\(\int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx\) [534]

Optimal result

Integrand size = 14, antiderivative size = 106 \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 b \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 (a+b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 b \sin (c+d x)}{(a-b) (a+b) d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(2*(a + b)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b) 
/(a + b)] - 2*b*Sin[c + d*x])/((a - b)*(a + b)*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3143, 27, 3042, 3134, 3042, 3132}

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.

\(\displaystyle \int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

Input:

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

Output:

(2*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/((a^2 - 
 b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*b*Sin[c + d*x])/((a^2 - b 
^2)*d*Sqrt[a + b*Cos[c + d*x]])
 

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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(105)=210\).

Time = 1.96 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.05

method	result	size
default	\(-\frac {2 \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{\left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\)	\(217\)

Input:

int(1/(a+cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-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)*s 
in(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)))/(a-b)/(a+b)/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 = 472, normalized size of antiderivative = 4.45 \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {b \cos \left (d x + c\right ) + a} b^{2} \sin \left (d x + c\right ) - \sqrt {\frac {1}{2}} {\left (-i \, a b \cos \left (d x + c\right ) - i \, a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - \sqrt {\frac {1}{2}} {\left (i \, a b \cos \left (d x + c\right ) + i \, a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (i \, b^{2} \cos \left (d x + c\right ) + i \, a b\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (-i \, b^{2} \cos \left (d x + c\right ) - i \, a b\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )\right )}}{3 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{3} b - a b^{3}\right )} d\right )}} \] Input:

integrate(1/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2/3*(3*sqrt(b*cos(d*x + c) + a)*b^2*sin(d*x + c) - sqrt(1/2)*(-I*a*b*cos( 
d*x + c) - I*a^2)*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*b*cos(d*x + c) + I*a^2)*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)*(I*b^2*cos(d*x + c) + I*a*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*sqrt(1/2)* 
(-I*b^2*cos(d*x + c) - I*a*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)))/((a^2*b^2 - b^4)*d*cos(d*x + c) + (a^3*b - a*b^3)*d)
 

Sympy [F]

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

integrate(1/(a+b*cos(d*x+c))**(3/2),x)
 

Output:

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

Maxima [F]

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

integrate(1/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(1/(a + b*cos(c + d*x))^(3/2),x)
 

Output:

int(1/(a + b*cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int(1/(a+b*cos(d*x+c))^(3/2),x)
 

Output:

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