\(\int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx\) [71]

Optimal result

Integrand size = 25, antiderivative size = 404 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=-\frac {3 a e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {3 a^2 e^3 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {3 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{b^2 d \sqrt {\sin (c+d x)}}+\frac {e (e \sin (c+d x))^{3/2}}{b d (a+b \cos (c+d x))} \] Output:

-3/2*a*e^(5/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2 
))/b^(5/2)/(-a^2+b^2)^(1/4)/d+3/2*a*e^(5/2)*arctanh(b^(1/2)*(e*sin(d*x+c)) 
^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^(1/4)/d-3/2*a^2*e^3*El 
lipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d 
*x+c)^(1/2)/b^3/(b-(-a^2+b^2)^(1/2))/d/(e*sin(d*x+c))^(1/2)-3/2*a^2*e^3*El 
lipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d 
*x+c)^(1/2)/b^3/(b+(-a^2+b^2)^(1/2))/d/(e*sin(d*x+c))^(1/2)+3*e^2*Elliptic 
E(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/b^2/d/sin(d*x+c) 
^(1/2)+e*(e*sin(d*x+c))^(3/2)/b/d/(a+b*cos(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 = 10.20 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.91 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {(e \sin (c+d x))^{5/2} \left (8 b^{3/2} \csc (c+d x)+\frac {\left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{\left (a^2-b^2\right ) \sin ^{\frac {5}{2}}(c+d x)}\right )}{8 b^{5/2} d (a+b \cos (c+d x))} \] Input:

Integrate[(e*Sin[c + d*x])^(5/2)/(a + b*Cos[c + d*x])^2,x]
 

Output:

((e*Sin[c + d*x])^(5/2)*(8*b^(3/2)*Csc[c + d*x] + ((a + b*Sqrt[Cos[c + d*x 
]^2])*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[S 
in[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[ 
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[Sin[c + d*x]] + b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + 
 Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]) + 
 8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2 
)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2)))/((a^2 - b^2)*Sin[c + d*x]^(5/2))))/(8 
*b^(5/2)*d*(a + b*Cos[c + d*x]))
 

Rubi [A] (warning: unable to verify)

Time = 1.66 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.97, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3172, 25, 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.

\(\displaystyle \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3172

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3180

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

Int[(e*Sin[c + d*x])^(5/2)/(a + b*Cos[c + d*x])^2,x]
 

Output:

(e*(e*Sin[c + d*x])^(3/2))/(b*d*(a + b*Cos[c + d*x])) - (3*e^2*((2*Ellipti 
cE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(b*d*Sqrt[Sin[c + d*x]]) - 
 (a*((-2*b*e*(ArcTan[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(2 
*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e]) - ArcTanh[(Sqrt[b]*Sqrt[e]*Sin[c + d* 
x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e])))/d + (a*e* 
EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c 
 + d*x]])/(b*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*e*Ellipti 
cPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x] 
])/(b*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]])))/b))/(2*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 3180

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

rule 3284

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

rule 3286

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

rule 3346

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1667\) vs. \(2(346)=692\).

Time = 15.64 (sec) , antiderivative size = 1668, normalized size of antiderivative = 4.13

method	result	size
default	\(\text {Expression too large to display}\)	\(1668\)

Input:

int((e*sin(d*x+c))^(5/2)/(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
 

Output:

(-2*e^3*a*b*(-1/2*(e*sin(d*x+c))^(3/2)/b^2/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2) 
+3/16/b^4/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((e*sin(d*x+c)-(e^2*(a^2-b^ 
2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*s 
in(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2 
-b^2)/b^2)^(1/2)))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c 
))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2 
)-1)))+1/4*e^3*a^2*(3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c 
))^(1/2)*sin(d*x+c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2 
)^(1/2)),1/2*2^(1/2))*b^2-3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2+2*sin 
(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),b/(b+(-a^2 
+b^2)^(1/2)),1/2*2^(1/2))*b^2-12*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/ 
2)*sin(d*x+c)^(5/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^3+6*(1-s 
in(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin( 
d*x+c))^(1/2),1/2*2^(1/2))*b^3+3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/ 
2)*sin(d*x+c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2 
)),1/2*2^(1/2))*b^3+3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+ 
c)^(5/2)*EllipticPi((1-sin(d*x+c))^(1/2),b/(b+(-a^2+b^2)^(1/2)),1/2*2^(1/2 
))*b^3+3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin( 
d*x+c)^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2* 
2^(1/2))*a^2-3*(-a^2+b^2)^(1/2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*sin(d*x+c))**(5/2)/(a+b*cos(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")
 

Output:

integrate((e*sin(d*x + c))^(5/2)/(b*cos(d*x + c) + a)^2, x)
 

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*sin(d*x + c))^(5/2)/(b*cos(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:

int((e*sin(c + d*x))^(5/2)/(a + b*cos(c + d*x))^2,x)
 

Output:

int((e*sin(c + d*x))^(5/2)/(a + b*cos(c + d*x))^2, x)
 

Reduce [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \right ) e^{2} \] Input:

int((e*sin(d*x+c))^(5/2)/(a+b*cos(d*x+c))^2,x)
 

Output:

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