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]

-16/21*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-1/7*csc(b*x+a)^3/b/sin(2*b*x+2*a)^(1/2)+32/21*sin(b*x+a)/b/sin(2*b*x+
2*a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.12, 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)

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fricas [A]  time = 0.51, size = 104, normalized size = 1.28 \[ \frac {32 \, \cos \left (b x + a\right )^{5} - 64 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 56 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )}{42 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]

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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [C]  time = 107.67, size = 222, normalized size = 2.74 \[ -\frac {\sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \left (-3 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+16 \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3\right )}{336 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, b} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}}\,{d x} \]

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)

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mupad [B]  time = 3.70, size = 302, normalized size = 3.73 \[ -\frac {10\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{21\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,12{}\mathrm {i}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {10{}\mathrm {i}}{21\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,32{}\mathrm {i}}{21\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^3*sin(2*a + 2*b*x)^(3/2)),x)

[Out]

(exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*12i)/(7*b*(exp(a*2i + b*x*
2i)*1i - 1i)^3) - (10*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(21*
b*(exp(a*2i + b*x*2i)*1i - 1i)^2) - (8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1
i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(10i/(21*b) - (exp(a*2i + b*x*2i)*32i)
/(21*b))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp(a*2i + b*x*2i) + 1)*(exp(a*2i
+ b*x*2i)*1i - 1i))

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

[Out]

Timed out

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