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

Optimal. Leaf size=58 \[ -\frac {\text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{2 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{2 b} \]

[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]
time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4390} \begin {gather*} \frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx &=-\frac {\sin ^{-1}(\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]
time = 0.06, size = 52, normalized size = 0.90 \begin {gather*} \frac {-\text {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} \end {gather*}

Antiderivative was successfully verified.

[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] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 2.80, size = 18450099, normalized size = 318105.16

method result size
default \(\text {Expression too large to display}\) \(18450099\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (52) = 104\).
time = 1.62, size = 242, normalized size = 4.17 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\cos \left (a+b\,x\right )}{\sqrt {\sin \left (2\,a+2\,b\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________