\(\int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx\) [55]

Optimal result

Integrand size = 14, antiderivative size = 221 \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {8 a b \sin (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {8 a (a+b)^2 \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 (a-b) (a+b)^2 \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+2 b \left (-5 a^2+b^2-4 a b \cos (c+d x)\right ) \sin (c+d x)}{3 (a-b)^2 (a+b)^2 d (a+b \cos (c+d x))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {3042, 3143, 27, 3042, 3233, 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.

\(\displaystyle \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(-2*b*Sin[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (((8*a* 
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*(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]]) 
)/(a^2 - b^2) - (8*a*b*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x 
]]))/(3*(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 3143

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp 
[1/((n + 1)*(a^2 - b^2))   Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) 
- b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

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 3233

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(488\) vs. \(2(210)=420\).

Time = 3.10 (sec) , antiderivative size = 489, normalized size of antiderivative = 2.21

method	result	size
default	\(-\frac {\sqrt {-\left (-2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 b \left (a -b \right ) \left (a +b \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a -b}{2 b}\right )^{2}}+\frac {16 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{3 \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-\left (-2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {2 \left (3 a -b \right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\left (3 a^{3}+3 a^{2} b -3 b^{2} a -3 b^{3}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {8 a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{3 \left (a -b \right ) \left (a +b \right )^{2} \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\)	\(489\)

Input:

int(1/(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)*(1/3/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+16/3*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)+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))-8/3*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(1/2*d*x+1/2*c),(-2*b 
/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))/sin(1/2* 
d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.00 \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-2/9*(sqrt(1/2)*(I*a^4 + 3*I*a^2*b^2 + (I*a^2*b^2 + 3*I*b^4)*cos(d*x + c)^ 
2 + 2*(I*a^3*b + 3*I*a*b^3)*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)*(-I*a^4 - 3*I*a^2*b^2 + (-I*a^2*b 
^2 - 3*I*b^4)*cos(d*x + c)^2 + 2*(-I*a^3*b - 3*I*a*b^3)*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) + 12*sqrt(1/2)*(- 
I*a*b^3*cos(d*x + c)^2 - 2*I*a^2*b^2*cos(d*x + c) - I*a^3*b)*sqrt(b)*weier 
strassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstra 
ssPInverse(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)) + 12*sqrt(1/2)*(I*a*b^3*cos(d 
*x + c)^2 + 2*I*a^2*b^2*cos(d*x + c) + I*a^3*b)*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*b^3*cos(d*x + c) + 5*a^2*b^2 - b^ 
4)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d*c 
os(d*x + c)^2 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^6*b - 
2*a^4*b^3 + a^2*b^5)*d)
 

Sympy [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+b*cos(d*x+c))**(5/2),x)
 

Output:

Integral((a + b*cos(c + d*x))**(-5/2), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(a + b*cos(c + d*x))^(5/2),x)
 

Output:

int(1/(a + b*cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(1/(a+b*cos(d*x+c))^(5/2),x)
 

Output:

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