\(\int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx\) [525]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-1)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 190 \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=-\frac {7 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}-\frac {14 \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{15 b}+\frac {4 \sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {11}{2}}(2 a+2 b x)}{5 b} \] Output:

-7/8*arcsin(cos(b*x+a)-sin(b*x+a))/b+7/8*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b* 
x+2*a)^(1/2))/b-7/4*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b+7/6*sin(b*x+a)*sin(2 
*b*x+2*a)^(3/2)/b-14/15*cos(b*x+a)*sin(2*b*x+2*a)^(5/2)/b+4/5*sin(b*x+a)*s 
in(2*b*x+2*a)^(7/2)/b+1/5*csc(b*x+a)^3*sin(2*b*x+2*a)^(11/2)/b
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.53 \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\frac {7 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )-\frac {2}{3} (10 \cos (a+b x)+9 \cos (3 (a+b x))+2 \cos (5 (a+b x))) \sqrt {\sin (2 (a+b x))}}{8 b} \] Input:

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2),x]
 

Output:

(7*(-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*(10*Cos[a + b*x] + 9*Cos[3*(a + b*x)] + 2 
*Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/3)/(8*b)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4788, 3042, 4796, 3042, 4789, 3042, 4790, 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.

\(\displaystyle \int \sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (2 a+2 b x)^{9/2}}{\sin (a+b x)^3}dx\)

\(\Big \downarrow \) 4788

\(\displaystyle \frac {16}{5} \int \csc (a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {16}{5} \int \frac {\sin (2 a+2 b x)^{9/2}}{\sin (a+b x)}dx+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4796

\(\displaystyle \frac {32}{5} \int \cos (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {32}{5} \int \cos (a+b x) \sin (2 a+2 b x)^{7/2}dx+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4789

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)dx+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \int \sin (a+b x) \sin (2 a+2 b x)^{5/2}dx+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4790

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \int \cos (a+b x) \sin (2 a+2 b x)^{3/2}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4789

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)}dx+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4790

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4793

\(\displaystyle \frac {32}{5} \left (\frac {7}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{8 b}\right )+\frac {\sin ^{\frac {11}{2}}(2 a+2 b x) \csc ^3(a+b x)}{5 b}\)

Input:

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2),x]
 

Output:

(Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(11/2))/(5*b) + (32*((Sin[a + b*x]*Sin[2* 
a + 2*b*x]^(7/2))/(8*b) + (7*(-1/6*(Cos[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/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 - (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)))/ 
6))/8))/5
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4788
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]
 

rule 4789
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]
 

rule 4790
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]
 

rule 4793
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]
 

rule 4796
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]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 108.87 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.32

method result size
default \(-\frac {64 \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 (\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 )^{6}-3 \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 )^{4}+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{7}+3 \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 )^{2}+10 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}-\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 )+10 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}+2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{21 \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 )}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )-1\right )^{2} \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1\right )^{2} b}\) \(441\)

Input:

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-64/21*(-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)+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)^6 
-3*(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)^4+2*tan(1/2*a+1/2*b*x)^7+3*(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)^2+10*tan(1/2*a+1/2*b 
*x)^5-(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))+1 
0*tan(1/2*a+1/2*b*x)^3+2*tan(1/2*a+1/2*b*x))/(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)-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)+1)^2/b
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.53 \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 4 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 42 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 42 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 21 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{96 \, b} \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")
 

Output:

-1/96*(8*sqrt(2)*(32*cos(b*x + a)^5 - 4*cos(b*x + a)^3 - 7*cos(b*x + a))*s 
qrt(cos(b*x + a)*sin(b*x + a)) - 42*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)) + 42*arctan(-(2*sqrt(2)*sq 
rt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) 
 - sin(b*x + a))) + 21*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
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(9/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{9/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

int(sin(2*a + 2*b*x)^(9/2)/sin(a + b*x)^3,x)
 

Output:

int(sin(2*a + 2*b*x)^(9/2)/sin(a + b*x)^3, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{4}d x \] Input:

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(9/2),x)
 

Output:

int(sqrt(sin(2*a + 2*b*x))*csc(a + b*x)**3*sin(2*a + 2*b*x)**4,x)