Integrand size = 33, antiderivative size = 500 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^4 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {a^4 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}-\frac {4 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g} \]
-2/3*a^2*(g*cos(f*x+e))^(3/2)/b^3/f/g-2/3*(g*cos(f*x+e))^(3/2)/b/f/g+2/7*( g*cos(f*x+e))^(7/2)/b/f/g^3+2/5*a*(g*cos(f*x+e))^(3/2)*sin(f*x+e)/b^2/f/g+ a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/ b^(9/2)/(-a^2+b^2)^(1/4)/f-a^4*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+ b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(9/2)/(-a^2+b^2)^(1/4)/f+a^5*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^5/f/(b-(-a^2+b^2)^(1/2))/(g*co s(f*x+e))^(1/2)+a^5*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Elli pticPi(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^5/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-2*a^3*(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*co s(f*x+e))^(1/2)/b^4/f/cos(f*x+e)^(1/2)-4/5*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 = 20.70 (sec) , antiderivative size = 816, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \sqrt {g \cos (e+f x)} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \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 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (5 a^2+2 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \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 ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {\sqrt {g \cos (e+f x)} \left (-\frac {\left (28 a^2+19 b^2\right ) \cos (e+f x)}{42 b^3}+\frac {\cos (3 (e+f x))}{14 b}+\frac {a \sin (2 (e+f x))}{5 b^2}\right )}{f} \]
-1/5*(a*Sqrt[g*Cos[e + f*x]]*((-4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a *AppellF1[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*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x] ] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x])/( Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) - ((5*a^2 + 2*b^2)*(a + b*S qrt[1 - Cos[e + f*x]^2])*(8*b^(5/2)*AppellF1[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]*Sqrt[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[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]]))*Sin[e + f*x]^2)/(12 *b^(3/2)*(-a^2 + b^2)*(1 - Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(b^3*f* Sqrt[Cos[e + f*x]]) + (Sqrt[g*Cos[e + f*x]]*(-1/42*((28*a^2 + 19*b^2)*Cos[ e + f*x])/b^3 + Cos[3*(e + f*x)]/(14*b) + (a*Sin[2*(e + f*x)])/(5*b^2)))/f
Time = 1.31 (sec) , antiderivative size = 500, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x) \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4 \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {a^4 \sqrt {g \cos (e+f x)}}{b^4 (a+b \sin (e+f x))}-\frac {a^3 \sqrt {g \cos (e+f x)}}{b^4}+\frac {a^2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{b^3}-\frac {a \sin ^2(e+f x) \sqrt {g \cos (e+f x)}}{b^2}+\frac {\sin ^3(e+f x) \sqrt {g \cos (e+f x)}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^3 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^4 f \sqrt {\cos (e+f x)}}-\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {a^4 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac {a^4 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}+\frac {2 a \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f g}-\frac {4 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}\) |
(a^4*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqr t[g])])/(b^(9/2)*(-a^2 + b^2)^(1/4)*f) - (a^4*Sqrt[g]*ArcTanh[(Sqrt[b]*Sqr t[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*(-a^2 + b^2)^(1 /4)*f) - (2*a^2*(g*Cos[e + f*x])^(3/2))/(3*b^3*f*g) - (2*(g*Cos[e + f*x])^ (3/2))/(3*b*f*g) + (2*(g*Cos[e + f*x])^(7/2))/(7*b*f*g^3) - (2*a^3*Sqrt[g* Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^4*f*Sqrt[Cos[e + f*x]]) - (4*a *Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x ]]) + (a^5*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), ( e + f*x)/2, 2])/(b^5*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (a^5 *g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2 , 2])/(b^5*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (2*a*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(5*b^2*f*g)
3.14.70.3.1 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 = 4.66 (sec) , antiderivative size = 1405, normalized size of antiderivative = 2.81
(1/4/b^5*g*a^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)))+1/2/b^5*g*a^4/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*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))+1/2/b^5*g*a^4/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*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))+2/b^3*a^2*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+16/7/b*cos(1 /2*f*x+1/2*e)^6*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-24/7/b*cos(1/2*f*x+1/2* e)^4*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+8/21/b*cos(1/2*f*x+1/2*e)^2*(2*g*c os(1/2*f*x+1/2*e)^2-g)^(1/2)+8/21/b*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-4/3 /b^3*cos(1/2*f*x+1/2*e)^2*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*a^2-4/3/b^3*( 2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*a^2-1/40*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*s in(1/2*f*x+1/2*e)^2)^(1/2)*a*g*(-128*b^4*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/ 2*e)^6+128*b^4*sin(1/2*f*x+1/2*e)^4*cos(1/2*f*x+1/2*e)-32*b^4*sin(1/2*f*x+ 1/2*e)^2*cos(1/2*f*x+1/2*e)+80*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x +1/2*e)^2-1)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*a^2*b^2+32*(sin(1 /2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticE(cos(1/2* f*x+1/2*e),2^(1/2))*b^4+5*a^2*sum(1/_alpha*(8*(g*(2*_alpha^2*b^2+a^2-2*...
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{4}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{4}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]