3.2.72 \(\int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\) [172]

3.2.72.1 Optimal result

Integrand size = 22, antiderivative size = 40 \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b} \]

output

-1/2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+ 
b*x),2^(1/2))/b+1/6*sin(2*b*x+2*a)^(3/2)/b
 

3.2.72.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sin ^{\frac {3}{2}}(2 (a+b x))}{6 b} \]

input

Integrate[Cos[a + b*x]^2*Sqrt[Sin[2*a + 2*b*x]],x]
 

output

(3*EllipticE[a - Pi/4 + b*x, 2] + Sin[2*(a + b*x)]^(3/2))/(6*b)
 

3.2.72.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4785, 3042, 3119}

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.

\(\displaystyle \int \sqrt {\sin (2 a+2 b x)} \cos ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\sin (2 a+2 b x)} \cos (a+b x)^2dx\)

\(\Big \downarrow \) 4785

\(\displaystyle \frac {1}{2} \int \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}+\frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b}\)

input

Int[Cos[a + b*x]^2*Sqrt[Sin[2*a + 2*b*x]],x]
 

output

EllipticE[a - Pi/4 + b*x, 2]/(2*b) + Sin[2*a + 2*b*x]^(3/2)/(6*b)
 

3.2.72.3.1 Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4785

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p 
_), x_Symbol] :> Simp[e^2*(e*Cos[a + b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1 
)/(2*b*g*(m + 2*p))), x] + Simp[e^2*((m + p - 1)/(m + 2*p))   Int[(e*Cos[a 
+ b*x])^(m - 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, 
x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && NeQ 
[m + 2*p, 0] && IntegersQ[2*m, 2*p]
 

3.2.72.4 Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 5.16 (sec) , antiderivative size = 26159851, normalized size of antiderivative = 653996.28

method	result	size
default	\(\text {Expression too large to display}\)	\(26159851\)

input

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

output

result too large to display
 

3.2.72.5 Fricas [F]

\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]

input

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

output

integral(cos(b*x + a)^2*sqrt(sin(2*b*x + 2*a)), x)
 

3.2.72.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.72.7 Maxima [F]

\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]

input

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

output

integrate(cos(b*x + a)^2*sqrt(sin(2*b*x + 2*a)), x)
 

3.2.72.8 Giac [F]

\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]

input

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

output

integrate(cos(b*x + a)^2*sqrt(sin(2*b*x + 2*a)), x)
 

3.2.72.9 Mupad [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int {\cos \left (a+b\,x\right )}^2\,\sqrt {\sin \left (2\,a+2\,b\,x\right )} \,d x \]

input

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

output

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