\(\int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [111]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {6 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b}-\frac {6 \cos (2 a+2 b x)}{5 b \sqrt {\sin (2 a+2 b x)}}-\frac {\csc ^2(a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

6/5*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b-6/5*cos(2*b*x+2*a)/b/
sin(2*b*x+2*a)^(1/2)-1/5*csc(b*x+a)^2/b/sin(2*b*x+2*a)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4385, 2716, 2719} \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {6 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b}-\frac {6 \cos (2 a+2 b x)}{5 b \sqrt {\sin (2 a+2 b x)}}-\frac {\csc ^2(a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \]

[In]

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

[Out]

(-6*EllipticE[a - Pi/4 + b*x, 2])/(5*b) - (6*Cos[2*a + 2*b*x])/(5*b*Sqrt[Sin[2*a + 2*b*x]]) - Csc[a + b*x]^2/(
5*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 4385

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\csc ^2(a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}}+\frac {6}{5} \int \frac {1}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {6 \cos (2 a+2 b x)}{5 b \sqrt {\sin (2 a+2 b x)}}-\frac {\csc ^2(a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}}-\frac {6}{5} \int \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {6 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b}-\frac {6 \cos (2 a+2 b x)}{5 b \sqrt {\sin (2 a+2 b x)}}-\frac {\csc ^2(a+b x)}{5 b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {-12 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\frac {2 (1-6 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \cot (a+b x)}{\sin ^{\frac {3}{2}}(2 (a+b x))}}{10 b} \]

[In]

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

[Out]

(-12*EllipticE[a - Pi/4 + b*x, 2] + (2*(1 - 6*Cos[2*(a + b*x)] + 3*Cos[4*(a + b*x)])*Cot[a + b*x])/Sin[2*(a +
b*x)]^(3/2))/(10*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(92)=184\).

Time = 21.94 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.95

method	result	size
default	\(\frac {\sqrt {2}\, \left (-\frac {8 \sqrt {2}}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}}}+\frac {4 \sqrt {2}\, \left (6 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \sin \left (2 x b +2 a \right )^{2} \operatorname {EllipticE}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \sin \left (2 x b +2 a \right )^{2} \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (2 x b +2 a \right )^{4}-4 \sin \left (2 x b +2 a \right )^{2}-2\right )}{5 \sin \left (2 x b +2 a \right )^{\frac {5}{2}} \cos \left (2 x b +2 a \right )}\right )}{8 b}\)	\(227\)

[In]

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

[Out]

1/8*2^(1/2)*(-8/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)+4/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)*(6*(sin(2*b*x+2*a)+1)^(1/2)*(-
2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1
/2))-3*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticF
((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+6*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-2)/cos(2*b*x+2*a))/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.45 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {6 \, \sqrt {2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + \sqrt {2} {\left (12 \, \cos \left (b x + a\right )^{4} - 18 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{10 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

-1/10*(6*sqrt(2*I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_e(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*s
in(b*x + a) + 6*sqrt(-2*I)*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a
)), -1)*sin(b*x + a) + 6*sqrt(2*I)*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*elliptic_f(arcsin(cos(b*x + a) + I*sin
(b*x + a)), -1)*sin(b*x + a) + 6*sqrt(-2*I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_f(arcsin(cos(b*x + a)
- I*sin(b*x + a)), -1)*sin(b*x + a) + sqrt(2)*(12*cos(b*x + a)^4 - 18*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*s
in(b*x + a)))/((b*cos(b*x + a)^3 - b*cos(b*x + a))*sin(b*x + a))

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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