3.3.37 \(\int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx\) [237]

3.3.37.1 Optimal result
3.3.37.2 Mathematica [C] (warning: unable to verify)
3.3.37.3 Rubi [A] (warning: unable to verify)
3.3.37.4 Maple [A] (verified)
3.3.37.5 Fricas [F(-1)]
3.3.37.6 Sympy [F]
3.3.37.7 Maxima [F]
3.3.37.8 Giac [F]
3.3.37.9 Mupad [F(-1)]

3.3.37.1 Optimal result

Integrand size = 25, antiderivative size = 370 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}-\frac {b \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\sqrt {a} \left (a^2-b^2\right )^{3/4} d \sqrt {e}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a d \sqrt {e \sin (c+d x)}}+\frac {b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}} \]

output
-b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/(a^2-b^2)^ 
(3/4)/d/a^(1/2)/e^(1/2)-b*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^( 
1/4)/e^(1/2))/(a^2-b^2)^(3/4)/d/a^(1/2)/e^(1/2)-2*(sin(1/2*c+1/4*Pi+1/2*d* 
x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x), 
2^(1/2))*sin(d*x+c)^(1/2)/a/d/(e*sin(d*x+c))^(1/2)-b^2*(sin(1/2*c+1/4*Pi+1 
/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2 
*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a/d/(a^2-b^2-a*(a^ 
2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-b^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2) 
/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a+(a^ 
2-b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/a/d/(a^2-b^2+a*(a^2-b^2)^(1/2))/(e 
*sin(d*x+c))^(1/2)
 
3.3.37.2 Mathematica [C] (warning: unable to verify)

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

Time = 5.37 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\frac {2 \left (b+a \sqrt {\cos ^2(c+d x)}\right ) \sqrt {\sin (c+d x)} \left (\frac {b \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {a} \left (-a^2+b^2\right )^{3/4}}-\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\sin (c+d x)}}{\left (-a^2+b^2+a^2 \sin ^2(c+d x)\right ) \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right )}\right )}{d (b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}} \]

input
Integrate[1/((a + b*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]
 
output
(2*(b + a*Sqrt[Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]]*((b*(-2*ArcTan[1 - (Sqr 
t[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[ 
2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] 
- Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]] 
+ Log[Sqrt[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d 
*x]] + a*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(3/4)) - (5*a*(a^ 
2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/ 
(a^2 - b^2)]*Sqrt[Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]])/((-a^2 + b^2 + a^2*S 
in[c + d*x]^2)*(5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, 
(a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4, -1/2, 2, 9/4, S 
in[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[5 
/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c 
+ d*x]^2))))/(d*(b + a*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])
 
3.3.37.3 Rubi [A] (warning: unable to verify)

Time = 1.67 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4360, 25, 25, 3042, 25, 3346, 3042, 3121, 3042, 3120, 3181, 25, 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.

\(\displaystyle \int \frac {1}{\sqrt {e \sin (c+d x)} (a+b \sec (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3181

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

input
Int[1/((a + b*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]]),x]
 
output
(2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a*d*Sqrt[e*Sin[c 
+ d*x]]) - (b*((2*a*e*(ArcTan[(Sqrt[a]*Sqrt[e]*Sin[c + d*x])/(a^2 - b^2)^( 
1/4)]/(2*Sqrt[a]*(a^2 - b^2)^(3/4)*e^(3/2)) + ArcTanh[(Sqrt[a]*Sqrt[e]*Sin 
[c + d*x])/(a^2 - b^2)^(1/4)]/(2*Sqrt[a]*(a^2 - b^2)^(3/4)*e^(3/2))))/d + 
(b*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin 
[c + d*x]])/(Sqrt[a^2 - b^2]*(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]]) 
 - (b*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[ 
Sin[c + d*x]])/(Sqrt[a^2 - b^2]*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x 
]])))/a
 

3.3.37.3.1 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 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 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]
 

rule 4360
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 
3.3.37.4 Maple [A] (verified)

Time = 7.54 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.34

method result size
default \(\frac {\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )\right )}{-2 a^{2} e^{2}+2 b^{2} e^{2}}-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \left (2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {a^{2}-b^{2}}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+\sqrt {a^{2}-b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{\sqrt {a^{2}-b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b^{2}+\sqrt {a^{2}-b^{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) b^{2}+\operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, -\frac {a}{\sqrt {a^{2}-b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}-\operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {a}{a +\sqrt {a^{2}-b^{2}}}, \frac {\sqrt {2}}{2}\right ) a \,b^{2}\right )}{2 a \sqrt {a^{2}-b^{2}}\, \left (\sqrt {a^{2}-b^{2}}-a \right ) \left (a +\sqrt {a^{2}-b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) \(494\)

input
int(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
(1/2*b*e*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*(ln(((e*sin(d*x+c))^ 
(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d*x+c))^(1/2)-(e^2*(a^2-b^2)/a^2) 
^(1/4)))+2*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4)))-1/2/a*( 
-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*(2*(a^2-b^2)^ 
(3/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*(a^2-b^2)^(1/2)*Ellip 
ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+(a^2-b^2)^(1/2)*EllipticPi((-s 
in(d*x+c)+1)^(1/2),-a/((a^2-b^2)^(1/2)-a),1/2*2^(1/2))*b^2+(a^2-b^2)^(1/2) 
*EllipticPi((-sin(d*x+c)+1)^(1/2),a/(a+(a^2-b^2)^(1/2)),1/2*2^(1/2))*b^2+E 
llipticPi((-sin(d*x+c)+1)^(1/2),-a/((a^2-b^2)^(1/2)-a),1/2*2^(1/2))*a*b^2- 
EllipticPi((-sin(d*x+c)+1)^(1/2),a/(a+(a^2-b^2)^(1/2)),1/2*2^(1/2))*a*b^2) 
/(a^2-b^2)^(1/2)/((a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin 
(d*x+c))^(1/2))/d
 
3.3.37.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]

input
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="fricas")
 
output
Timed out
 
3.3.37.6 Sympy [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]

input
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))**(1/2),x)
 
output
Integral(1/(sqrt(e*sin(c + d*x))*(a + b*sec(c + d*x))), x)
 
3.3.37.7 Maxima [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]

input
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")
 
output
integrate(1/((b*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))), x)
 
3.3.37.8 Giac [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]

input
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="giac")
 
output
integrate(1/((b*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))), x)
 
3.3.37.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{\sqrt {e\,\sin \left (c+d\,x\right )}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]

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