\(\int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [530]

Optimal result

Integrand size = 23, antiderivative size = 326 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{5 b^4 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {8 a \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{5 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \] Output:

2/5*(16*a^4-8*a^2*b^2-3*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^4/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^( 
1/2)-8/5*a*(4*a^2+b^2)*((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))/b^4/d/(a+b*cos(d*x+c))^(1/2)-2*a^2*cos( 
d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-2/5*a*(8*a^2-3*b^ 
2)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^3/(a^2-b^2)/d+2/5*(6*a^2-b^2)*cos(d 
*x+c)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/(a^2-b^2)/d
 

Mathematica [A] (verified)

Time = 1.62 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (16 a^5+16 a^4 b-8 a^3 b^2-8 a^2 b^3-3 a b^4-3 b^5\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-8 a \left (4 a^4-3 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )-b \left (16 a^4-7 a^2 b^2+b^4+4 a b \left (a^2-b^2\right ) \cos (c+d x)+\left (-a^2 b^2+b^4\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{5 (a-b) b^4 (a+b) d \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^(3/2),x]
 

Output:

(2*(16*a^5 + 16*a^4*b - 8*a^3*b^2 - 8*a^2*b^3 - 3*a*b^4 - 3*b^5)*Sqrt[(a + 
 b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 8*a*(4*a 
^4 - 3*a^2*b^2 - b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d* 
x)/2, (2*b)/(a + b)] - b*(16*a^4 - 7*a^2*b^2 + b^4 + 4*a*b*(a^2 - b^2)*Cos 
[c + d*x] + (-(a^2*b^2) + b^4)*Cos[2*(c + d*x)])*Sin[c + d*x])/(5*(a - b)* 
b^4*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.73 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 3271, 27, 3042, 3528, 27, 3042, 3502, 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.

\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

\(\displaystyle -\frac {2 \int \frac {\cos (c+d x) \left (4 a^2-b \cos (c+d x) a-\left (6 a^2-b^2\right ) \cos ^2(c+d x)\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (4 a^2-b \cos (c+d x) a-\left (6 a^2-b^2\right ) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle -\frac {-\frac {\frac {2 \int \frac {3 \left (a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \cos (c+d x)\right )}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3231

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3134

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3132

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3142

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3140

\(\displaystyle -\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {-\frac {2 \left (6 a^2-b^2\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}-\frac {\frac {\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {8 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d}}{5 b}}{b \left (a^2-b^2\right )}\)

Input:

Int[Cos[c + d*x]^4/(a + b*Cos[c + d*x])^(3/2),x]
 

Output:

(-2*a^2*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d 
*x]]) - ((-2*(6*a^2 - b^2)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d 
*x])/(5*b*d) - (((2*(16*a^4 - 8*a^2*b^2 - 3*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)]) - (8*a*(a^2 - b^2)*(4*a^2 + b^2)*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]]))/b - 
 (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d))/(5*b)) 
/(b*(a^2 - b^2))
 

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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((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 - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*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] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

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]
 

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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1284\) vs. \(2(313)=626\).

Time = 8.42 (sec) , antiderivative size = 1285, normalized size of antiderivative = 3.94

method	result	size
default	\(\text {Expression too large to display}\)	\(1285\)

Input:

int(cos(d*x+c)^4/(a+cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/5*(-8*(-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^3* 
(a^2-b^2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-8*(-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^2*(a^3-a^2*b-a*b^2+b^3)*sin(1/2*d 
*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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*(8*a^4+2*a^3*b-4*a^2*b^2-2*a*b^3+b^4)*sin(1/2*d*x+1/ 
2*c)^2*cos(1/2*d*x+1/2*c)-16*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+ 
1/2*c)^2)^(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))*a^ 
5+12*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(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)*E 
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2+4*(-2*b*sin(1/2*d* 
x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(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))*a*b^4+16*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1 
/2*d*x+1/2*c)^2)^(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1 
/2))*a^5-16*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4*b-8*(-2*b*sin(1 
/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 667, normalized size of antiderivative = 2.05 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

2/15*(sqrt(1/2)*(32*I*a^6 - 28*I*a^4*b^2 - 9*I*a^2*b^4 + (32*I*a^5*b - 28* 
I*a^3*b^3 - 9*I*a*b^5)*cos(d*x + c))*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) + sqrt(1/2)*(-32*I*a^6 + 28*I*a^4*b^2 + 9*I*a^2*b^ 
4 + (-32*I*a^5*b + 28*I*a^3*b^3 + 9*I*a*b^5)*cos(d*x + c))*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) + 3*sqrt(1/2)*(16*I*a^5*b - 
8*I*a^3*b^3 - 3*I*a*b^5 + (16*I*a^4*b^2 - 8*I*a^2*b^4 - 3*I*b^6)*cos(d*x + 
 c))*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*sqrt(1/ 
2)*(-16*I*a^5*b + 8*I*a^3*b^3 + 3*I*a*b^5 + (-16*I*a^4*b^2 + 8*I*a^2*b^4 + 
 3*I*b^6)*cos(d*x + c))*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*(8*a^4*b^2 - 3*a^2*b^4 - (a^2*b^4 - b^6)*cos(d*x + c)^2 + 2*(a 
^3*b^3 - a*b^5)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^2 
*b^6 - b^8)*d*cos(d*x + c) + (a^3*b^5 - a*b^7)*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**4/(a+b*cos(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(3/2), x)
 

Giac [F]

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

integrate(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(3/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int(cos(c + d*x)^4/(a + b*cos(c + d*x))^(3/2),x)
 

Output:

int(cos(c + d*x)^4/(a + b*cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^4/(a+b*cos(d*x+c))^(3/2),x)
 

Output:

int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**4)/(cos(c + d*x)**2*b**2 + 2*c 
os(c + d*x)*a*b + a**2),x)