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

Optimal result

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

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4382, 2720} \[ \int \frac {\cos ^2(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {\sqrt {\sin (2 a+2 b x)}}{2 b}+\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b} \]

[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*} \text {integral}& = \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 {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{2 b}+\frac {\sqrt {\sin (2 a+2 b x)}}{2 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^2(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=\frac {\left (1+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {\sin (2 (a+b x))}}{2 b} \]

[In]

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

[Out]

((1 + Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*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 = 9.33 (sec) , antiderivative size = 67488705, normalized size of antiderivative = 1687217.62

method	result	size
default	\(\text {Expression too large to display}\)	\(67488705\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)