\(\int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [537]

Optimal result

Integrand size = 23, antiderivative size = 345 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {5 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2+15 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^3 d \sqrt {a+b \cos (c+d x)}}-\frac {b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {5 b \tan (c+d x)}{4 a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}} \] Output:

1/4*b*(7*a^2-15*b^2)*(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))/a^3/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-5/4 
*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))/a^2/d/(a+b*cos(d*x+c))^(1/2)+1/4*(4*a^2+15*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 
))/a^3/d/(a+b*cos(d*x+c))^(1/2)-1/4*b^2*(7*a^2-15*b^2)*sin(d*x+c)/a^3/(a^2 
-b^2)/d/(a+b*cos(d*x+c))^(1/2)-5/4*b*tan(d*x+c)/a^2/d/(a+b*cos(d*x+c))^(1/ 
2)+1/2*sec(d*x+c)*tan(d*x+c)/a/d/(a+b*cos(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.70 (sec) , antiderivative size = 597, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {2 \left (4 a^3 b-20 a b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^4+29 a^2 b^2-45 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (7 a^2 b^2-15 b^4\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}}{16 a^3 (-a+b) (a+b) d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {2 b^4 \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {7 b \tan (c+d x)}{4 a^3}+\frac {\sec (c+d x) \tan (c+d x)}{2 a^2}\right )}{d} \] Input:

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

Output:

-1/16*((2*(4*a^3*b - 20*a*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elliptic 
F[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^4 + 29*a 
^2*b^2 - 45*b^4)*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]] - ((2*I)*(7*a^2*b^2 - 15*b^4 
)*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)]*Sqr 
t[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[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*Cos[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))) 
/(a^3*(-a + b)*(a + b)*d) + (Sqrt[a + b*Cos[c + d*x]]*((2*b^4*Sin[c + d*x] 
)/(a^3*(a^2 - b^2)*(a + b*Cos[c + d*x])) - (7*b*Tan[c + d*x])/(4*a^3) + (S 
ec[c + d*x]*Tan[c + d*x])/(2*a^2)))/d
 

Rubi [A] (verified)

Time = 3.02 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.08, number of steps used = 24, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.043, Rules used = {3042, 3281, 27, 3042, 3534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\int -\frac {\left (4 a^2+6 b \cos (c+d x) a+15 b^2-5 b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{2 (a+b \cos (c+d x))^{3/2}}dx}{a}+\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {\left (4 a^2+6 b \cos (c+d x) a+15 b^2-5 b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {4 a^2+6 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 b^2-5 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \int \frac {\left (4 a^4+11 b^2 a^2+2 b \left (a^2-5 b^2\right ) \cos (c+d x) a-15 b^4+b^2 \left (7 a^2-15 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\int \frac {\left (4 a^4+11 b^2 a^2+2 b \left (a^2-5 b^2\right ) \cos (c+d x) a-15 b^4+b^2 \left (7 a^2-15 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\int \frac {4 a^4+11 b^2 a^2+2 b \left (a^2-5 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-15 b^4+b^2 \left (7 a^2-15 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {b \left (7 a^2-15 b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {b \left (7 a^2-15 b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx+\frac {\int \frac {\left (b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {b \left (7 a^2-15 b^2\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\int \frac {b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\int \frac {b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {\int \frac {b \left (4 a^4+11 b^2 a^2-15 b^4\right )-5 a b^2 \left (a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-5 a b^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a b^2 \left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {5 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {5 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {10 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {\frac {b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {10 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {5 b \tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\frac {2 b \left (7 a^2-15 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\frac {2 b \left (4 a^4+11 a^2 b^2-15 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {10 a b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}}{a \left (a^2-b^2\right )}-\frac {2 b^2 \left (7 a^2-15 b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{2 a}}{4 a}\)

Input:

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

Output:

(Sec[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + b*Cos[c + d*x]]) - (-1/2*(((2* 
b*(7*a^2 - 15*b^2)*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)]) + ((-10*a*b^2*(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*b*(4*a^4 + 11*a^2*b^2 - 15*b^4)*Sqrt[(a 
+ b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*S 
qrt[a + b*Cos[c + d*x]]))/b)/(a*(a^2 - b^2)) - (2*b^2*(7*a^2 - 15*b^2)*Sin 
[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/a + (5*b*Tan[c + d* 
x])/(a*d*Sqrt[a + b*Cos[c + d*x]]))/(4*a)
 

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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 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. \(1545\) vs. \(2(329)=658\).

Time = 9.04 (sec) , antiderivative size = 1546, normalized size of antiderivative = 4.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1546\)

Input:

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

Output:

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/a*(-1/2* 
cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2 
)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*si 
n(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c) 
^2-1)-1/8*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b) 
/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2) 
*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+ 
1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c), 
(-2*b/(a-b))^(1/2))-3/8*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2 
*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d 
*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(- 
2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos( 
1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a 
+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b 
))^(1/2))*b^2)-2*b^2/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/ 
2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2* 
c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-2*b/a^2...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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