Integrand size = 17, antiderivative size = 226 \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\sqrt {2} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}} \] Output:
2^(1/2)*(1-b/(-4*a*c+b^2)^(1/2))*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))/(b^2-2*c*(a+ c)-b*(-4*a*c+b^2)^(1/2))^(1/2)+2^(1/2)*(1+b/(-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))/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 1.53 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.19 \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\frac {\left (i b+\sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {\left (-i b+\sqrt {-b^2+4 a c}\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}}{\sqrt {-\frac {b^2}{2}+2 a c}} \] Input:
Integrate[Sin[x]/(a + b*Sin[x] + c*Sin[x]^2),x]
Output:
(((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]] + (((-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 = 0.72 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, 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.
\(\displaystyle \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{a+b \sin (x)+c \sin (x)^2}dx\) |
\(\Big \downarrow \) 3737 |
\(\displaystyle \int \left (\frac {1-\frac {b}{\sqrt {b^2-4 a c}}}{-\sqrt {b^2-4 a c}+b+2 c \sin (x)}+\frac {\frac {b}{\sqrt {b^2-4 a c}}+1}{\sqrt {b^2-4 a c}+b+2 c \sin (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {2} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\sqrt {2} \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\) |
Input:
Int[Sin[x]/(a + b*Sin[x] + c*Sin[x]^2),x]
Output:
(Sqrt[2]*(1 - b/Sqrt[b^2 - 4*a*c])*ArcTan[(2*c + (b - Sqrt[b^2 - 4*a*c])*T an[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]])])/Sqrt[b^ 2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]] + (Sqrt[2]*(1 + 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]])])/Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4* a*c]]
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 = 0.98 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.96
method | result | size |
default | \(4 a \left (\frac {2 \sqrt {-4 a c +b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {2 \sqrt {-4 a c +b^{2}}\, \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (8 a c -2 b^{2}\right ) \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )\) | \(217\) |
risch | \(-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{4} c^{2}-8 a^{3} b^{2} c +32 a^{3} c^{3}+b^{4} a^{2}-32 a^{2} b^{2} c^{2}+16 a^{2} c^{4}+10 a \,b^{4} c -8 a \,b^{2} c^{3}-b^{6}+b^{4} c^{2}\right ) \textit {\_Z}^{4}+\left (32 a^{3} c -8 a^{2} b^{2}+32 a^{2} c^{2}-24 a \,b^{2} c +4 b^{4}\right ) \textit {\_Z}^{2}+16 a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i x}+\left (-\frac {i a^{3}}{2}+\frac {i a^{2} b^{2}}{8 c}-\frac {i a^{2} c}{2}+\frac {5 i b^{2} a}{8}+\frac {i a \,c^{2}}{2}-\frac {i b^{4}}{8 c}-\frac {5 i b^{2} c}{8}+\frac {i c^{3}}{2}+\frac {i b^{4}}{8 a}-\frac {i c^{2} b^{2}}{8 a}\right ) \textit {\_R}^{3}+\left (\frac {i a^{3}}{b}-\frac {i a^{2} b}{4 c}+\frac {2 i a^{2} c}{b}-\frac {3 i a b}{2}+\frac {i a \,c^{2}}{b}+\frac {i b^{3}}{4 c}-\frac {i b c}{4}\right ) \textit {\_R}^{2}+\left (-\frac {i a^{2}}{2 c}+i a +\frac {3 i c}{2}-\frac {i b^{2}}{2 a}\right ) \textit {\_R} +\frac {i a^{2}}{b c}+\frac {i a}{b}\right )\right )}{2}\) | \(344\) |
Input:
int(sin(x)/(a+b*sin(x)+c*sin(x)^2),x,method=_RETURNVERBOSE)
Output:
4*a*(2*(-4*a*c+b^2)^(1/2)/(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*(-4*a*c+b^2)^(1/2)/(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)))
Leaf count of result is larger than twice the leaf count of optimal. 3519 vs. \(2 (192) = 384\).
Time = 0.38 (sec) , antiderivative size = 3519, normalized size of antiderivative = 15.57 \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(sin(x)/(a+b*sin(x)+c*sin(x)**2),x)
Output:
Timed out
\[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {\sin \left (x\right )}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \] Input:
integrate(sin(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="maxima")
Output:
integrate(sin(x)/(c*sin(x)^2 + b*sin(x) + a), x)
Timed out. \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(sin(x)/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")
Output:
Timed out
Time = 44.47 (sec) , antiderivative size = 5048, normalized size of antiderivative = 22.34 \[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:
int(sin(x)/(a + c*sin(x)^2 + b*sin(x)),x)
Output:
atan(-((((8*a^3*c + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 - 2*a^2*b^2 + 8*a^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)*(256*a^3*c - 64*a^2*b^2 + 256*a^2*c^2 - 64*a*b^2*c) - 32*a*b^3 + ((8*a^3*c + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 - 2*a^2*b^2 + 8*a^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)*(96*a*b^4 + 256*a^4*c - 64*a^3*b^2 + 512*a^2*c^3 + 768*a^3*c^2 - 12 8*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 - 1 28*a^3*b*c) + 128*a^2*b*c) - tan(x/2)*(128*a^2*c - 64*a*b^2 + 64*a^3) + 32 *a^2*b)*((8*a^3*c + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 - 2*a^2*b^2 + 8*a^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 )*1i - (((8*a^3*c + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 - 2*a^2*b^2 + 8*a^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)*(256*a^3*c - 64*a^2*b^2 + 256*a^2*c^2 - 64*a*b^2*c) - 32*a*b^3 - ((8*a^3*c + b*(-(4*a*c - b^2)^3)^(1/2) + b^4 - 2*a^2*b^2 + 8*a^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)*...
\[ \int \frac {\sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int \frac {\sin \left (x \right )}{\sin \left (x \right )^{2} c +\sin \left (x \right ) b +a}d x \] Input:
int(sin(x)/(a+b*sin(x)+c*sin(x)^2),x)
Output:
int(sin(x)/(sin(x)**2*c + sin(x)*b + a),x)