\(\int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx\) [518]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{\sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \] Output:

-1/2*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2)*(b/(2*a+b))^(1/2))*(2*a+b*sin(2*d 
*x+2*c))^(1/2)*2^(1/2)/d/((2*a+b*sin(2*d*x+2*c))/(2*a+b))^(1/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\frac {(2 a+b) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}}{d \sqrt {4 a+2 b \sin (2 (c+d x))}} \] Input:

Integrate[Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]
 

Output:

((2*a + b)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2* 
(c + d*x)])/(2*a + b)])/(d*Sqrt[4*a + 2*b*Sin[2*(c + d*x)]])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3145, 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.

\(\displaystyle \int \sqrt {a+b \sin (c+d x) \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+b \sin (c+d x) \cos (c+d x)}dx\)

\(\Big \downarrow \) 3145

\(\displaystyle \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}dx\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {2 a+b \sin (2 c+2 d x)} \int \sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}dx}{\sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{\sqrt {2} d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}\)

Input:

Int[Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]
 

Output:

(EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x]] 
)/(Sqrt[2]*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])
 

Defintions of rubi rules used

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 3145

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ 
Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, 
 x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(75)=150\).

Time = 0.40 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.11

method	result	size
default	\(-\frac {\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, \left (2 a -b \right ) \left (2 \operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) a +\operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b -2 a \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )-\operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) b \right )}{b \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d}\)	\(312\)

Input:

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

Output:

-((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/(2*a+b))^(1 
/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)/b*(2*a-b)*(2*EllipticE(((2*a+b*s 
in(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a+EllipticE(((2*a+b 
*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b-2*a*EllipticF(( 
(2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))-EllipticF(( 
(2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b)/cos(2*d* 
x+2*c)/(4*a+2*b*sin(2*d*x+2*c))^(1/2)/d
 

Fricas [F]

\[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

integral(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
 

Sympy [F]

\[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\int \sqrt {a + b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(sqrt(a + b*sin(c + d*x)*cos(c + d*x)), x)
 

Maxima [F]

\[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

integrate(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
 

Giac [F]

\[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\int { \sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

integrate(sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {a+b \cos (c+d x) \sin (c+d x)} \, dx=\int \sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}d x \] Input:

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

Output:

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