\(\int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [164]

Optimal result

Integrand size = 20, antiderivative size = 58 \[ \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{2 b} \]

[Out]

-1/2*arcsin(cos(b*x+a)-sin(b*x+a))/b+1/2*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4390} \[ \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b} \]

[In]

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

[Out]

-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b)

Rule 4390

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-ArcSin[Cos[a + b*x] - Sin[a + b*
x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[
b*c - a*d, 0] && EqQ[d/b, 2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )}{2 b} \]

[In]

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

[Out]

(-ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]])/(2*b)

Maple [B] (warning: unable to verify)

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

Time = 2.07 (sec) , antiderivative size = 18441891, normalized size of antiderivative = 317963.64

method	result	size
default	\(\text {Expression too large to display}\)	\(18441891\)

[In]

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

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (52) = 104\).

Time = 0.26 (sec) , antiderivative size = 242, normalized size of antiderivative = 4.17 \[ \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{8 \, b} \]

[In]

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

[Out]

1/8*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x +
a))/(cos(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) - 2*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)
) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a))) - log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*
x + a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x
+ a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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