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

Optimal. Leaf size=104 \[ -\frac{4 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{b}+\frac{2 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b}-\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}+\frac{2 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{b} \]

[Out]

(2*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/b + (2*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])/b -
(4*Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/b - (Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(5/2))/b

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Rubi [A]  time = 0.102062, antiderivative size = 104, 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, 4301, 4306} \[ -\frac{4 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{b}+\frac{2 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b}-\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}+\frac{2 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/b + (2*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])/b -
(4*Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/b - (Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(5/2))/b

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 4301

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

Rule 4306

Int[sin[(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 ^3(a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx &=-\frac{\csc ^3(a+b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{b}-4 \int \csc (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx\\ &=-\frac{\csc ^3(a+b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{b}-8 \int \cos (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\\ &=-\frac{4 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{b}-\frac{\csc ^3(a+b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{b}-4 \int \frac{\sin (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=\frac{2 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{b}+\frac{2 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{b}-\frac{4 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{b}-\frac{\csc ^3(a+b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0983066, size = 68, normalized size = 0.65 \[ \frac{2 \left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x))-2 \sqrt{\sin (2 (a+b x))} \csc (a+b x)+\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 6.01, size = 542, normalized size = 5.2 \begin{align*} \text{result too large to display} \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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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)

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Fricas [B]  time = 0.530154, size = 814, normalized size = 7.83 \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 ) \sin \left (b x + a\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 ) \sin \left (b x + a\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 ) \sin \left (b x + a\right ) + 8 \, \sqrt{2} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 8 \, \sin \left (b x + a\right )}{2 \, b \sin \left (b x + a\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]

-1/2*(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))*sin(b*x + a) - 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)))*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)*sin(b*x + a) + 8*sqrt(2)*sqrt(cos(b*x +
a)*sin(b*x + a)) + 8*sin(b*x + a))/(b*sin(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)**3*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 \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)