Integrand size = 14, antiderivative size = 221 \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {2 \sqrt {2} c \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}-\frac {2 \sqrt {2} c \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}} \]
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)/(-4*a*c+b^2)^(1/2)/(b^2-2*c*(a+c) -b*(-4*a*c+b^2)^(1/2))^(1/2)-2*c*arctan(1/2*(2*c+(b+(-4*a*c+b^2)^(1/2))*ta n(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)/(-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 = 1.00 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.05 \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=-\frac {2 i c \left (\frac {\arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}-\frac {\arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {-\frac {b^2}{2}+2 a c}} \]
((-2*I)*c*(ArcTan[(2*c + (b - I*Sqrt[-b^2 + 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqr t[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]])]/Sqrt[b^2 - 2*c*(a + c) - I *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[b^2 - 2 *c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]]))/Sqrt[-1/2*b^2 + 2*a*c]
Time = 0.49 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3729, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a+b \sin (x)+c \sin (x)^2}dx\) |
| \(\Big \downarrow \) 3729 |
| \(\displaystyle \frac {2 c \int \frac {1}{b+2 c \sin (x)-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{b+2 c \sin (x)+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \int \frac {1}{b+2 c \sin (x)-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{b+2 c \sin (x)+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {4 c \int \frac {1}{\left (b-\sqrt {b^2-4 a c}\right ) \tan ^2\left (\frac {x}{2}\right )+4 c \tan \left (\frac {x}{2}\right )+b-\sqrt {b^2-4 a c}}d\tan \left (\frac {x}{2}\right )}{\sqrt {b^2-4 a c}}-\frac {4 c \int \frac {1}{\left (b+\sqrt {b^2-4 a c}\right ) \tan ^2\left (\frac {x}{2}\right )+4 c \tan \left (\frac {x}{2}\right )+b+\sqrt {b^2-4 a c}}d\tan \left (\frac {x}{2}\right )}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {8 c \int \frac {1}{4 \left (4 c^2-\left (b+\sqrt {b^2-4 a c}\right )^2\right )-\left (4 c+2 \left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )^2}d\left (4 c+2 \left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )}{\sqrt {b^2-4 a c}}-\frac {8 c \int \frac {1}{-\left (4 c+2 \left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )^2-8 \left (b^2-\sqrt {b^2-4 a c} b-2 c (a+c)\right )}d\left (4 c+2 \left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \sqrt {2} c \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+4 c}{2 \sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}-\frac {2 \sqrt {2} c \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+4 c}{2 \sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {b^2-4 a c} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\) |
(2*Sqrt[2]*c*ArcTan[(4*c + 2*(b - Sqrt[b^2 - 4*a*c])*Tan[x/2])/(2*Sqrt[2]* Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/(Sqrt[b^2 - 4*a*c]*Sqrt[b ^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]]) - (2*Sqrt[2]*c*ArcTan[(4*c + 2*(b + Sqrt[b^2 - 4*a*c])*Tan[x/2])/(2*Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt [b^2 - 4*a*c]])])/(Sqrt[b^2 - 4*a*c]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])
3.1.5.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n_.) + (c_.)*sin[(d_.) + (e_.)* (x_)]^(n2_.))^(-1), x_Symbol] :> Module[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c /q) Int[1/(b - q + 2*c*Sin[d + e*x]^n), x], x] - Simp[2*(c/q) Int[1/(b + q + 2*c*Sin[d + e*x]^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Time = 1.02 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.12
| method | result | size |
| default | \(2 a \left (-\frac {\left (b \sqrt {-4 a c +b^{2}}+4 a c -b^{2}\right ) \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {\left (-b \sqrt {-4 a c +b^{2}}+4 a c -b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{a \left (4 a c -b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )\) | \(247\) |
| risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{4} c^{2}-8 a^{3} b^{2} c +32 a^{3} c^{3}+a^{2} b^{4}-32 a^{2} b^{2} c^{2}+16 a^{2} c^{4}+10 a \,b^{4} c -8 a \,b^{2} c^{3}-b^{6}+b^{4} c^{2}\right ) \textit {\_Z}^{4}+\left (8 a^{2} c^{2}-6 a \,b^{2} c +8 a \,c^{3}+b^{4}-2 b^{2} c^{2}\right ) \textit {\_Z}^{2}+c^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i x}+\left (\frac {b^{5}}{c^{2}}+2 b \,c^{2}-\frac {8 a^{4}}{b}+22 a^{2} b +\frac {6 b \,a^{3}}{c}-\frac {24 c \,a^{3}}{b}-\frac {b^{3} a^{2}}{c^{2}}-\frac {24 c^{2} a^{2}}{b}-\frac {8 b^{3} a}{c}+18 b c a -\frac {8 c^{3} a}{b}-3 b^{3}\right ) \textit {\_R}^{3}+\left (-\frac {4 i c^{2} a}{b}+\frac {i b \,a^{2}}{c}-\frac {8 i c \,a^{2}}{b}+6 i a b -\frac {4 i a^{3}}{b}-\frac {i b^{3}}{c}+i b c \right ) \textit {\_R}^{2}+\left (-\frac {b^{3}}{c^{2}}-\frac {2 c^{2}}{b}-\frac {2 a^{2}}{b}+\frac {4 b a}{c}-\frac {4 a c}{b}+2 b \right ) \textit {\_R} -\frac {i a}{b}+\frac {i b}{c}-\frac {i c}{b}\right )\) | \(367\) |
2*a*(-(b*(-4*a*c+b^2)^(1/2)+4*a*c-b^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))+(-b*(-4*a*c+b^2)^(1/2)+4 *a*c-b^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)*a rctan((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)))
Leaf count of result is larger than twice the leaf count of optimal. 3495 vs. \(2 (187) = 374\).
Time = 0.51 (sec) , antiderivative size = 3495, normalized size of antiderivative = 15.81 \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
-1/4*sqrt(2)*sqrt(-(b^2 - 2*a*c - 2*c^2 + (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^ 2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^ 2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a *c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))*log(2*b^2*c*sin(x) + 4* b*c^2 + 2*(4*a*c^4 + (8*a^2 - b^2)*c^3 + 2*(2*a^3 - 3*a*b^2)*c^2 - (a^2*b^ 2 - b^4)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2) *c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*sin(x) - sqrt(2)*((a^2*b^4 - b^6 + 8*a*c^5 + 2* (12*a^2 - b^2)*c^4 + 6*(4*a^3 - 3*a*b^2)*c^3 + (8*a^4 - 22*a^2*b^2 + 3*b^4 )*c^2 - 2*(3*a^3*b^2 - 4*a*b^4)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4 *a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2*b ^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c))*cos(x) - (b^4 - 4*a*b^2* c)*cos(x))*sqrt(-(b^2 - 2*a*c - 2*c^2 + (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 - b^2)*c^4 - 12*(2*a^3 - a*b^2)*c^3 - 2*(8*a^4 - 11*a^2* b^2 + b^4)*c^2 - 4*(a^5 - 3*a^3*b^2 + 2*a*b^4)*c)))/(a^2*b^2 - b^4 - 4*a*c ^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c))) + 1/4*sqrt(2)*sqrt(-(b^2 - 2*a*c - 2*c^2 - (a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)*sqrt(b^2/(a^4*b^2 - 2*a^2*b^4 + b^6 - 4*a*c^5 - (16*a^2 ...
Timed out. \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {1}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \]
Timed out. \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 24.61 (sec) , antiderivative size = 5064, normalized size of antiderivative = 22.91 \[ \int \frac {1}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \]
atan(((-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^ 2 - 6*a*b^2*c)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2) *((-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2*c)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4* c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2)*(ta n(x/2)*(64*a*b^3 - 256*a^2*b*c) - 128*a^3*c + (-(8*a*c^3 + b*(-(4*a*c - b^ 2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2*c)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^3*b^2* c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2)*(tan(x/2)*(96*a*b^4 + 256*a^4*c - 64*a^3*b^2 + 512*a^2*c^3 + 768*a^3*c^2 - 128*a*b^2*c^2 - 576*a^2*b^2*c) + 32*a^2*b^3 + 128*a^2*b*c^2 - 32*a*b^3*c - 128*a^3*b*c) + 32*a^2*b^2 - 128 *a^2*c^2 + 32*a*b^2*c) + tan(x/2)*(128*a*c^2 - 32*a*b^2 + 64*a^2*c) + 32*a *b*c)*1i + (-(8*a*c^3 + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b ^2*c^2 - 6*a*b^2*c)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c ^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^ (1/2)*(tan(x/2)*(128*a*c^2 - 32*a*b^2 + 64*a^2*c) - (-(8*a*c^3 + b*(-(4*a* c - b^2)^3)^(1/2) + b^4 + 8*a^2*c^2 - 2*b^2*c^2 - 6*a*b^2*c)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^ 3*b^2*c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2)*(tan(x/2)*(64*a*b^3 - 25...