\(\int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1392]

Optimal result

Integrand size = 33, antiderivative size = 584 \[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {a^4 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{5/2} \left (-a^2+b^2\right )^{5/4} f g^{3/2}}-\frac {2 b}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b \left (a^2-b^2\right ) f g^3}-\frac {2 b (g \cos (e+f x))^{3/2}}{3 \left (a^2-b^2\right ) f g^3}-\frac {4 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}-\frac {a^5 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^3 \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a \sin (e+f x)}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \] Output:

a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(5/2)/ 
(-a^2+b^2)^(5/4)/f/g^(3/2)-a^4*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+ 
b^2)^(1/4)/g^(1/2))/b^(5/2)/(-a^2+b^2)^(5/4)/f/g^(3/2)-2*b/(a^2-b^2)/f/g/( 
g*cos(f*x+e))^(1/2)+2/3*a^2*(g*cos(f*x+e))^(3/2)/b/(a^2-b^2)/f/g^3-2/3*b*( 
g*cos(f*x+e))^(3/2)/(a^2-b^2)/f/g^3-4*a*(g*cos(f*x+e))^(1/2)*EllipticE(sin 
(1/2*f*x+1/2*e),2^(1/2))/(a^2-b^2)/f/g^2/cos(f*x+e)^(1/2)+2*a^3*(g*cos(f*x 
+e))^(1/2)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^2/(a^2-b^2)/f/g^2/cos(f 
*x+e)^(1/2)-a^5*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^3/(a^2-b^2)/(b-(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e 
))^(1/2)-a^5*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^3/(a^2-b^2)/(b+(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^ 
(1/2)+2*a*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 = 26.49 (sec) , antiderivative size = 820, normalized size of antiderivative = 1.40 \[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\cos ^2(e+f x) \left (\frac {2 \cos (e+f x)}{3 b}+\frac {2 \sec (e+f x) (-b+a \sin (e+f x))}{a^2-b^2}\right )}{f (g \cos (e+f x))^{3/2}}+\frac {a \cos ^{\frac {3}{2}}(e+f x) \left (\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (a^2-2 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{(a-b) b (a+b) f (g \cos (e+f x))^{3/2}} \] Input:

Integrate[Sin[e + f*x]^4/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(Cos[e + f*x]^2*((2*Cos[e + f*x])/(3*b) + (2*Sec[e + f*x]*(-b + a*Sin[e + 
f*x]))/(a^2 - b^2)))/(f*(g*Cos[e + f*x])^(3/2)) + (a*Cos[e + f*x]^(3/2)*(( 
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]]))/(Sq 
rt[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 - (S 
qrt[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)*b*(a + b)*f*(g*Cos[e + f*x])^(3/ 
2))
 

Rubi [A] (verified)

Time = 2.26 (sec) , antiderivative size = 523, normalized size of antiderivative = 0.90, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3381, 3042, 3045, 27, 244, 2009, 3046, 3042, 3121, 3042, 3119, 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.

\(\displaystyle \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^4}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}dx\)

\(\Big \downarrow \) 3381

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (\frac {2 \sin (e+f x)}{f g \sqrt {g \cos (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\cos (e+f x)}}\right )}{a^2-b^2}+\frac {b \left (-\frac {2 g^2}{\sqrt {g \cos (e+f x)}}-\frac {2}{3} (g \cos (e+f x))^{3/2}\right )}{f g^3 \left (a^2-b^2\right )}-\frac {a^2 \left (\frac {a^2 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} f \sqrt [4]{b^2-a^2}}-\frac {a^2 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} f \sqrt [4]{b^2-a^2}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^3 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^3 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^3 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^2 f \sqrt {\cos (e+f x)}}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}\right )}{g^2 \left (a^2-b^2\right )}\)

Input:

Int[Sin[e + f*x]^4/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(b*((-2*g^2)/Sqrt[g*Cos[e + f*x]] - (2*(g*Cos[e + f*x])^(3/2))/3))/((a^2 - 
 b^2)*f*g^3) - (a^2*((a^2*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/(( 
-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(5/2)*(-a^2 + b^2)^(1/4)*f) - (a^2*Sqrt[g] 
*ArcTanh[(Sqrt[b]*Sqrt[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]]*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^2 - b^2)*g^2) 
 + (a*((-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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 3377

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])
 

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 = 5.67 (sec) , antiderivative size = 1290, normalized size of antiderivative = 2.21

method	result	size
default	\(\text {Expression too large to display}\)	\(1290\)

Input:

int(sin(f*x+e)^4/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
                                                                                    
                                                                                    
 

Output:

(64*b/g*(-1/128/(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)+1/128/(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)+1/96/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-1/256*a^4/(a-b)/(a+b)/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)/(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)))-1/8*(-32*cos(1/2*f*x+1/2*e) 
*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2* 
e)^2*b^4-16*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2) 
*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(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)*(-1+2*sin(1/ 
2*f*x+1/2*e)^2)^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^( 
1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*b^4+sin(1/2*f*x+1/2*e)^2*g*(-1+ 
2*sin(1/2*f*x+1/2*e)^2)*sum(1/_alpha*(8*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-1+2 
*sin(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^ 
2+4*b^2)/a^2,2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*_alpha^3...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**4/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

integrate(sin(f*x + e)^4/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Giac [F]

\[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

integrate(sin(f*x + e)^4/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int(sin(e + f*x)^4/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x)
 

Output:

int(sin(e + f*x)^4/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {\sin ^4(e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \left (\int \frac {\sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )^{4}}{\cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) b +\cos \left (f x +e \right )^{2} a}d x \right )}{g^{2}} \] Input:

int(sin(f*x+e)^4/(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)**4)/(cos(e + f*x)**2*sin(e + 
 f*x)*b + cos(e + f*x)**2*a),x))/g**2