Integrand size = 22, antiderivative size = 104 \[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \arcsin (\cos (a+b x)-\sin (a+b x))}{b}+\frac {2 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b}-\frac {4 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}-\frac {\csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b} \] Output:
2*arcsin(cos(b*x+a)-sin(b*x+a))/b+2*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)^(1/2)/b-csc(b*x+a)^3*sin(2*b*x+2*a) ^(5/2)/b
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {2 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-2 \csc (a+b x) \sqrt {\sin (2 (a+b x))}\right )}{b} \] Input:
Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(3/2),x]
Output:
(2*(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*Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]]))/b
Time = 0.49 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4788, 3042, 4796, 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.
\(\displaystyle \int \sin ^{\frac {3}{2}}(2 a+2 b x) \csc ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (2 a+2 b x)^{3/2}}{\sin (a+b x)^3}dx\) |
\(\Big \downarrow \) 4788 |
\(\displaystyle -4 \int \csc (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -4 \int \frac {\sin (2 a+2 b x)^{3/2}}{\sin (a+b x)}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4796 |
\(\displaystyle -8 \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -8 \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4789 |
\(\displaystyle -8 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -8 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
\(\Big \downarrow \) 4794 |
\(\displaystyle -8 \left (\frac {1}{2} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\) |
Input:
Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(3/2),x]
Output:
-8*((-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 + (Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)) - (Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(5/2))/b
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_)]/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 = 9.77 (sec) , antiderivative size = 542, normalized size of antiderivative = 5.21
method | result | size |
default | \(\frac {4 \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 (4 \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 )}\, \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 ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )-1\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-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 )}\, \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 ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )-1\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )+\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\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 )-1\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}\, \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2} \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 )}-\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\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 )-1\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\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 )-1\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )}\, b}\) | \(542\) |
Input:
int(csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
4*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(4*(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)*( 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)*(ta n(1/2*a+1/2*b*x)-1)*(tan(1/2*a+1/2*b*x)+1))^(1/2)*EllipticE((tan(1/2*a+1/2 *b*x)+1)^(1/2),1/2*2^(1/2))-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)*(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)*(tan(1/2*a+1/2*b*x)-1)*(tan(1/2 *a+1/2*b*x)+1))^(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)^3-tan(1/2*a+1/2*b*x))^(1/2)*(tan(1/2*a+1/2*b*x)*(tan(1 /2*a+1/2*b*x)-1)*(tan(1/2*a+1/2*b*x)+1))^(1/2)*tan(1/2*a+1/2*b*x)^2+2*tan( 1/2*a+1/2*b*x)^2*(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)-(tan(1/2*a+1/2*b*x)^3-tan(1/2*a +1/2*b*x))^(1/2)*(tan(1/2*a+1/2*b*x)*(tan(1/2*a+1/2*b*x)-1)*(tan(1/2*a+1/2 *b*x)+1))^(1/2))/tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^3-tan(1/2*a+1/2*b* x))^(1/2)/(tan(1/2*a+1/2*b*x)*(tan(1/2*a+1/2*b*x)-1)*(tan(1/2*a+1/2*b*x)+1 ))^(1/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (98) = 196\).
Time = 0.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.84 \[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) \sin \left (b x + a\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) \sin \left (b x + a\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 8 \, \sin \left (b x + a\right )}{2 \, b \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")
Output:
-1/2*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - s in(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))*sin(b*x + a) - 2*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)) )*sin(b*x + a) + 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)*sin(b*x + a) + 8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) + 8*sin(b*x + a))/(b*sin( b*x + a))
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**(3/2),x)
Output:
Timed out
\[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")
Output:
integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(3/2), x)
\[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(3/2), x)
Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{3/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:
int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^3,x)
Output:
int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^3, x)
\[ \int \csc ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )d x \] Input:
int(csc(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x)
Output:
int(sqrt(sin(2*a + 2*b*x))*csc(a + b*x)**3*sin(2*a + 2*b*x),x)