\(\int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1393]

Optimal result

Integrand size = 33, antiderivative size = 509 \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {a^3 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {2 a}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 a^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b \left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {4 b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {a^4 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 b \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \] Output:

-a^3*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(3/2) 
/(-a^2+b^2)^(5/4)/f/g^(3/2)+a^3*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2 
+b^2)^(1/4)/g^(1/2))/b^(3/2)/(-a^2+b^2)^(5/4)/f/g^(3/2)+2*a/(a^2-b^2)/f/g/ 
(g*cos(f*x+e))^(1/2)-2*a^2*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2*f*x+1/2* 
e),2^(1/2))/b/(a^2-b^2)/f/g^2/cos(f*x+e)^(1/2)+4*b*(g*cos(f*x+e))^(1/2)*El 
lipticE(sin(1/2*f*x+1/2*e),2^(1/2))/(a^2-b^2)/f/g^2/cos(f*x+e)^(1/2)+a^4*c 
os(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^( 
1/2))/b^2/(a^2-b^2)/(b-(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)+a^4*cos( 
f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2 
))/b^2/(a^2-b^2)/(b+(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)-2*b*sin(f*x 
+e)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 26.38 (sec) , antiderivative size = 793, normalized size of antiderivative = 1.56 \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {2 \cos (e+f x) (a-b \sin (e+f x))}{\left (a^2-b^2\right ) f (g \cos (e+f x))^{3/2}}-\frac {\cos ^{\frac {3}{2}}(e+f x) \left (\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (a^2-2 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{(a-b) (a+b) f (g \cos (e+f x))^{3/2}} \] Input:

Integrate[Sin[e + f*x]^3/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(2*Cos[e + f*x]*(a - b*Sin[e + f*x]))/((a^2 - b^2)*f*(g*Cos[e + f*x])^(3/2 
)) - (Cos[e + f*x]^(3/2)*((4*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*Appe 
llF1[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)*Sq 
rt[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]] + L 
og[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])) - ((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]))))/((a - b)*(a + 
b)*f*(g*Cos[e + f*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 2.67 (sec) , antiderivative size = 469, normalized size of antiderivative = 0.92, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {3042, 3381, 3042, 3045, 15, 3046, 3042, 3121, 3042, 3119, 3346, 3042, 3121, 3042, 3119, 3180, 266, 827, 218, 221, 3042, 3286, 3042, 3284}

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.

\(\displaystyle \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^3}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3381

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \int \frac {\sin ^2(e+f x)}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}+\frac {a \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}+\frac {a \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {a \int \frac {1}{(g \cos (e+f x))^{3/2}}d(g \cos (e+f x))}{f g \left (a^2-b^2\right )}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \int \frac {\sin (e+f x)^2}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3046

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {2 \int \sqrt {g \cos (e+f x)}dx}{g^2}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {2 \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{g^2}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {a^2 \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)}dx}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3346

\(\displaystyle -\frac {a^2 \left (\frac {\int \sqrt {g \cos (e+f x)}dx}{b}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \left (\frac {\int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {a^2 \left (\frac {\sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{b \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \left (\frac {\sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)}dx}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3180

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {b g \int \frac {\sqrt {g \cos (e+f x)}}{b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2}d(g \cos (e+f x))}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \int \frac {g^2 \cos ^2(e+f x)}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 b}\right )}{f}-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b}+\frac {a g \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3286

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (-\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )} \left (b \sin \left (e+f x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {g \cos (e+f x)}}+\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {a^2 \left (\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{f}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}\right )}{b}\right )}{g^2 \left (a^2-b^2\right )}-\frac {b \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {2 a}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\)

Input:

Int[Sin[e + f*x]^3/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(2*a)/((a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]]) - (a^2*((2*Sqrt[g*Cos[e + f*x 
]]*EllipticE[(e + f*x)/2, 2])/(b*f*Sqrt[Cos[e + f*x]]) - (a*((2*b*g*(ArcTa 
n[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^ 
2)^(1/4)*Sqrt[g]) - ArcTanh[(Sqrt[b]*Sqrt[g]*Cos[e + f*x])/(-a^2 + b^2)^(1 
/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[g])))/f + (a*g*Sqrt[Cos[e + f*x]]* 
EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(b - Sqrt[-a^ 
2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (a*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2* 
b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b*(b + Sqrt[-a^2 + b^2])*f*Sq 
rt[g*Cos[e + f*x]])))/b))/((a^2 - b^2)*g^2) - (b*((-4*Sqrt[g*Cos[e + f*x]] 
*EllipticE[(e + f*x)/2, 2])/(f*g^2*Sqrt[Cos[e + f*x]]) + (2*Sin[e + f*x])/ 
(f*g*Sqrt[g*Cos[e + f*x]])))/(a^2 - b^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3046

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 
1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1)))   Int[(a*Sin[e + f 
*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && 
GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3180

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ 
)]), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[a*(g/(2*b))   Int[1/(S 
qrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Simp[a*(g/(2*b))   In 
t[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Simp[b*(g/f)   Su 
bst[Int[Sqrt[x]/(g^2*(a^2 - b^2) + b^2*x^2), x], x, g*Cos[e + f*x]], x])] / 
; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 3346

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* 
(x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b   Int 
[(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b   Int[(g*Cos[e + f*x])^p/( 
a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - 
 b^2, 0]
 

rule 3381

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a*(d^2/(a^2 
- b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ 
b*(d/(a^2 - b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] 
 - Simp[a^2*(d^2/(g^2*(a^2 - b^2)))   Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[ 
e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g 
}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1 
]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.16 (sec) , antiderivative size = 1102, normalized size of antiderivative = 2.17

method	result	size
default	\(\text {Expression too large to display}\)	\(1102\)

Input:

int(sin(f*x+e)^3/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

(16/g*a*(1/32/(a^2-b^2)*2^(1/2)/g/(cos(1/2*f*x+1/2*e)-1/2*2^(1/2))*(-2*g*s 
in(1/2*f*x+1/2*e)^2+g)^(1/2)-1/32/(a^2-b^2)*2^(1/2)/g/(cos(1/2*f*x+1/2*e)+ 
1/2*2^(1/2))*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+1/64*a^2/(a-b)/(a+b)/b^2/ 
(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*c 
os(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)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)-2* 
arctan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/ 
2)+1)))+32*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^(1/2)*b/g* 
(-1/8*(sin(1/2*f*x+1/2*e)^2-1)/(a^2-b^2)/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)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2) 
*(2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2-(-1+2*sin(1/2*f*x+1/2*e)^2)^(1 
/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2)))+1/ 
8*(sin(1/2*f*x+1/2*e)^2-1)/b^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)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2 
)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))-1/128*a^2/(a-b)/(a+b)/b^4*sum((-2* 
sin(1/2*f*x+1/2*e)^2*_alpha^2*b^2+sin(1/2*f*x+1/2*e)^2*a^2+2*b^2*_alpha^2- 
a^2)/_alpha/(2*_alpha^2-1)*(2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^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...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**3/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

integrate(sin(f*x + e)^3/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Giac [F]

\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(sin(f*x+e)^3/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

integrate(sin(f*x + e)^3/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int(sin(e + f*x)^3/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x)
 

Output:

int(sin(e + f*x)^3/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {\sin ^3(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )^{3}}{\cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{2} a}d x \right )}{g^{2}} \] Input:

int(sin(f*x+e)^3/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

(sqrt(g)*int((sqrt(cos(e + f*x))*sin(e + f*x)**3)/(cos(e + f*x)**2*sin(e + 
 f*x)*b + cos(e + f*x)**2*a),x))/g**2