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

Optimal. Leaf size=24 \[ -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, 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 = {4292} \[ -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/b)

Rule 4292

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[((e*Sin[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin {align*} \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx &=-\frac {\csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.96 \[ -\frac {\sqrt {\sin (2 (a+b x))} \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/b)

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fricas [A]  time = 0.45, size = 39, normalized size = 1.62 \[ -\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \sin \left (b x + a\right )}{b \sin \left (b x + a\right )} \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )}{\sqrt {\sin \left (2 \, b x + 2 \, a\right )}}\,{d x} \]

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)

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maple [C]  time = 9.96, size = 308, normalized size = 12.83 \[ \frac {\sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (2 \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticE \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )+\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}-\sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{b \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}} \]

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )}{\sqrt {\sin \left (2 \, b x + 2 \, a\right )}}\,{d x} \]

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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mupad [B]  time = 0.31, size = 24, normalized size = 1.00 \[ -\frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{b\,\sin \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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