Integrand size = 22, antiderivative size = 77 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {10 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]
-10/21*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*P i+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)
Time = 0.91 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {40 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )+\left (-13 \csc ^2(a+b x)-3 \csc ^4(a+b x)+7 \sec ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{84 b} \]
(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)
Time = 0.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4788, 3042, 3116, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^2 \sin (2 a+2 b x)^{5/2}}dx\) |
\(\Big \downarrow \) 4788 |
\(\displaystyle \frac {10}{7} \int \frac {1}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{7} \int \frac {1}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {10}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\cos (2 a+2 b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {10}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\cos (2 a+2 b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {10}{7} \left (\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b}-\frac {\cos (2 a+2 b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\) |
(10*(EllipticF[a - Pi/4 + b*x, 2]/(3*b) - Cos[2*a + 2*b*x]/(3*b*Sin[2*a + 2*b*x]^(3/2))))/7 - Csc[a + b*x]^2/(7*b*Sin[2*a + 2*b*x]^(3/2))
3.2.12.3.1 Defintions of rubi rules used
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] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
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] + Simp[(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]
Time = 39.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.00
method | result | size |
default | \(\frac {\sqrt {2}\, \left (-\frac {16 \sqrt {2}}{7 \sin \left (2 x b +2 a \right )^{\frac {7}{2}}}+\frac {8 \sqrt {2}\, \left (5 \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (2 x b +2 a \right )^{3}+10 \sin \left (2 x b +2 a \right )^{4}-4 \sin \left (2 x b +2 a \right )^{2}-6\right )}{21 \sin \left (2 x b +2 a \right )^{\frac {7}{2}} \cos \left (2 x b +2 a \right )}\right )}{16 b}\) | \(154\) |
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
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.32 \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {20 \, \sqrt {2 i} {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 20 \, \sqrt {-2 i} {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - \sqrt {2} {\left (20 \, \cos \left (b x + a\right )^{4} - 30 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{84 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )}} \]
-1/84*(20*sqrt(2*I)*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*e lliptic_f(arcsin(cos(b*x + a) + I*sin(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)
Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2}} \,d x \]