Integrand size = 20, antiderivative size = 53 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {2 \cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}} \] Output:
-2/3*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)+4/3*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/ 2)
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\left (-\frac {1}{6} \cot (a+b x) \csc (a+b x)+\frac {1}{2} \sec (a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{b} \] Input:
Integrate[Csc[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]
Output:
((-1/6*(Cot[a + b*x]*Csc[a + b*x]) + Sec[a + b*x]/2)*Sqrt[Sin[2*(a + b*x)] ])/b
Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4796, 3042, 4791, 3042, 4780}
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.
\(\displaystyle \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{3/2}}dx\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx\) |
\(\Big \downarrow \) 4791 |
\(\displaystyle 2 \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {2}{3} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )\) |
\(\Big \downarrow \) 4780 |
\(\displaystyle 2 \left (\frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )\) |
Input:
Int[Csc[a + b*x]/Sin[2*a + 2*b*x]^(3/2),x]
Output:
2*(-1/3*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(3/2)) + (2*Sin[a + b*x])/(3*b*Sq rt[Sin[2*a + 2*b*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)/(b*g*m) ), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b , 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
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] + Simp [(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] && !Int egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[2*g Int[Cos[a + b*x]*(g*Sin[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]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 8.52 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.66
method | result | size |
default | \(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right ) \left (2 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+1\right )}{12 b \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}}\) | \(194\) |
Input:
int(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/12/b*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(tan(1/2*a+1/ 2*b*x)^2-1)/tan(1/2*a+1/2*b*x)*(2*(tan(1/2*a+1/2*b*x)+1)^(1/2)*(-2*tan(1/2 *a+1/2*b*x)+2)^(1/2)*(-tan(1/2*a+1/2*b*x))^(1/2)*EllipticF((tan(1/2*a+1/2* b*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*b*x)-tan(1/2*a+1/2*b*x)^4+1)/(tan (1/2*a+1/2*b*x)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)/(tan(1/2*a+1/2*b*x)^3-tan( 1/2*a+1/2*b*x))^(1/2)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {4 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 4 \, \cos \left (b x + a\right )}{6 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
1/6*(4*cos(b*x + a)^3 + sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*s in(b*x + a)) - 4*cos(b*x + a))/(b*cos(b*x + a)^3 - b*cos(b*x + a))
Timed out. \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)/sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)/sin(2*b*x + 2*a)^(3/2), x)
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\csc \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)/sin(2*b*x + 2*a)^(3/2), x)
Time = 22.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.94 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (1+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}-{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\right )}{3\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )} \] Input:
int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^(3/2)),x)
Output:
(4*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)*(exp(a*4i + b*x*4i) - exp(a*2i + b*x*2i) + 1))/(3*b*(exp(a*2i + b*x*2i) - 1)^2*(exp(a*2i + b*x*2i) + 1))
\[ \int \frac {\csc (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {\sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{2}}d x \] Input:
int(csc(b*x+a)/sin(2*b*x+2*a)^(3/2),x)
Output:
int((sqrt(sin(2*a + 2*b*x))*csc(a + b*x))/sin(2*a + 2*b*x)**2,x)