Integrand size = 22, antiderivative size = 127 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b}-\frac {6 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b} \]
-3*arcsin(cos(b*x+a)-sin(b*x+a))/b+3*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2* a)^(1/2))/b+4*sin(b*x+a)*sin(2*b*x+2*a)^(3/2)/b+csc(b*x+a)^3*sin(2*b*x+2*a )^(7/2)/b-6*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b
Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.55 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {-3 \arcsin (\cos (a+b x)-\sin (a+b x))+3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+\csc (a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{b} \]
(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]] + 3*Log[Cos[a + b*x] + Sin[a + b*x ] + Sqrt[Sin[2*(a + b*x)]]] + Csc[a + b*x]*Sin[2*(a + b*x)]^(3/2))/b
Time = 0.61 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4788, 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.
\(\displaystyle \int \sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (2 a+2 b x)^{5/2}}{\sin (a+b x)^3}dx\) |
\(\Big \downarrow \) 4788 |
\(\displaystyle 8 \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 8 \int \frac {\sin (2 a+2 b x)^{5/2}}{\sin (a+b x)}dx+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle 16 \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \int \cos (a+b x) \sin (2 a+2 b x)^{3/2}dx+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle 16 \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4790 |
\(\displaystyle 16 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4793 |
\(\displaystyle 16 \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
(Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(7/2))/b + 16*((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 - (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))
3.2.17.3.1 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[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 contains higher order function than in optimal. Order 4 vs. order 3.
Time = 107.88 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.91
method | result | size |
default | \(\frac {16 \sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )\right )}{3 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, b}\) | \(243\) |
16/3*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(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*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^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)^3-tan(1/2*a+1/2*x*b))/(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)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (119) = 238\).
Time = 0.27 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.11 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{4 \, b} \]
1/4*(8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*cos(b*x + a) + 6*arctan(-(s qrt(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*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*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
\[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
\[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]