\(\int \frac {\sin ^4(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) [1385]

Optimal result

Integrand size = 33, antiderivative size = 509 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {a^4 \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {2 a^2 \sqrt {g \cos (e+f x)}}{b^3 f g}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}+\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g^3}-\frac {2 a^3 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {4 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^5 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f g} \] Output:

-a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2) 
/(-a^2+b^2)^(3/4)/f/g^(1/2)-a^4*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2 
+b^2)^(1/4)/g^(1/2))/b^(7/2)/(-a^2+b^2)^(3/4)/f/g^(1/2)-2*a^2*(g*cos(f*x+e 
))^(1/2)/b^3/f/g-2*(g*cos(f*x+e))^(1/2)/b/f/g+2/5*(g*cos(f*x+e))^(5/2)/b/f 
/g^3-2*a^3*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^4/f/( 
g*cos(f*x+e))^(1/2)-4/3*a*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2 
^(1/2))/b^2/f/(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^4/(a^2-b*(b-(-a^2+b^2)^(1/ 
2)))/f/(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^4/(a^2-b*(b+(-a^2+b^2)^(1/2)))/f/ 
(g*cos(f*x+e))^(1/2)+2/3*a*(g*cos(f*x+e))^(1/2)*sin(f*x+e)/b^2/f/g
 

Mathematica [C] (warning: unable to verify)

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

Time = 26.77 (sec) , antiderivative size = 1953, normalized size of antiderivative = 3.84 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[e + f*x]*(Cos[2*(e + f*x)]/(5*b) + (2*a*Sin[e + f*x])/(3*b^2)))/(f*Sq 
rt[g*Cos[e + f*x]]) - (Sqrt[Cos[e + f*x]]*((-2*(10*a^2 - 27*b^2)*(a + b*Sq 
rt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e 
 + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 
- Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2 
, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, 
 Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF 
1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Co 
s[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2 
*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*A 
rcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[S 
qrt[-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]]))/(-a^2 + b^2)^(3/4))*Sin[e + f 
*x])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((30*a^2 + 27*b^2)* 
(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 + 
b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/ 
(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ((1 
+ I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2 
)^(3/4)) + (4*Sqrt[Cos[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, C...
 

Rubi [F]

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

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

\(\Big \downarrow \) 3388

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3045

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a \int \frac {\sin (e+f x)^3}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3388

\(\displaystyle -\frac {a \left (\frac {\int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin ^2(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3048

\(\displaystyle -\frac {a \left (\frac {\frac {2}{3} \int \frac {1}{\sqrt {g \cos (e+f x)}}dx-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\frac {2}{3} \int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {a \left (\frac {\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\frac {2 \sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \int \frac {\sin (e+f x)^2}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3388

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (\frac {\int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3045

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}-\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}}d(g \cos (e+f x))}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \int \frac {\sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3346

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\int \frac {1}{\sqrt {g \cos (e+f x)}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\int \frac {1}{\sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {\sqrt {\cos (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{b \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 3181

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {b g \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b^2 \cos ^2(e+f x) g^2+\left (a^2-b^2\right ) g^2\right )}d(g \cos (e+f x))}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \int \frac {1}{b^2 g^4 \cos ^4(e+f x)+\left (a^2-b^2\right ) g^2}d\sqrt {g \cos (e+f x)}}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b g^2 \cos ^2(e+f x)+\sqrt {b^2-a^2} g}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a \left (\frac {\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 f \sqrt {g \cos (e+f x)}}-\frac {2 \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 f g}}{b}-\frac {a \left (-\frac {a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b f \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 b g \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} g-b g^2 \cos ^2(e+f x)}d\sqrt {g \cos (e+f x)}}{2 g \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {g} \cos (e+f x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} g^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{f}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {b^2-a^2}-b \cos (e+f x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {g \cos (e+f x)} \left (b \cos (e+f x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 \sqrt {g \cos (e+f x)}}{b f g}\right )}{b}\right )}{b}-\frac {2 g^2 \sqrt {g \cos (e+f x)}-\frac {2}{5} (g \cos (e+f x))^{5/2}}{b f g^3}\)

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

Input:

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

Output:

$Aborted
 

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 27

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

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 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 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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 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 3048

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

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 3181

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q)   Int[1/( 
Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f)   Subst[ 
Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S 
imp[a/(2*q)   Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / 
; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 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 3388

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1158\) vs. \(2(449)=898\).

Time = 4.66 (sec) , antiderivative size = 1159, normalized size of antiderivative = 2.28

method	result	size
default	\(\text {Expression too large to display}\)	\(1159\)

Input:

int(sin(f*x+e)^4/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

(64*b*(1/160/b^4/g*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(4*sin(1/2*f*x+1/2* 
e)^4*b^2-4*sin(1/2*f*x+1/2*e)^2*b^2-5*a^2-4*b^2)-1/16*a^4/b^4*(g^2*(a^2-b^ 
2)/b^2)^(1/4)*2^(1/2)*(-ln((-(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)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2) 
+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)- 
2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+g))-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))/(16 
*a^2-16*b^2)/g)-1/24*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2)^ 
(1/2)*a*(64*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)-48*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-1+2*sin(1/2 
*f*x+1/2*e)^2)^(1/2)*EllipticF(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)*EllipticF(cos(1 
/2*f*x+1/2*e),2^(1/2))*b^4+3*a^2*sum(1/_alpha/(2*_alpha^2-1)*(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*b^2-8*b^2*_alpha*(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)-a^2*2^(1/2)* 
arctanh(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)/(g...
 

Fricas [F(-1)]

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

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate(sin(f*x+e)**4/(g*cos(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

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

integrate(sin(f*x+e)^4/(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm= 
"giac")
 

Output:

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

Mupad [F(-1)]

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

int(sin(e + f*x)^4/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))),x)
                                                                                    
                                                                                    
 

Output:

int(sin(e + f*x)^4/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))), x)
 

Reduce [F]

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

int(sin(f*x+e)^4/(g*cos(f*x+e))^(1/2)/(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)*sin(e + f* 
x)*b + cos(e + f*x)*a),x))/g