Integrand size = 20, antiderivative size = 110 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b} \] Output:
-3/8*arcsin(cos(b*x+a)-sin(b*x+a))/b+3/8*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b* x+2*a)^(1/2))/b-3/4*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b+1/2*sin(b*x+a)*sin(2 *b*x+2*a)^(3/2)/b
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {3 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )-2 (2 \cos (a+b x)+\cos (3 (a+b x))) \sqrt {\sin (2 (a+b x))}}{8 b} \] Input:
Integrate[Csc[a + b*x]*Sin[2*a + 2*b*x]^(5/2),x]
Output:
(3*(-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*(2*Cos[a + b*x] + Cos[3*(a + b*x)])*Sqrt[S in[2*(a + b*x)]])/(8*b)
Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4796, 3042, 4789, 3042, 4790, 3042, 4793}
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.
\(\displaystyle \int \sin ^{\frac {5}{2}}(2 a+2 b x) \csc (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (2 a+2 b x)^{5/2}}{\sin (a+b x)}dx\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle 2 \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \cos (a+b x) \sin (2 a+2 b x)^{3/2}dx\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle 2 \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle 2 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )\) |
\(\Big \downarrow \) 4793 |
\(\displaystyle 2 \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )\) |
Input:
Int[Csc[a + b*x]*Sin[2*a + 2*b*x]^(5/2),x]
Output:
2*((3*((-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b + Log[Cos[a + b*x] + Si n[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 - (Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 + (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b))
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] + Simp[2*p*( g/(2*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] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[-2*Cos[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p* (g/(2*p + 1)) Int[Cos[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] && !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]
Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim p[-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]
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 has leaf size over 500,000. Avoiding possible recursion issues.
Time = 45.38 (sec) , antiderivative size = 99228795, normalized size of antiderivative = 902079.95
Input:
int(csc(b*x+a)*sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (96) = 192\).
Time = 0.09 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.55 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")
Output:
-1/32*(8*sqrt(2)*(4*cos(b*x + a)^3 - cos(b*x + a))*sqrt(cos(b*x + a)*sin(b *x + a)) - 6*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)) + 6*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))) + 3*lo g(-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))/b
Timed out. \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)**(5/2),x)
Output:
Timed out
\[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)*sin(2*b*x + 2*a)^(5/2), x)
\[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(csc(b*x+a)*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)*sin(2*b*x + 2*a)^(5/2), x)
Timed out. \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \] Input:
int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x),x)
Output:
int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x), x)
\[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}d x \] Input:
int(csc(b*x+a)*sin(2*b*x+2*a)^(5/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*csc(a + b*x)*sin(2*a + 2*b*x)**2,x)