3.102 \(\int \frac{\csc (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=53 \[ \frac{4 \sin (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)} \]

[Out]

(-2*Cos[a + b*x])/(3*b*Sin[2*a + 2*b*x]^(3/2)) + (4*Sin[a + b*x])/(3*b*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0642377, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4308, 4303, 4292} \[ \frac{4 \sin (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(-2*Cos[a + b*x])/(3*b*Sin[2*a + 2*b*x]^(3/2)) + (4*Sin[a + b*x])/(3*b*Sqrt[Sin[2*a + 2*b*x]])

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 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[a + b*x]*(g*Sin[c + d
*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x
], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

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)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx &=2 \int \frac{\cos (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{4}{3} \int \frac{\sin (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{2 \cos (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{4 \sin (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.100315, size = 43, normalized size = 0.81 \[ \frac{\sqrt{\sin (2 (a+b x))} \left (\frac{1}{2} \sec (a+b x)-\frac{1}{6} \cot (a+b x) \csc (a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]

[Out]

((-(Cot[a + b*x]*Csc[a + b*x])/6 + Sec[a + b*x]/2)*Sqrt[Sin[2*(a + b*x)]])/b

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Maple [C]  time = 2.469, size = 194, normalized size = 3.7 \begin{align*} -{\frac{1}{12\,b}\sqrt{-{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) ^{-1}}} \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) \left ( 2\,\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 ) \tan \left ( 1/2\,bx+a/2 \right ) - \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{4}+1 \right ) \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) }}}{\frac{1}{\sqrt{ \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x)

[Out]

-1/12/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)*(2*(t
an(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)-tan(1/2*b*x+1/2*a)^4+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)^3-tan(1/2*b*x+1/2*a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)/sin(2*b*x + 2*a)^(3/2), x)

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Fricas [A]  time = 0.498337, size = 192, normalized size = 3.62 \begin{align*} \frac{4 \, \cos \left (b x + a\right )^{3} + \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 4 \, \cos \left (b x + a\right )}{6 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")

[Out]

1/6*(4*cos(b*x + a)^3 + sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*sin(b*x + a)) - 4*cos(b*x + a))/(b*co
s(b*x + a)^3 - b*cos(b*x + a))

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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)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/sin(2*b*x + 2*a)^(3/2), x)