Integrand size = 33, antiderivative size = 369 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^2 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \sqrt [4]{-a^2+b^2} f}-\frac {a^2 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \sqrt [4]{-a^2+b^2} f}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}-\frac {2 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \sqrt {\cos (e+f x)}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]
-2/3*(g*cos(f*x+e))^(3/2)/b/f/g+a^2*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(- a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(5/2)/(-a^2+b^2)^(1/4)/f-a^2*arctanh(b^( 1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(5/2)/(-a^2+ b^2)^(1/4)/f+a^3*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ellipti cPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/ b^3/f/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+a^3*g*(cos(1/2*f*x+1/2*e)^ 2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2 )^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^3/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e ))^(1/2)-2*a*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin (1/2*f*x+1/2*e),2^(1/2))*(g*cos(f*x+e))^(1/2)/b^2/f/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 12.01 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {\sqrt {g \cos (e+f x)} \left (-8 b^{3/2} \cos ^{\frac {3}{2}}(e+f x)-\frac {a \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{\left (a^2-b^2\right ) (a+b \sin (e+f x))}\right )}{12 b^{5/2} f \sqrt {\cos (e+f x)}} \]
(Sqrt[g*Cos[e + f*x]]*(-8*b^(3/2)*Cos[e + f*x]^(3/2) - (a*(8*b^(5/2)*Appel lF1[3/4, -1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]* Cos[e + f*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]* Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqr t[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2 ]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] + Log[Sqr t[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Co s[e + f*x]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/((a^2 - b^2)*(a + b*Sin[e + f* x]))))/(12*b^(5/2)*f*Sqrt[Cos[e + f*x]])
Time = 1.04 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 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.
| \(\displaystyle \int \frac {\sin ^2(e+f x) \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^2 \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {a^2 \sqrt {g \cos (e+f x)}}{b^2 (a+b \sin (e+f x))}-\frac {a \sqrt {g \cos (e+f x)}}{b^2}+\frac {\sin (e+f x) \sqrt {g \cos (e+f x)}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} f \sqrt [4]{b^2-a^2}}-\frac {a^2 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} f \sqrt [4]{b^2-a^2}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^3 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^3 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^2 f \sqrt {\cos (e+f x)}}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}\) |
(a^2*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqr t[g])])/(b^(5/2)*(-a^2 + b^2)^(1/4)*f) - (a^2*Sqrt[g]*ArcTanh[(Sqrt[b]*Sqr t[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(5/2)*(-a^2 + b^2)^(1 /4)*f) - (2*(g*Cos[e + f*x])^(3/2))/(3*b*f*g) - (2*a*Sqrt[g*Cos[e + f*x]]* EllipticE[(e + f*x)/2, 2])/(b^2*f*Sqrt[Cos[e + f*x]]) + (a^3*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^3*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (a^3*g*Sqrt[Cos[e + f*x]]*El lipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^3*(b + Sqrt[-a^ 2 + b^2])*f*Sqrt[g*Cos[e + f*x]])
3.14.72.3.1 Defintions of rubi rules used
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.49 (sec) , antiderivative size = 1002, normalized size of antiderivative = 2.72
(16*b*g*(-1/24/b^2*(-2*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2 *e)^2+(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2))/g+1/64*a^2/b^4/(g^2*(a^2-b^2)/b ^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^(1/4 )*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/(2 *g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e )^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*g*co s(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2) ^(1/4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2*(a^2-b^2 )/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))))-1/8*(g*(2*cos(1/2*f*x+1/2*e)^2- 1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a*g*(16*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*c os(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*b^2+sum(1 /_alpha*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2) ^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b ^2/a^2*(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e) ^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),- 4*b^2/a^2*(_alpha^2-1),2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a ^2*2^(1/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2* cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g*(2 *_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f* x+1/2*e)^2))^(1/2))*(-g*sin(1/2*f*x+1/2*e)^2*(2*sin(1/2*f*x+1/2*e)^2-1)...
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]