3.2.73 \(\int \frac {\cos ^2(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [173]

Optimal. Leaf size=40 \[ \frac {F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sqrt {\sin (2 a+2 b x)}}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4382, 2720} \begin {gather*} \frac {\sqrt {\sin (2 a+2 b x)}}{2 b}+\frac {F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2720

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

Rule 4382

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] + Dist[e^2*((m + p - 1)/(m + 2*p)), Int[(e*Co
s[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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx &=\frac {\sqrt {\sin (2 a+2 b x)}}{2 b}+\frac {1}{2} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sqrt {\sin (2 a+2 b x)}}{2 b}\\ \end {align*}

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Mathematica [A]
time = 1.00, size = 76, normalized size = 1.90 \begin {gather*} \frac {2 \sqrt {\sin (2 (a+b x))}-\frac {\sqrt {2} F\left (\text {ArcSin}(\cos (a+b x)-\sin (a+b x))\left |\frac {1}{2}\right .\right ) (\cos (a+b x)+\sin (a+b x))}{\sqrt {1+\sin (2 (a+b x))}}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Sin[2*(a + b*x)]] - (Sqrt[2]*EllipticF[ArcSin[Cos[a + b*x] - Sin[a + b*x]], 1/2]*(Cos[a + b*x] + Sin[a
+ b*x]))/Sqrt[1 + Sin[2*(a + b*x)]])/(4*b)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 15.17, size = 66249372, normalized size = 1656234.30

method	result	size
default	\(\text {Expression too large to display}\)	\(66249372\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

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

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

[Out]

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

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

[Out]

Timed out

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

[Out]

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

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

[Out]

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

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