\(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [315]

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

Optimal result

Integrand size = 33, antiderivative size = 296 \[ \int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=-\frac {\left (3 a^2 A-6 A b^2-14 a b B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^3 A+12 a A b^2+4 a^2 b B+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 (5 A b+2 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 a A-2 b B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {a A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d} \] Output:

-1/3*(3*A*a^2-6*A*b^2-14*B*a*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)+1/3*(3 
*A*a^3+12*A*a*b^2+4*B*a^2*b+2*B*b^3)*((a+b*cos(d*x+c))/(a+b))^(1/2)*Invers 
eJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+ 
a^2*(5*A*b+2*B*a)*((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)-1/3*b*(3*A*a-2*B* 
b)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d+a*A*(a+b*cos(d*x+c))^(3/2)*tan(d*x+ 
c)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.91 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.49 \[ \int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {\frac {8 b \left (9 a A b+9 a^2 B+b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (27 a^2 A b+6 A b^3+12 a^3 B+14 a b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (-3 a^2 A+6 A b^2+14 a b B\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} \left (3 a^2 A+2 b^2 B \cos (c+d x)\right ) \tan (c+d x)}{12 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x 
]
 

Output:

((8*b*(9*a*A*b + 9*a^2*B + b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellip 
ticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(27*a^2*A* 
b + 6*A*b^3 + 12*a^3*B + 14*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*El 
lipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + ((2*I) 
*(-3*a^2*A + 6*A*b^2 + 14*a*b*B)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]* 
Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*Csc[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*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)])))/(a*b*Sqrt[-(a + b)^(-1)] 
) + 4*Sqrt[a + b*Cos[c + d*x]]*(3*a^2*A + 2*b^2*B*Cos[c + d*x])*Tan[c + d* 
x])/(12*d)
 

Rubi [A] (verified)

Time = 2.64 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 3468, 27, 3042, 3528, 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.

\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\)

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {\left (3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \cos ^2(c+d x)+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {\left (3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \cos ^2(c+d x)+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 A b+2 a B) a^2-b \left (3 A a^2-14 b B a-6 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (9 B a^2+9 A b a+b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-\left (\left (3 a^2 A-14 a b B-6 A b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {\int -\frac {\left (3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {\left (3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\left (3 a^2 A-14 a b B-6 A b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\left (3 a^2 A-14 a b B-6 A b^2\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 A b+2 a B) a^2+b \left (3 A a^3+4 b B a^2+12 A b^2 a+2 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a B+5 A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a B+5 A b) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {6 a^2 b (2 a B+5 A b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 A+4 a^2 b B+12 a A b^2+2 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 A-14 a b B-6 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a A-2 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\)

Input:

Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^2,x]
 

Output:

(((-2*(3*a^2*A - 6*A*b^2 - 14*a*b*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)]) + ((2*b* 
(3*a^3*A + 12*a*A*b^2 + 4*a^2*b*B + 2*b^3*B)*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]]) 
+ (6*a^2*b*(5*A*b + 2*a*B)*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)/3 - (2*b*( 
3*a*A - 2*b*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/2 + (a*A*(a + 
 b*Cos[c + d*x])^(3/2)*Tan[c + d*x])/d
 

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 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 3468
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c 
 + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 
1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] 
/; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

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 3528
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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*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] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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. \(1490\) vs. \(2(288)=576\).

Time = 25.99 (sec) , antiderivative size = 1491, normalized size of antiderivative = 5.04

method result size
parts \(\text {Expression too large to display}\) \(1491\)
default \(\text {Expression too large to display}\) \(1563\)

Input:

int((a+cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^2,x,method=_RETURNV 
ERBOSE)
 

Output:

-A*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2*b+(-2*a^3-2*a^2*b)*sin(1/2*d*x+1/2*c)^ 
2*cos(1/2*d*x+1/2*c)+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))*a^3-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-2*Ellipti 
cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+2*EllipticE(cos(1/2*d*x+1/ 
2*c),(-2*b/(a-b))^(1/2))*b^3+5*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b) 
)^(1/2))*a^2*b-EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-4*Elli 
pticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2)*sin(1/2*d*x+1/2*c)^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))*a^3+(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)*Elliptic 
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*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))*a*b^2-2*(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))*b^3-5*a^2*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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(- 
2*b/(a-b))^(1/2))+(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2...
 

Fricas [F(-1)]

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

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorith 
m="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((a+b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorith 
m="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^2, 
x)
 

Giac [F]

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

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorith 
m="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^2, 
x)
 

Mupad [F(-1)]

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

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^2,x)
 

Output:

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^2, x)
 

Reduce [F]

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

int((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^2,x)
 

Output:

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