3.100 \(\int \csc (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\)

Optimal. Leaf size=53 \[ \frac{\log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{b}-\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b} \]

[Out]

-(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]]]/b

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Rubi [A]  time = 0.0472142, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4308, 4305} \[ \frac{\log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{b}-\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]]]/b

Rule 4308

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[a + b*x]*(g*S
in[c + d*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ
[p] && IntegerQ[2*p]

Rule 4305

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 \csc (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx &=2 \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))}{b}+\frac{\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0406555, size = 52, normalized size = 0.98 \[ \frac{\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )}{b}-\frac{\sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 1.042, size = 157, normalized size = 3. \begin{align*} 2\,{\frac{ \left ( \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ( 1/2\,bx+a/2 \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1},1/2\,\sqrt{2} \right ) }{b\sqrt{\tan \left ( 1/2\,bx+a/2 \right ) \left ( \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }\sqrt{ \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-\tan \left ( 1/2\,bx+a/2 \right ) }}\sqrt{-{\frac{\tan \left ( 1/2\,bx+a/2 \right ) }{ \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)^2-1)/(tan(1/2*b*x+1/2*a)*(tan(1/2
*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/
2)/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right ) \sqrt{\sin \left (2 \, b x + 2 \, a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.520019, size = 657, normalized size = 12.4 \begin{align*} \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 )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right ) \sqrt{\sin \left (2 \, b x + 2 \, a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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