\(\int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1395]

Optimal result

Integrand size = 31, antiderivative size = 413 \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{\left (-a^2+b^2\right )^{5/4} f g^{3/2}}+\frac {2 b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt {\cos (e+f x)}}+\frac {a^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {a^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{\left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 (a-b \sin (e+f x))}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}} \] Output:

-a*b^(1/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/( 
-a^2+b^2)^(5/4)/f/g^(3/2)+a*b^(1/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/( 
-a^2+b^2)^(1/4)/g^(1/2))/(-a^2+b^2)^(5/4)/f/g^(3/2)+2*b*(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)+ 
a^2*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))/(a^2-b^2)/(b-(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)+a^2*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))/(a^2-b^2)/(b+(-a^2+b^2)^(1/2))/f/g/(g*cos(f*x+e))^(1/2)+2*(a-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 = 15.11 (sec) , antiderivative size = 783, normalized size of antiderivative = 1.90 \[ \int \frac {\sin (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {2 \cos (e+f x) (a-b \sin (e+f x))}{\left (a^2-b^2\right ) f (g \cos (e+f x))^{3/2}}+\frac {b \cos ^{\frac {3}{2}}(e+f x) \left (-\frac {4 a \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \sin ^2(e+f x)}{12 \sqrt {b} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{(a-b) (a+b) f (g \cos (e+f x))^{3/2}} \] Input:

Integrate[Sin[e + f*x]/((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 
)) + (b*Cos[e + f*x]^(3/2)*((-4*a*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*App 
ellF1[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)*S 
qrt[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])) - ((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*Ar 
cTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - Log[Sqr 
t[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Co 
s[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*Sqrt[b]*(-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 = 1.88 (sec) , antiderivative size = 388, normalized size of antiderivative = 0.94, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.581, Rules used = {3042, 3345, 27, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3180

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((2*b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x] 
]) + a*b*((2*b*g*(ArcTan[(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*Sqrt[g*Cos[e + f*x]])))/((a^2 - b^2)*g^2) + (2*(a - b 
*Sin[e + f*x]))/((a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]])
 

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 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 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 3345

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co 
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 
Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 
 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 
 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], 
x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[p, -1] && IntegerQ[2*m]
 

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]
 

Maple [C] (warning: unable to verify)

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

Time = 4.29 (sec) , antiderivative size = 1209, normalized size of antiderivative = 2.93

method	result	size
default	\(\text {Expression too large to display}\)	\(1209\)

Input:

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

Output:

(-4/g*a*(-1/8/(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)+1/8/(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)-1/16/(a-b)/(a+b)/(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 
/4*(32*cos(1/2*f*x+1/2*e)^3*(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e) 
^2))^(1/2)*a^2*b^2+16*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)^2) 
^(1/2)*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)*a^2*b^2-32*cos(1/2*f*x+1/2*e)*(-g*(2*sin( 
1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*a^2*b^2-sum((2*_alpha^2*b^2- 
a^2)/_alpha/(2*_alpha^2-1)*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(si 
n(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(cos( 
1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha* 
(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*EllipticPi(c 
os(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),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*g*(4*_alpha^2-3)/(4*a^2-3*b^2...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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