\(\int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx\) [493]

Optimal result

Integrand size = 23, antiderivative size = 314 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx=\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{315 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{315 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \left (8 a^2+39 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{315 b^2 d}+\frac {2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac {8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac {2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d} \] Output:

2/315*(8*a^4+33*a^2*b^2+147*b^4)*(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))/b^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2 
/315*a*(8*a^4+31*a^2*b^2-39*b^4)*((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJac 
obiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/b^3/d/(a+b*cos(d*x+c))^(1/2)+ 
2/315*a*(8*a^2+39*b^2)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/d+2/315*(8*a^ 
2+49*b^2)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/d-8/63*a*(a+b*cos(d*x+c))^ 
(5/2)*sin(d*x+c)/b^2/d+2/9*cos(d*x+c)*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/ 
d
 

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx=\frac {8 \left (8 a^5+8 a^4 b+33 a^3 b^2+33 a^2 b^3+147 a b^4+147 b^5\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-8 a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+b \left (-32 a^4+916 a^2 b^2+301 b^4+\left (-8 a^3 b+1606 a b^3\right ) \cos (c+d x)+4 \left (53 a^2 b^2+84 b^4\right ) \cos (2 (c+d x))+170 a b^3 \cos (3 (c+d x))+35 b^4 \cos (4 (c+d x))\right ) \sin (c+d x)}{1260 b^3 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(8*(8*a^5 + 8*a^4*b + 33*a^3*b^2 + 33*a^2*b^3 + 147*a*b^4 + 147*b^5)*Sqrt[ 
(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 8*a* 
(8*a^4 + 31*a^2*b^2 - 39*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF 
[(c + d*x)/2, (2*b)/(a + b)] + b*(-32*a^4 + 916*a^2*b^2 + 301*b^4 + (-8*a^ 
3*b + 1606*a*b^3)*Cos[c + d*x] + 4*(53*a^2*b^2 + 84*b^4)*Cos[2*(c + d*x)] 
+ 170*a*b^3*Cos[3*(c + d*x)] + 35*b^4*Cos[4*(c + d*x)])*Sin[c + d*x])/(126 
0*b^3*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.913, Rules used = {3042, 3272, 27, 3042, 3502, 27, 3042, 3232, 27, 3042, 3232, 27, 3042, 3231, 3042, 3134, 3042, 3132, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3272

\(\displaystyle \frac {2 \int \frac {1}{2} (a+b \cos (c+d x))^{3/2} \left (-4 a \cos ^2(c+d x)+7 b \cos (c+d x)+2 a\right )dx}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a+b \cos (c+d x))^{3/2} \left (-4 a \cos ^2(c+d x)+7 b \cos (c+d x)+2 a\right )dx}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-4 a \sin \left (c+d x+\frac {\pi }{2}\right )^2+7 b \sin \left (c+d x+\frac {\pi }{2}\right )+2 a\right )dx}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int -\frac {1}{2} (a+b \cos (c+d x))^{3/2} \left (6 a b-\left (8 a^2+49 b^2\right ) \cos (c+d x)\right )dx}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int (a+b \cos (c+d x))^{3/2} \left (6 a b-\left (8 a^2+49 b^2\right ) \cos (c+d x)\right )dx}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (6 a b+\left (-8 a^2-49 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {-\frac {\frac {2}{5} \int \frac {3}{2} \sqrt {a+b \cos (c+d x)} \left (b \left (2 a^2-49 b^2\right )-a \left (8 a^2+39 b^2\right ) \cos (c+d x)\right )dx-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {3}{5} \int \sqrt {a+b \cos (c+d x)} \left (b \left (2 a^2-49 b^2\right )-a \left (8 a^2+39 b^2\right ) \cos (c+d x)\right )dx-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3}{5} \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (2 a^2-49 b^2\right )-a \left (8 a^2+39 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {2}{3} \int -\frac {2 a b \left (a^2+93 b^2\right )+\left (8 a^4+33 b^2 a^2+147 b^4\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (-\frac {1}{3} \int \frac {2 a b \left (a^2+93 b^2\right )+\left (8 a^4+33 b^2 a^2+147 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (-\frac {1}{3} \int \frac {2 a b \left (a^2+93 b^2\right )+\left (8 a^4+33 b^2 a^2+147 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {\left (8 a^4+33 a^2 b^2+147 b^4\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (8 a^4+33 a^2 b^2+147 b^4\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {-\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d}\)

Input:

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

Output:

(2*Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(9*b*d) + ((-8*a* 
(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d) - ((-2*(8*a^2 + 49*b^2)*( 
a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (3*(((-2*(8*a^4 + 33*a^2*b 
^2 + 147*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b 
)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*a*(8*a^4 + 31*a^2*b^2 - 
39*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a 
 + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/3 - (2*a*(8*a^2 + 39*b^2)*Sqrt[a + 
 b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5)/(7*b))/(9*b)
 

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 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*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]
 

rule 3232

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

rule 3272

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 - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
 + n))   Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d 
*(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 
 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*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] && GtQ[m, 2] && (IntegerQ[m 
] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] 
&& NeQ[c, 0])))
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(994\) vs. \(2(295)=590\).

Time = 11.61 (sec) , antiderivative size = 995, normalized size of antiderivative = 3.17

method	result	size
default	\(\text {Expression too large to display}\)	\(995\)

Input:

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

Output:

-2/315*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10*b^5+(1360*a*b^4+2240*b^5)*sin(1/2 
*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-424*a^2*b^3-2040*a*b^4-2072*b^5)*sin(1/ 
2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(-4*a^3*b^2+424*a^2*b^3+1568*a*b^4+952*b 
^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(8*a^4*b+2*a^3*b^2-282*a^2*b^3 
-444*a*b^4-168*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-8*(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)*Ellip 
ticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5-31*(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^3*b^2+39*(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*b^4+8*(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^5-8*(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^4*b+33*(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*b 
^2-33*(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^2*b^3+147* 
(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-...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.64 \[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/945*(4*sqrt(1/2)*(-4*I*a^5 - 15*I*a^3*b^2 + 66*I*a*b^4)*sqrt(b)*weierst 
rassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3* 
b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + 4*sqrt(1/2)*(4*I*a^5 + 15* 
I*a^3*b^2 - 66*I*a*b^4)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^ 
2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) 
 + 2*a)/b) + 3*sqrt(1/2)*(-8*I*a^4*b - 33*I*a^2*b^3 - 147*I*b^5)*sqrt(b)*w 
eierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weier 
strassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*( 
3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) + 3*sqrt(1/2)*(8*I*a^4*b 
+ 33*I*a^2*b^3 + 147*I*b^5)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^ 
2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^ 
2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) 
 + 2*a)/b)) - 3*(35*b^5*cos(d*x + c)^3 + 50*a*b^4*cos(d*x + c)^2 - 4*a^3*b 
^2 + 88*a*b^4 + (3*a^2*b^3 + 49*b^5)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a 
)*sin(d*x + c))/(b^4*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3}d x \right ) a \] Input:

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

Output:

int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**4,x)*b + int(sqrt(cos(c + d*x)* 
b + a)*cos(c + d*x)**3,x)*a