\(\int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [112]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {10 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-10/21*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-10/21*cos(2*b*x+2*
a)/b/sin(2*b*x+2*a)^(3/2)-1/7*csc(b*x+a)^2/b/sin(2*b*x+2*a)^(3/2)

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, 2720} \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {10 \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[In]

Int[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(10*EllipticF[a - Pi/4 + b*x, 2])/(21*b) - (10*Cos[2*a + 2*b*x])/(21*b*Sin[2*a + 2*b*x]^(3/2)) - Csc[a + b*x]^
2/(7*b*Sin[2*a + 2*b*x]^(3/2))

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {10}{7} \int \frac {1}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {10}{21} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {10 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {40 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )+\left (-13 \csc ^2(a+b x)-3 \csc ^4(a+b x)+7 \sec ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{84 b} \]

[In]

Integrate[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(40*EllipticF[a - Pi/4 + b*x, 2] + (-13*Csc[a + b*x]^2 - 3*Csc[a + b*x]^4 + 7*Sec[a + b*x]^2)*Sqrt[Sin[2*(a +
b*x)]])/(84*b)

Maple [A] (verified)

Time = 39.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.00

method	result	size
default	\(\frac {\sqrt {2}\, \left (-\frac {16 \sqrt {2}}{7 \sin \left (2 x b +2 a \right )^{\frac {7}{2}}}+\frac {8 \sqrt {2}\, \left (5 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (2 x b +2 a \right )^{3}+10 \sin \left (2 x b +2 a \right )^{4}-4 \sin \left (2 x b +2 a \right )^{2}-6\right )}{21 \sin \left (2 x b +2 a \right )^{\frac {7}{2}} \cos \left (2 x b +2 a \right )}\right )}{16 b}\)	\(154\)

[In]

int(csc(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16*2^(1/2)*(-16/7*2^(1/2)/sin(2*b*x+2*a)^(7/2)+8/21*2^(1/2)/sin(2*b*x+2*a)^(7/2)*(5*(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)*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))*sin(2*b*x
+2*a)^3+10*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-6)/cos(2*b*x+2*a))/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.32 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {20 \, \sqrt {2 i} {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 20 \, \sqrt {-2 i} {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {2} {\left (20 \, \cos \left (b x + a\right )^{4} - 30 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{84 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )}} \]

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

-1/84*(20*sqrt(2*I)*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*elliptic_f(arcsin(cos(b*x + a) + I*si
n(b*x + a)), -1) + 20*sqrt(-2*I)*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*elliptic_f(arcsin(cos(b*
x + a) - I*sin(b*x + a)), -1) - sqrt(2)*(20*cos(b*x + a)^4 - 30*cos(b*x + a)^2 + 7)*sqrt(cos(b*x + a)*sin(b*x
+ a)))/(b*cos(b*x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(csc(b*x+a)**2/sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

Giac [F]

\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^2/sin(2*b*x + 2*a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2}} \,d x \]

[In]

int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^(5/2)),x)

[Out]

int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^(5/2)), x)