\(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx\) [579]

Optimal result

Integrand size = 25, antiderivative size = 434 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d e^2 \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{\left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d e^2 \sqrt {e \cos (c+d x)}}-\frac {2 (b-a \sin (c+d x))}{3 \left (a^2-b^2\right ) d e (e \cos (c+d x))^{3/2}} \] Output:

-b^(5/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a 
^2+b^2)^(7/4)/d/e^(5/2)-b^(5/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2 
+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(7/4)/d/e^(5/2)+2/3*a*cos(d*x+c)^(1/2)*Inv 
erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/(a^2-b^2)/d/e^2/(e*cos(d*x+c))^(1/2)-a 
*b^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))/(a^2-b^2)/(a^2-b*(b-(-a^2+b^2)^(1/2)))/d/e^2/(e*cos(d*x+c))^(1 
/2)-a*b^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))/(a^2-b^2)/(a^2-b*(b+(-a^2+b^2)^(1/2)))/d/e^2/(e*cos(d*x+c 
))^(1/2)-2/3*(b-a*sin(d*x+c))/(a^2-b^2)/d/e/(e*cos(d*x+c))^(3/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 23.15 (sec) , antiderivative size = 1192, normalized size of antiderivative = 2.75 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx =\text {Too large to display} \] Input:

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])),x]
 

Output:

(2*Cos[c + d*x]*(-b + a*Sin[c + d*x]))/(3*(a^2 - b^2)*d*(e*Cos[c + d*x])^( 
5/2)) + (Cos[c + d*x]^(5/2)*((-2*(a^2 - 3*b^2)*(a + b*Sqrt[1 - Cos[c + d*x 
]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Co 
s[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, 3/2, 1, 9/4 
, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 
 + b^2*(-1 + Cos[c + d*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + 
I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I) 
*Sqrt[b]*Sqrt[Cos[c + d*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[c + d*x]] + I*b*Cos[c + d*x]] 
- Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d 
*x]] + I*b*Cos[c + d*x]]))/(-a^2 + b^2)^(3/4))*Sin[c + d*x])/(Sqrt[1 - Cos 
[c + d*x]^2]*(a + b*Sin[c + d*x])) - (2*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2 
])*((5*b*(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)]*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, ...
 

Rubi [A] (warning: unable to verify)

Time = 1.86 (sec) , antiderivative size = 414, normalized size of antiderivative = 0.95, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3175, 27, 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.

\(\displaystyle \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3175

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3181

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

Int[1/((e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])),x]
 

Output:

((2*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x] 
]) - 3*b^2*((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]*Sqrt[e]*C 
os[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]])) 
)/(3*(a^2 - b^2)*e^2) - (2*(b - a*Sin[c + d*x]))/(3*(a^2 - b^2)*d*e*(e*Cos 
[c + d*x])^(3/2))
 

Defintions of rubi rules used

rule 27

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

rule 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 3175

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^ 
(m + 1)*((b - a*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* 
(a^2*(p + 2) - b^2*(m + p + 2) + a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegersQ 
[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. \(1019\) vs. \(2(385)=770\).

Time = 4.05 (sec) , antiderivative size = 1020, normalized size of antiderivative = 2.35

method	result	size
default	\(\text {Expression too large to display}\)	\(1020\)

Input:

int(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

(4/e^2*b*(-1/24*2^(1/2)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-2^ 
(1/2)+cos(1/2*d*x+1/2*c))/e/(2*2^(1/2)*cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/ 
2*c)^2-3)-1/24*2^(1/2)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(2^(1 
/2)+cos(1/2*d*x+1/2*c))/e/(2*2^(1/2)*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2* 
c)^2+3)-b^2/(a-b)/(a+b)*(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*co 
s(1/2*d*x+1/2*c)^2-e)^(1/2)+1))/(16*a^2-16*b^2)/e)-2*(e*(2*cos(1/2*d*x+1/2 
*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a/e^2*(1/(a^2-b^2)*(-1/6*cos(1/2*d*x+ 
1/2*c)/e*(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2 
*d*x+1/2*c)^2-1/2)^2+1/3*(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)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-1/16/(a-b)/(a+b)*sum(1/_alpha/(2*_alph 
a^2-1)*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*e*(4* 
_alpha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*b^2*cos(1/2*d*x+1/ 
2*c)^2+_alpha^2*b^2-3*a^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2 
)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))+8*b^2...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c)),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cos(d*x+c))**(5/2)/(a+b*sin(d*x+c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}} \,d x } \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c)),x, algorithm="maxima")
 

Output:

integrate(1/((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)), x)
 

Giac [F]

\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}} \,d x } \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c)),x, algorithm="giac")
 

Output:

integrate(1/((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \,d x \] Input:

int(1/((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))),x)
 

Output:

int(1/((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))), x)
 

Reduce [F]

\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) b +\cos \left (d x +c \right )^{3} a}d x \right )}{e^{3}} \] Input:

int(1/(e*cos(d*x+c))^(5/2)/(a+b*sin(d*x+c)),x)
 

Output:

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