\(\int (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \, dx\) [499]

Optimal result

Integrand size = 23, antiderivative size = 255 \[ \int (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \, dx=-\frac {5 b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {7 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2+3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {5 b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

-5/4*b*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b 
))^(1/2))/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+7/4*a*b*((a+b*cos(d*x+c))/(a+b) 
)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos( 
d*x+c))^(1/2)+1/4*(4*a^2+3*b^2)*((a+b*cos(d*x+c))/(a+b))^(1/2)*EllipticPi( 
sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+5/4 
*b*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d+1/2*a*(a+b*cos(d*x+c))^(1/2)*sec(d* 
x+c)*tan(d*x+c)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.67 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.99 \[ \int (a+b \cos (c+d x))^{3/2} \sec ^3(c+d x) \, dx=\frac {\frac {8 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {10 i b^2 \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}}{16 d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {5}{4} b \tan (c+d x)+\frac {1}{2} a \sec (c+d x) \tan (c+d x)\right )}{d} \] Input:

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

Output:

((8*a*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a 
 + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^2 + b^2)*Sqrt[(a + b*Cos[c + d* 
x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + 
 d*x]] + ((10*I)*b^2*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[ 
c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt 
[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*Ellip 
ticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - 
 b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Co 
s[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt 
[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b 
*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b 
*Cos[c + d*x])^2)))/(16*d) + (Sqrt[a + b*Cos[c + d*x]]*((5*b*Tan[c + d*x]) 
/4 + (a*Sec[c + d*x]*Tan[c + d*x])/2))/d
 

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.07, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.913, Rules used = {3042, 3278, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3278

\(\displaystyle \frac {1}{2} \int \frac {\left (a b \cos ^2(c+d x)+2 \left (a^2+2 b^2\right ) \cos (c+d x)+5 a b\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\left (a b \cos ^2(c+d x)+2 \left (a^2+2 b^2\right ) \cos (c+d x)+5 a b\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {a b \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a b}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\left (2 b \cos (c+d x) a^2-5 b^2 \cos ^2(c+d x) a+\left (4 a^2+3 b^2\right ) a\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\left (2 b \cos (c+d x) a^2-5 b^2 \cos ^2(c+d x) a+\left (4 a^2+3 b^2\right ) a\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a^2-5 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+\left (4 a^2+3 b^2\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int -\frac {\left (7 a^2 \cos (c+d x) b^2+a \left (4 a^2+3 b^2\right ) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-5 a b \int \sqrt {a+b \cos (c+d x)}dx}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {\left (7 a^2 \cos (c+d x) b^2+a \left (4 a^2+3 b^2\right ) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-5 a b \int \sqrt {a+b \cos (c+d x)}dx}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {7 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (4 a^2+3 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-5 a b \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {7 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (4 a^2+3 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {5 a b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {7 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (4 a^2+3 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {5 a b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {7 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+a \left (4 a^2+3 b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {1}{4} \left (\frac {\frac {7 a^2 b^2 \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+a b \left (4 a^2+3 b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {7 a^2 b^2 \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b \left (4 a^2+3 b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2+3 b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {7 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2+3 b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {7 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2+3 b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {14 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {a b \left (4 a^2+3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {14 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {a b \left (4 a^2+3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {14 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {14 a^2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (4 a^2+3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {10 a b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {5 b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

Input:

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

Output:

(a*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((-10*a*b* 
Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a 
 + b*Cos[c + d*x])/(a + b)]) + ((14*a^2*b^2*Sqrt[(a + b*Cos[c + d*x])/(a + 
 b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + 
 (2*a*b*(4*a^2 + 3*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, ( 
c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/(2*a) + (5*b* 
Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/d)/4
 

Defintions of rubi rules used

rule 25

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

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 3278

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*(a + b*Si 
n[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 - b^2))), 
 x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + 
d*Sin[e + f*x])^(n - 2)*Simp[c*(a*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) 
+ (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d*(b*c - a 
*d)*(m + n + 1)*Sin[e + f*x]^2, 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] && LtQ[m, -1 
] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]
 

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 3481

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

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(979\) vs. \(2(243)=486\).

Time = 6.62 (sec) , antiderivative size = 980, normalized size of antiderivative = 3.84

method	result	size
default	\(\text {Expression too large to display}\)	\(980\)

Input:

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

Output:

-1/4*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-40*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^2+(28*a*b+40*b^2)*sin(1/2*d*x+1/2*c) 
^4*cos(1/2*d*x+1/2*c)+(-4*a^2-14*a*b-10*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2* 
d*x+1/2*c)+4*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2* 
d*x+1/2*c)^2)^(1/2)*(7*b*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))* 
a-5*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+5*b^2*EllipticE(c 
os(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4*EllipticPi(cos(1/2*d*x+1/2*c),2,(- 
2*b/(a-b))^(1/2))*a^2-3*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2) 
)*b^2)*sin(1/2*d*x+1/2*c)^4-4*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b) 
)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(7*b*EllipticF(cos(1/2*d*x+1/2*c),(-2 
*b/(a-b))^(1/2))*a-5*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+ 
5*b^2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4*EllipticPi(cos(1/ 
2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2-3*EllipticPi(cos(1/2*d*x+1/2*c),2,( 
-2*b/(a-b))^(1/2))*b^2)*sin(1/2*d*x+1/2*c)^2+7*b*(sin(1/2*d*x+1/2*c)^2)^(1 
/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2* 
d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b 
)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),( 
-2*b/(a-b))^(1/2))*a+5*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/ 
2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b) 
)^(1/2))-4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**3,x)*b + int(sqrt( 
cos(c + d*x)*b + a)*sec(c + d*x)**3,x)*a