\(\int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) [540]

Optimal result

Integrand size = 23, antiderivative size = 281 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (8 a^2-9 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:

2/3*(8*a^4-15*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^3/(a^2-b^2)^2/d/((a+b*cos(d*x+c))/(a+b)) 
^(1/2)-2/3*a*(8*a^2-9*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^3/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1 
/2)-2/3*a^2*cos(d*x+c)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-8/3 
*a^2*(a^2-2*b^2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (8 a^4-15 a^2 b^2+3 b^4\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+a \left (-8 a^3+8 a^2 b+9 a b^2-9 b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {a^2 b \left (-4 a^3+8 a b^2+\left (-5 a^2 b+9 b^3\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2}\right )}{3 b^3 d (a+b \cos (c+d x))^{3/2}} \] Input:

Integrate[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^(5/2),x]
 

Output:

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((8*a^4 - 15*a^2*b^2 + 3*b^4)*El 
lipticE[(c + d*x)/2, (2*b)/(a + b)] + a*(-8*a^3 + 8*a^2*b + 9*a*b^2 - 9*b^ 
3)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^2 + (a^2*b*(-4*a^3 + 8* 
a*b^2 + (-5*a^2*b + 9*b^3)*Cos[c + d*x])*Sin[c + d*x])/(a^2 - b^2)^2))/(3* 
b^3*d*(a + b*Cos[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3271, 27, 3042, 3500, 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.

\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \int \frac {2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3231

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3134

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3132

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3142

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3140

\(\displaystyle -\frac {2 a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (8 a^4-17 a^2 b^2+9 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\)

Input:

Int[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^(5/2),x]
 

Output:

(-2*a^2*Cos[c + d*x]*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]) 
^(3/2)) - (-(((2*(8*a^4 - 15*a^2*b^2 + 3*b^4)*Sqrt[a + b*Cos[c + d*x]]*Ell 
ipticE[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b) 
]) - (2*a*(8*a^4 - 17*a^2*b^2 + 9*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]]))/(b* 
(a^2 - b^2))) + (8*a^2*(a^2 - 2*b^2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a 
 + b*Cos[c + d*x]]))/(3*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 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(910\) vs. \(2(270)=540\).

Time = 7.50 (sec) , antiderivative size = 911, normalized size of antiderivative = 3.24

method	result	size
default	\(\text {Expression too large to display}\)	\(911\)

Input:

int(cos(d*x+c)^3/(a+cos(d*x+c)*b)^(5/2),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/b^3/(-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*b/(a-b)*sin( 
1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*Ellipt 
icF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-EllipticE(cos(1/2*d*x+1/2*c), 
(-2*b/(a-b))^(1/2))*a+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)- 
6/b^3*a^2/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2*b-a-b)/(a^2-b^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)*(2*cos(1/2*d*x+ 
1/2*c)*sin(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a 
-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^ 
2+(a+b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))-2 
/b^3*a^3*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+ 
(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+8/3 
*sin(1/2*d*x+1/2*c)^2*b/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*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)+(3*a-b)/(3*a^3+3*a^2*b-3* 
a*b^2-3*b^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)* 
EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^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)*(EllipticF(cos(...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 835, normalized size of antiderivative = 2.97 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3/(a+b*cos(d*x+c))**(5/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)
 

Giac [F]

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

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate(cos(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int(cos(c + d*x)^3/(a + b*cos(c + d*x))^(5/2),x)
 

Output:

int(cos(c + d*x)^3/(a + b*cos(c + d*x))^(5/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x)
 

Output:

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