\(\int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [516]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{3 b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{3 b}-\frac {2 \cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{5 b}+\frac {\csc ^2(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x)}{5 b} \] Output:

2/3*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2))/b-2/3*cos(2*b*x+2*a)*sin(2*b*x+2 
*a)^(1/2)/b-2/5*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(5/2)/b+1/5*csc(b*x+a)^2*sin 
(2*b*x+2*a)^(9/2)/b
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {20 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 (a+b x))}+9 \sin (2 (a+b x))-10 \sin (4 (a+b x))-3 \sin (6 (a+b x))}{30 b \sqrt {\sin (2 (a+b x))}} \] Input:

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(7/2),x]
 

Output:

(20*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] + 9*Sin[2*(a + b*x 
)] - 10*Sin[4*(a + b*x)] - 3*Sin[6*(a + b*x)])/(30*b*Sqrt[Sin[2*(a + b*x)] 
])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.09, 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, 3115, 3042, 3115, 3042, 3120}

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.

\(\displaystyle \int \sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (2 a+2 b x)^{7/2}}{\sin (a+b x)^2}dx\)

\(\Big \downarrow \) 4788

\(\displaystyle \frac {14}{5} \int \sin ^{\frac {7}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {14}{5} \int \sin (2 a+2 b x)^{7/2}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {14}{5} \left (\frac {5}{7} \int \sin ^{\frac {3}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {14}{5} \left (\frac {5}{7} \int \sin (2 a+2 b x)^{3/2}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {14}{5} \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {14}{5} \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b}+\frac {14}{5} \left (\frac {5}{7} \left (\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )\)

Input:

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(7/2),x]
 

Output:

(Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(9/2))/(5*b) + (14*((5*(EllipticF[a - Pi/ 
4 + b*x, 2]/(3*b) - (Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b)))/7 - 
(Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^(5/2))/(7*b)))/5
 

Defintions of rubi rules used

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, 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]
 

Maple [A] (verified)

Time = 32.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {4 \sqrt {2}\, \left (\frac {\sqrt {2}\, \sin \left (2 b x +2 a \right )^{\frac {5}{2}}}{20}+\frac {\sqrt {2}\, \left (\sqrt {\sin \left (2 b x +2 a \right )+1}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 b x +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (2 b x +2 a \right )^{3}-2 \sin \left (2 b x +2 a \right )\right )}{24 \cos \left (2 b x +2 a \right ) \sqrt {\sin \left (2 b x +2 a \right )}}\right )}{b}\)	\(139\)

Input:

int(csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

4*2^(1/2)*(1/20*2^(1/2)*sin(2*b*x+2*a)^(5/2)+1/24*2^(1/2)*((sin(2*b*x+2*a) 
+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*EllipticF((s 
in(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+2*sin(2*b*x+2*a)^3-2*sin(2*b*x+2*a))/c 
os(2*b*x+2*a)/sin(2*b*x+2*a)^(1/2))/b
 

Fricas [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")
 

Output:

integral(-(cos(2*b*x + 2*a)^2 - 1)*csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), 
x)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(7/2), x)
 

Giac [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="giac")
 

Output:

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{7/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \] Input:

int(sin(2*a + 2*b*x)^(7/2)/sin(a + b*x)^2,x)
 

Output:

int(sin(2*a + 2*b*x)^(7/2)/sin(a + b*x)^2, x)
 

Reduce [F]

\[ \int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \csc \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{3}d x \] Input:

int(csc(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x)
 

Output:

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