Integrand size = 33, antiderivative size = 448 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a^3 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} \sqrt [4]{-a^2+b^2} f}+\frac {a^3 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} \sqrt [4]{-a^2+b^2} f}+\frac {2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}+\frac {2 a^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^3 f \sqrt {\cos (e+f x)}}+\frac {4 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b f \sqrt {\cos (e+f x)}}-\frac {a^4 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^4 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b f g} \]
2/3*a*(g*cos(f*x+e))^(3/2)/b^2/f/g-2/5*(g*cos(f*x+e))^(3/2)*sin(f*x+e)/b/f /g-a^3*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/ 2)/b^(7/2)/(-a^2+b^2)^(1/4)/f+a^3*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a ^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(7/2)/(-a^2+b^2)^(1/4)/f-a^4*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^4/f/(b-(-a^2+b^2)^(1/2))/(g *cos(f*x+e))^(1/2)-a^4*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*E llipticPi(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^4/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+2*a^2*(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^3/f/cos(f*x+e)^(1/2)+4/5*(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/f/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 20.81 (sec) , antiderivative size = 789, normalized size of antiderivative = 1.76 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {\sqrt {g \cos (e+f x)} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (5 a^2+2 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {\sqrt {g \cos (e+f x)} \left (\frac {2 a \cos (e+f x)}{3 b^2}-\frac {\sin (2 (e+f x))}{5 b}\right )}{f} \]
(Sqrt[g*Cos[e + f*x]]*((-4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Appell F1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Co s[e + f*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sq rt[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*Sqrt[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]))))/(5*b^2*f*Sqrt[ Cos[e + f*x]]) + (Sqrt[g*Cos[e + f*x]]*((2*a*Cos[e + f*x])/(3*b^2) - Sin[2 *(e + f*x)]/(5*b)))/f
Time = 1.14 (sec) , antiderivative size = 448, 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.
| \(\displaystyle \int \frac {\sin ^3(e+f x) \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3 \sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (-\frac {a^3 \sqrt {g \cos (e+f x)}}{b^3 (a+b \sin (e+f x))}+\frac {a^2 \sqrt {g \cos (e+f x)}}{b^3}-\frac {a \sin (e+f x) \sqrt {g \cos (e+f x)}}{b^2}+\frac {\sin ^2(e+f x) \sqrt {g \cos (e+f x)}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^3 f \sqrt {\cos (e+f x)}}-\frac {a^4 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^4 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac {a^3 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f \sqrt [4]{b^2-a^2}}+\frac {2 a (g \cos (e+f x))^{3/2}}{3 b^2 f g}-\frac {2 \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b f g}+\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b f \sqrt {\cos (e+f x)}}\) |
-((a^3*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*S qrt[g])])/(b^(7/2)*(-a^2 + b^2)^(1/4)*f)) + (a^3*Sqrt[g]*ArcTanh[(Sqrt[b]* Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(7/2)*(-a^2 + b^2) ^(1/4)*f) + (2*a*(g*Cos[e + f*x])^(3/2))/(3*b^2*f*g) + (2*a^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^3*f*Sqrt[Cos[e + f*x]]) + (4*Sqrt[g *Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(5*b*f*Sqrt[Cos[e + f*x]]) - (a^ 4*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/ 2, 2])/(b^4*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a^4*g*Sqrt[C os[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^ 4*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (2*(g*Cos[e + f*x])^(3/ 2)*Sin[e + f*x])/(5*b*f*g)
3.14.71.3.1 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 = 3.91 (sec) , antiderivative size = 1462, normalized size of antiderivative = 3.26
(-16*g*a*(-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*c os(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))))+1/12*(g*(2*cos(1/2*f*x+1/2*e)^ 2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g*(64*cos(1/2*f*x+1/2*e)^5*sin(1/2*f*x+1/ 2*e)^2*b^4-64*cos(1/2*f*x+1/2*e)^5*b^4-96*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x +1/2*e)^2*b^4-48*EllipticF(cos(1/2*f*x+1/2*e),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)*sin(1/2*f*x+1/2*e)^2*a^2*b^2+16 *EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(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)*sin(1/2*f*x+1/2*e)^2*b^4-48*EllipticE(cos(1/2*f *x+1/2*e),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)*sin(1/2*f*x+1/2*e)^2*b^4+96*cos(1/2*f*x+1/2*e)^3*b^4+32*b^4*sin(1/2 *f*x+1/2*e)^2*cos(1/2*f*x+1/2*e)+48*EllipticF(cos(1/2*f*x+1/2*e),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)*a^2*b^2-16*E llipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*...
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]