\(\int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) [527]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{b}-\frac {6 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {4 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{b}+\frac {\csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)}{b} \] Output:

-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-6*cos(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
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.55 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {-3 \arcsin (\cos (a+b x)-\sin (a+b x))+3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+\csc (a+b x) \sin ^{\frac {3}{2}}(2 (a+b x))}{b} \] Input:

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(5/2),x]
 

Output:

(-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
 

Rubi [A] (verified)

Time = 0.60 (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

\(\displaystyle 16 \left (\frac {3}{4} \left (\frac {1}{2} \left (\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}\right )-\frac {\sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{2 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\right )+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x)}{b}\)

Input:

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(5/2),x]
 

Output:

(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))
 

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 = 103.65 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.91

method	result	size
default	\(\frac {16 \sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1}}\, \left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{3 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, b}\)	\(243\)

Input:

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

16/3*(-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)^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)^3-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)^2-1))^(1/2)/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (119) = 238\).

Time = 0.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.11 \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{4 \, b} \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")
 

Output:

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
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(5/2), x)
 

Giac [F]

\[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")
 

Output:

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{5/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \] Input:

int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x)^3,x)
 

Output:

int(sin(2*a + 2*b*x)^(5/2)/sin(a + b*x)^3, x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \csc ^3(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{2}d x \] Input:

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(5/2),x)
 

Output:

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