -(3*(-4*a*c+b^2)^(1/2)*a*b*c-(-4*a*c+b^2)^(1/2)*b^3+(-4*a*c+b^2)^(1/2)*b*c ^2+4*a^2*c^2-5*a*b^2*c+4*a*c^3+b^4-b^2*c^2)/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/(2*a+2*b+2*c)/(t an(1/2*x)-1)-2/(2*a-2*b+2*c)/(tan(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 16739 vs. \(2 (282) = 564\).
Time = 3.95 (sec) , antiderivative size = 16739, normalized size of antiderivative = 51.66 \[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int \frac {\sec ^{2}{\left (x \right )}}{a + b \sin {\left (x \right )} + c \sin ^{2}{\left (x \right )}}\, dx \]
\[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\sec \left (x\right )^{2}}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \]
-(2*b*cos(2*x)*cos(x) + 2*b*cos(x) + ((a^2 - b^2 + 2*a*c + c^2)*cos(2*x)^2 + (a^2 - b^2 + 2*a*c + c^2)*sin(2*x)^2 + a^2 - b^2 + 2*a*c + c^2 + 2*(a^2 - b^2 + 2*a*c + c^2)*cos(2*x))*integrate(2*(2*b^2*c*cos(3*x)^2 + 2*b^2*c* cos(x)^2 + 2*b^2*c*sin(3*x)^2 + 2*b^2*c*sin(x)^2 + b*c^2*sin(x) + 4*(2*a*b ^2 - 3*a*c^2 - c^3 - (2*a^2 - b^2)*c)*cos(2*x)^2 + 2*(2*b^3 - b*c^2)*cos(x )*sin(2*x) + 4*(2*a*b^2 - 3*a*c^2 - c^3 - (2*a^2 - b^2)*c)*sin(2*x)^2 - (b *c^2*sin(3*x) - b*c^2*sin(x) + 2*(b^2*c - a*c^2 - c^3)*cos(2*x))*cos(4*x) - 2*(2*b^2*c*cos(x) + (2*b^3 - b*c^2)*sin(2*x))*cos(3*x) - 2*(b^2*c - a*c^ 2 - c^3 + (2*b^3 - b*c^2)*sin(x))*cos(2*x) + (b*c^2*cos(3*x) - b*c^2*cos(x ) - 2*(b^2*c - a*c^2 - c^3)*sin(2*x))*sin(4*x) - (4*b^2*c*sin(x) + b*c^2 - 2*(2*b^3 - b*c^2)*cos(2*x))*sin(3*x))/(2*a*c^3 + c^4 + (a^2 - b^2)*c^2 + (2*a*c^3 + c^4 + (a^2 - b^2)*c^2)*cos(4*x)^2 + 4*(a^2*b^2 - b^4 + 2*a*b^2* c + b^2*c^2)*cos(3*x)^2 + 4*(4*a^4 - 4*a^2*b^2 + 6*a*c^3 + c^4 + (13*a^2 - b^2)*c^2 + 4*(3*a^3 - a*b^2)*c)*cos(2*x)^2 + 4*(a^2*b^2 - b^4 + 2*a*b^2*c + b^2*c^2)*cos(x)^2 + (2*a*c^3 + c^4 + (a^2 - b^2)*c^2)*sin(4*x)^2 + 4*(a ^2*b^2 - b^4 + 2*a*b^2*c + b^2*c^2)*sin(3*x)^2 + 8*(2*a^3*b - 2*a*b^3 + 4* a*b*c^2 + b*c^3 + (5*a^2*b - b^3)*c)*cos(x)*sin(2*x) + 4*(4*a^4 - 4*a^2*b^ 2 + 6*a*c^3 + c^4 + (13*a^2 - b^2)*c^2 + 4*(3*a^3 - a*b^2)*c)*sin(2*x)^2 + 4*(a^2*b^2 - b^4 + 2*a*b^2*c + b^2*c^2)*sin(x)^2 + 2*(2*a*c^3 + c^4 + (a^ 2 - b^2)*c^2 - 2*(4*a*c^3 + c^4 + (5*a^2 - b^2)*c^2 + 2*(a^3 - a*b^2)*c...
Timed out. \[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 28.27 (sec) , antiderivative size = 37118, normalized size of antiderivative = 114.56 \[ \int \frac {\sec ^2(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
atan(((-(8*a*c^7 + b^8 + 24*a^2*c^6 + 24*a^3*c^5 + 8*a^4*c^4 + b^5*(-(4*a* c - b^2)^3)^(1/2) - 2*b^2*c^6 + 3*b^4*c^4 - 3*b^6*c^2 - 18*a*b^2*c^5 + 24* a*b^4*c^3 + 3*b*c^4*(-(4*a*c - b^2)^3)^(1/2) - 54*a^2*b^2*c^4 + 33*a^2*b^4 *c^2 - 38*a^3*b^2*c^3 - 3*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 10*a*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) + 6*a*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b^3*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(3*a^2*b^8 - b^10 - 3*a^4*b^6 + a ^6*b^4 + 16*a^2*c^8 + 96*a^3*c^7 + 240*a^4*c^6 + 320*a^5*c^5 + 240*a^6*c^4 + 96*a^7*c^3 + 16*a^8*c^2 + b^4*c^6 - 3*b^6*c^4 + 3*b^8*c^2 - 8*a*b^2*c^7 + 30*a*b^4*c^5 - 36*a*b^6*c^3 - 36*a^3*b^6*c + 30*a^5*b^4*c - 8*a^7*b^2*c - 96*a^2*b^2*c^6 + 159*a^2*b^4*c^4 - 82*a^2*b^6*c^2 - 312*a^3*b^2*c^5 + 2 60*a^3*b^4*c^3 - 448*a^4*b^2*c^4 + 159*a^4*b^4*c^2 - 312*a^5*b^2*c^3 - 96* a^6*b^2*c^2 + 14*a*b^8*c)))^(1/2)*((-(8*a*c^7 + b^8 + 24*a^2*c^6 + 24*a^3* c^5 + 8*a^4*c^4 + b^5*(-(4*a*c - b^2)^3)^(1/2) - 2*b^2*c^6 + 3*b^4*c^4 - 3 *b^6*c^2 - 18*a*b^2*c^5 + 24*a*b^4*c^3 + 3*b*c^4*(-(4*a*c - b^2)^3)^(1/2) - 54*a^2*b^2*c^4 + 33*a^2*b^4*c^2 - 38*a^3*b^2*c^3 - 3*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 10*a*b^6*c + 3*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) + 6*a*b* c^3*(-(4*a*c - b^2)^3)^(1/2) - 4*a*b^3*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(3*a ^2*b^8 - b^10 - 3*a^4*b^6 + a^6*b^4 + 16*a^2*c^8 + 96*a^3*c^7 + 240*a^4*c^ 6 + 320*a^5*c^5 + 240*a^6*c^4 + 96*a^7*c^3 + 16*a^8*c^2 + b^4*c^6 - 3*b^6* c^4 + 3*b^8*c^2 - 8*a*b^2*c^7 + 30*a*b^4*c^5 - 36*a*b^6*c^3 - 36*a^3*b^...