\(\int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx\) [585]

Optimal result

Integrand size = 25, antiderivative size = 404 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=-\frac {a e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))} \] Output:

-1/2*a*e^(3/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2 
))/b^(3/2)/(-a^2+b^2)^(3/4)/d-1/2*a*e^(3/2)*arctanh(b^(1/2)*(e*cos(d*x+c)) 
^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d-e^2*cos(d*x+c) 
^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^2/d/(e*cos(d*x+c))^(1/2)+1 
/2*a^2*e^2*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a^2+b^2 
)^(1/2)),2^(1/2))/b^2/(a^2-b*(b-(-a^2+b^2)^(1/2)))/d/(e*cos(d*x+c))^(1/2)+ 
1/2*a^2*e^2*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^ 
2)^(1/2)),2^(1/2))/b^2/(a^2-b*(b+(-a^2+b^2)^(1/2)))/d/(e*cos(d*x+c))^(1/2) 
-e*(e*cos(d*x+c))^(1/2)/b/d/(a+b*sin(d*x+c))
 

Mathematica [C] (warning: unable to verify)

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

Time = 12.49 (sec) , antiderivative size = 614, normalized size of antiderivative = 1.52 \[ \int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx=-\frac {(e \cos (c+d x))^{3/2} \sec (c+d x)}{b d (a+b \sin (c+d x))}+\frac {(e \cos (c+d x))^{3/2} \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^2(c+d x)}}{\left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(c+d x)\right )\right )}+\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{b d \cos ^{\frac {3}{2}}(c+d x) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))} \] Input:

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

Output:

-(((e*Cos[c + d*x])^(3/2)*Sec[c + d*x])/(b*d*(a + b*Sin[c + d*x]))) + ((e* 
Cos[c + d*x])^(3/2)*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*b*(a^2 - b^2)*App 
ellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2) 
]*Sqrt[Cos[c + d*x]]*Sqrt[1 - Cos[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1 
/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*( 
2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a 
^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[c + d*x]^2, (b^2*C 
os[c + d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x 
]^2))) + (a*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2 
)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*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[c 
 + d*x]] + b*Cos[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - 
b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 
- b^2)^(3/4)))*Sin[c + d*x]^2)/(b*d*Cos[c + d*x]^(3/2)*(1 - Cos[c + d*x]^2 
)*(a + b*Sin[c + d*x]))
 

Rubi [A] (warning: unable to verify)

Time = 1.72 (sec) , antiderivative size = 398, normalized size of antiderivative = 0.99, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 3172, 3042, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3172

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{2 b}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 3181

\(\displaystyle -\frac {e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {a \left (\frac {b e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b^2 \cos ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \cos (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

\(\displaystyle -\frac {e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {a \left (\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \cos ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{2 b}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}\)

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

-1/2*(e^2*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(b*d*Sqrt[e*Co 
s[c + d*x]]) - (a*((2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e]*Cos[c + d*x])/(-a^ 
2 + b^2)^(1/4)]/(Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2)) - ArcTanh[(Sqrt[b]*Sq 
rt[e]*Cos[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3 
/2))))/d + (a*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), 
(c + d*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + 
 d*x]]) - (a*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), ( 
c + d*x)/2, 2])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + 
d*x]])))/b))/b - (e*Sqrt[e*Cos[c + d*x]])/(b*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

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 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 3042

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

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 3172

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre 
eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I 
ntegersQ[2*m, 2*p]
 

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 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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1765\) vs. \(2(351)=702\).

Time = 6.06 (sec) , antiderivative size = 1766, normalized size of antiderivative = 4.37

method	result	size
default	\(\text {Expression too large to display}\)	\(1766\)

Input:

int((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

(8*e^2*a*b*(1/b^2*(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((-(e^2*(a^2-b^2)/b 
^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)-2*e*cos(1/2*d*x+1/2*c 
)^2-(e^2*(a^2-b^2)/b^2)^(1/2)+e)/((e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d 
*x+1/2*c)^2-e)^(1/2)*2^(1/2)-2*e*cos(1/2*d*x+1/2*c)^2-(e^2*(a^2-b^2)/b^2)^ 
(1/2)+e))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2* 
c)^2-e)^(1/2)+1)-2*arctan(-2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2* 
d*x+1/2*c)^2-e)^(1/2)+1))/(16*a^2-16*b^2)/e-1/64*(a^2-b^2)/b^2*(6*(arctan( 
(2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^ 
2*(a^2-b^2)/b^2)^(1/4))+arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2) 
+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+1/2*ln((2*e*cos(1/2 
*d*x+1/2*c)^2-e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/ 
2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^ 
2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2 
)/b^2)^(1/2))))*(4*b^2*cos(1/2*d*x+1/2*c)^4-4*b^2*cos(1/2*d*x+1/2*c)^2+a^2 
)*2^(1/2)*(e^2*(a^2-b^2)/b^2)^(1/4)+(8*a^2-8*b^2)*(2*e*cos(1/2*d*x+1/2*c)^ 
2-e)^(1/2))/e/(a-b)^2/(a+b)^2/(4*b^2*cos(1/2*d*x+1/2*c)^4-4*b^2*cos(1/2*d* 
x+1/2*c)^2+a^2))-2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*e^2*(-1/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1 
/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos 
(1/2*d*x+1/2*c),2^(1/2))+1/16*(3*a^2-b^2)/b^4*sum(1/_alpha/(2*_alpha^2-...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((e*cos(d*x+c))**(3/2)/(a+b*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*e*( - 2*sqrt(cos(c + d*x)) - int((sqrt(cos(c + d*x))*sin(c + d*x) 
)/(cos(c + d*x)*sin(c + d*x)*b + cos(c + d*x)*a),x)*sin(c + d*x)*b*d - int 
((sqrt(cos(c + d*x))*sin(c + d*x))/(cos(c + d*x)*sin(c + d*x)*b + cos(c + 
d*x)*a),x)*a*d))/(2*b*d*(sin(c + d*x)*b + a))