Integrand size = 22, antiderivative size = 164 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b}+\frac {4 \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x)}{3 b} \]
-5/4*arcsin(cos(b*x+a)-sin(b*x+a))/b-5/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b* x+2*a)^(1/2))/b-5/3*cos(b*x+a)*sin(2*b*x+2*a)^(3/2)/b+4/3*sin(b*x+a)*sin(2 *b*x+2*a)^(5/2)/b+1/3*csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2)/b+5/2*sin(b*x+a)*s in(2*b*x+2*a)^(1/2)/b
Time = 0.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {-5 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 \sqrt {\sin (2 (a+b x))} (6 \sin (a+b x)+\sin (3 (a+b x)))}{4 b} \]
(-5*(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*Sqrt[Sin[2*(a + b*x)]]*(6*Sin[a + b*x] + S in[3*(a + b*x)]))/(4*b)
Time = 0.74 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4788, 3042, 4796, 3042, 4789, 3042, 4790, 3042, 4789, 3042, 4794}
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.
| \(\displaystyle \int \sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (2 a+2 b x)^{7/2}}{\sin (a+b x)^3}dx\) |
| \(\Big \downarrow \) 4788 |
| \(\displaystyle 4 \int \csc (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 4 \int \frac {\sin (2 a+2 b x)^{7/2}}{\sin (a+b x)}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 4796 |
| \(\displaystyle 8 \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 8 \int \cos (a+b x) \sin (2 a+2 b x)^{5/2}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 4789 |
| \(\displaystyle 8 \left (\frac {5}{6} \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 8 \left (\frac {5}{6} \int \sin (a+b x) \sin (2 a+2 b x)^{3/2}dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 4790 |
| \(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 4789 |
| \(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
| \(\Big \downarrow \) 4794 |
| \(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\) |
(Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2))/(3*b) + 8*((Sin[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(6*b) + (5*((3*((-1/2*ArcSin[Cos[a + b*x] - Sin[a + b*x]]/b - Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 + (S in[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 - (Cos[a + b*x]*Sin[2*a + 2* b*x]^(3/2))/(4*b)))/6)
3.2.16.3.1 Defintions of rubi rules used
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]
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[sin[(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 contains higher order function than in optimal. Order 4 vs. order 3.
Time = 102.09 (sec) , antiderivative size = 973, normalized size of antiderivative = 5.93
32/5*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(2*(tan(1/2*a+1/ 2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2 )*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*(tan(1/2*a+1/2*x*b)* (tan(1/2*a+1/2*x*b)-1)*(tan(1/2*a+1/2*x*b)+1))^(1/2)*tan(1/2*a+1/2*x*b)^4- (tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1 /2*x*b))^(1/2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*(tan(1/ 2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)-1)*(tan(1/2*a+1/2*x*b)+1))^(1/2)*tan(1/2* a+1/2*x*b)^4+2*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1 /2*x*b)^6-4*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*( -tan(1/2*a+1/2*x*b))^(1/2)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1 /2))*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)-1)*(tan(1/2*a+1/2*x*b)+1))^(1 /2)*tan(1/2*a+1/2*x*b)^2+2*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2* x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticF((tan(1/2*a+1/2*x*b)+1) ^(1/2),1/2*2^(1/2))*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)-1)*(tan(1/2*a+ 1/2*x*b)+1))^(1/2)*tan(1/2*a+1/2*x*b)^2-4*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+ 1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^4+2*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*ta n(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticE((tan(1/2*a +1/2*x*b)+1)^(1/2),1/2*2^(1/2))*(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)-1) *(tan(1/2*a+1/2*x*b)+1))^(1/2)-(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+ 1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1/2)*EllipticF((tan(1/2*a+1/2*...
Time = 0.27 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.71 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) + 10 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 10 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 5 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{16 \, b} \]
1/16*(8*sqrt(2)*(4*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a))*sin (b*x + a) + 10*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)) - 10*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))) + 5 *log(-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*c os(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
\[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \]
\[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{7/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]