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

Optimal. Leaf size=81 \[ \frac{32 \sin (a+b x)}{21 b \sqrt{\sin (2 a+2 b x)}}-\frac{16 \cos (a+b x)}{21 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

(-16*Cos[a + b*x])/(21*b*Sin[2*a + 2*b*x]^(3/2)) - Csc[a + b*x]^3/(7*b*Sqrt[Sin[2*a + 2*b*x]]) + (32*Sin[a + b
*x])/(21*b*Sqrt[Sin[2*a + 2*b*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0918782, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4300, 4308, 4303, 4292} \[ \frac{32 \sin (a+b x)}{21 b \sqrt{\sin (2 a+2 b x)}}-\frac{16 \cos (a+b x)}{21 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*Cos[a + b*x])/(21*b*Sin[2*a + 2*b*x]^(3/2)) - Csc[a + b*x]^3/(7*b*Sqrt[Sin[2*a + 2*b*x]]) + (32*Sin[a + b
*x])/(21*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4300

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))/(2*b*g*(m + p + 1)), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Sin[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] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

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 ^3(a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx &=-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}}+\frac{8}{7} \int \frac{\csc (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}}+\frac{16}{7} \int \frac{\cos (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{16 \cos (a+b x)}{21 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}}+\frac{32}{21} \int \frac{\sin (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{16 \cos (a+b x)}{21 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\csc ^3(a+b x)}{7 b \sqrt{\sin (2 a+2 b x)}}+\frac{32 \sin (a+b x)}{21 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.116, size = 55, normalized size = 0.68 \[ \frac{\sqrt{\sin (2 (a+b x))} (-12 \cos (2 (a+b x))+4 \cos (4 (a+b x))+5) \csc ^4(a+b x) \sec (a+b x)}{42 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5 - 12*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/(42*b)

________________________________________________________________________________________

Maple [C]  time = 9.663, size = 222, normalized size = 2.7 \begin{align*} -{\frac{1}{336\,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 ( -3\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{8}+16\,\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 ) \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-2\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}+2\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+3 \right ) \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-3}{\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)^3/sin(2*b*x+2*a)^(3/2),x)

[Out]

-1/336*(-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)^3*(-3*
tan(1/2*b*x+1/2*a)^8+16*(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)^3-2*tan(1/2*b*x+1/2*a)^6+2*tan(1/2*b
*x+1/2*a)^2+3)/(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)/b

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.506828, size = 281, normalized size = 3.47 \begin{align*} \frac{32 \, \cos \left (b x + a\right )^{5} - 64 \, \cos \left (b x + a\right )^{3} + \sqrt{2}{\left (32 \, \cos \left (b x + a\right )^{4} - 56 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )}{42 \,{\left (b \cos \left (b x + a\right )^{5} - 2 \, 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)^3/sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")

[Out]

1/42*(32*cos(b*x + a)^5 - 64*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 56*cos(b*x + a)^2 + 21)*sqrt(cos(b*
x + a)*sin(b*x + a)) + 32*cos(b*x + a))/(b*cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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