\(\int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [443]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 272 \[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {\left (2 a A b+2 a^2 B+b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{d \sqrt {a+b \sec (c+d x)}}+\frac {b (2 A b+3 a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{d \sqrt {a+b \sec (c+d x)}}+\frac {(2 a A-b B) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {b B \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{d} \] Output:

(2*A*a*b+2*B*a^2+B*b^2)*((b+a*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2 
*d*x+1/2*c,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/d/(a+b*sec(d*x+c))^(1 
/2)+b*(2*A*b+3*B*a)*((b+a*cos(d*x+c))/(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+ 
1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/d/(a+b*sec(d*x+c))^(1/2 
)+(2*A*a-B*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*s 
ec(d*x+c))^(1/2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)+b*B*sec 
(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.18 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {(a+b \sec (c+d x))^{3/2} \left (\frac {8 a (2 A b+a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{(b+a \cos (c+d x))^2}+\frac {2 \left (2 a^2 A+4 A b^2+5 a b B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{(b+a \cos (c+d x))^2}+\frac {2 i (2 a A-b B) \sqrt {-\frac {a (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {a (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 b (a+b) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )+a \operatorname {EllipticPi}\left (1-\frac {a}{b},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )\right )\right )}{a \sqrt {\frac {1}{a-b}} b (b+a \cos (c+d x))^{3/2}}+\frac {4 b B \tan (c+d x)}{b+a \cos (c+d x)}\right )}{4 d \sec ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[((a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d 
*x]],x]
 

Output:

((a + b*Sec[c + d*x])^(3/2)*((8*a*(2*A*b + a*B)*Sqrt[(b + a*Cos[c + d*x])/ 
(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(b + a*Cos[c + d*x])^2 + ( 
2*(2*a^2*A + 4*A*b^2 + 5*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti 
cPi[2, (c + d*x)/2, (2*a)/(a + b)])/(b + a*Cos[c + d*x])^2 + ((2*I)*(2*a*A 
 - b*B)*Sqrt[-((a*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(a*(1 + Cos[c + d*x] 
))/(a - b)]*Csc[c + d*x]*(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(- 
1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*(2*b*EllipticF[I*ArcS 
inh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*El 
lipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], 
(-a + b)/(a + b)])))/(a*Sqrt[(a - b)^(-1)]*b*(b + a*Cos[c + d*x])^(3/2)) + 
 (4*b*B*Tan[c + d*x])/(b + a*Cos[c + d*x])))/(4*d*Sec[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 2.84 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.03, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.657, Rules used = {3042, 4514, 27, 3042, 4596, 3042, 4346, 3042, 3286, 3042, 3284, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}

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.

\(\displaystyle \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4514

\(\displaystyle \int \frac {b (2 A b+3 a B) \sec ^2(c+d x)+2 a (2 A b+a B) \sec (c+d x)+a (2 a A-b B)}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {b (2 A b+3 a B) \sec ^2(c+d x)+2 a (2 A b+a B) \sec (c+d x)+a (2 a A-b B)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {b (2 A b+3 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )+a (2 a A-b B)}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4596

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4346

\(\displaystyle \frac {1}{2} \left (\int \frac {a (2 a A-b B)+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}}dx}{\sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\int \frac {a (2 a A-b B)+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {1}{2} \left (\int \frac {a (2 a A-b B)+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\int \frac {a (2 a A-b B)+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {1}{2} \left (\int \frac {a (2 a A-b B)+2 a (2 A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4523

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx+(2 a A-b B) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(2 a A-b B) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4343

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(2 a A-b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(2 a A-b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(2 a A-b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {(2 a A-b B) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{2} \left (\left (2 a^2 B+2 a A b+b^2 B\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4345

\(\displaystyle \frac {1}{2} \left (\frac {\left (2 a^2 B+2 a A b+b^2 B\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {\left (2 a^2 B+2 a A b+b^2 B\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{2} \left (\frac {\left (2 a^2 B+2 a A b+b^2 B\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {\left (2 a^2 B+2 a A b+b^2 B\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{2} \left (\frac {2 \left (2 a^2 B+2 a A b+b^2 B\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 (2 a A-b B) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 b (3 a B+2 A b) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}\right )+\frac {b B \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}\)

Input:

Int[((a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x 
]
 

Output:

((2*(2*a*A*b + 2*a^2*B + b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti 
cF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a + b*Sec[c + d 
*x]]) + (2*b*(2*A*b + 3*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi 
[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a + b*Sec[c + 
d*x]]) + (2*(2*a*A - b*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b 
*Sec[c + d*x]])/(d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]])) 
/2 + (b*B*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d
 

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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3132
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 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 4343
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S 
qrt[b + a*Sin[e + f*x]])   Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4345
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S 
qrt[a + b*Csc[e + f*x]])   Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[ 
{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4346
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)], x_Symbol] :> Simp[d*Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x 
]]/Sqrt[a + b*Csc[e + f*x]])   Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f*x]] 
), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4514
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(m + n)   Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n* 
Simp[a^2*A*(m + n) + a*b*B*n + (a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1) 
)*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2, x], x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 
- b^2, 0] && GtQ[m, 1] &&  !(IGtQ[n, 1] &&  !IntegerQ[m])
 

rule 4523
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d 
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Simp[A/a   I 
nt[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Simp[(A*b - a*B) 
/(a*d)   Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ 
[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
 

rule 4596
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)]), x_Symbol] :> Simp[C/d^2   Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*C 
sc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[ 
a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - 
 b^2, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 20.29 (sec) , antiderivative size = 1389, normalized size of antiderivative = 5.11

method result size
default \(\text {Expression too large to display}\) \(1389\)
parts \(\text {Expression too large to display}\) \(1458\)

Input:

int((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/d/((a-b)/(a+b))^(1/2)*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c 
)+b*cos(d*x+c)+b)/sec(d*x+c)^(1/2)*((-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*A*(1/ 
(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2* 
EllipticPi(((a-b)/(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(a+b)/(a-b),I/((a- 
b)/(a+b))^(1/2))+(-6*cos(d*x+c)^2-12*cos(d*x+c)-6)*B*(1/(1+cos(d*x+c)))^(1 
/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticPi(((a-b)/ 
(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))+( 
-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*c 
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticE(((a-b)/(a+b))^(1/2)*(-csc(d 
*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*A* 
(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a 
*b*EllipticE(((a-b)/(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^( 
1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b 
+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticE(((a-b)/(a+b))^(1/2)*(-c 
sc(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1) 
*B*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 
)*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b) 
)^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*A*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+ 
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticF(((a-b)/(a+b))^(1/2 
)*(-csc(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-4*cos(d*x+c)^2-8*cos...
 

Fricas [F(-1)]

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

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \] Input:

integrate((a+b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)
 

Output:

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**(3/2)/sqrt(sec(c + d*x 
)), x)
 

Maxima [F]

\[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)/sqrt(sec(d*x + c 
)), x)
 

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algo 
rithm="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)/sqrt(sec(d*x + c 
)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(1/ 
2),x)
 

Output:

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(1/ 
2), x)
 

Reduce [F]

\[ \int \frac {(a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right ) b +a}}{\sec \left (d x +c \right )}d x \right ) a^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )d x \right ) b^{2}+2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right ) b +a}d x \right ) a b \] Input:

int((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x)
 

Output:

int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a))/sec(c + d*x),x)*a**2 + i 
nt(sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a)*sec(c + d*x),x)*b**2 + 2*in 
t(sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a),x)*a*b