\(\int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [526]

Optimal result

Integrand size = 22, antiderivative size = 164 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{3 b}+\frac {4 \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{3 b}+\frac {\csc ^3(a+b x) \sin ^{\frac {9}{2}}(2 a+2 b x)}{3 b} \] Output:

-5/4*arcsin(cos(b*x+a)-sin(b*x+a))/b-5/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b* 
x+2*a)^(1/2))/b+5/2*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b-5/3*cos(b*x+a)*sin(2 
*b*x+2*a)^(3/2)/b+4/3*sin(b*x+a)*sin(2*b*x+2*a)^(5/2)/b+1/3*csc(b*x+a)^3*s 
in(2*b*x+2*a)^(9/2)/b
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {-5 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 \sqrt {\sin (2 (a+b x))} (6 \sin (a+b x)+\sin (3 (a+b x)))}{4 b} \] Input:

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(7/2),x]
 

Output:

(-5*(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*Sqrt[Sin[2*(a + b*x)]]*(6*Sin[a + b*x] + S 
in[3*(a + b*x)]))/(4*b)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4788, 3042, 4796, 3042, 4789, 3042, 4790, 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.

\(\displaystyle \int \sin ^{\frac {7}{2}}(2 a+2 b x) \csc ^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (2 a+2 b x)^{7/2}}{\sin (a+b x)^3}dx\)

\(\Big \downarrow \) 4788

\(\displaystyle 4 \int \csc (a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 \int \frac {\sin (2 a+2 b x)^{7/2}}{\sin (a+b x)}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 4796

\(\displaystyle 8 \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle 8 \int \cos (a+b x) \sin (2 a+2 b x)^{5/2}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 4789

\(\displaystyle 8 \left (\frac {5}{6} \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle 8 \left (\frac {5}{6} \int \sin (a+b x) \sin (2 a+2 b x)^{3/2}dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 4790

\(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 4789

\(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

\(\Big \downarrow \) 4794

\(\displaystyle 8 \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x) \csc ^3(a+b x)}{3 b}\)

Input:

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(7/2),x]
 

Output:

(Csc[a + b*x]^3*Sin[2*a + 2*b*x]^(9/2))/(3*b) + 8*((Sin[a + b*x]*Sin[2*a + 
 2*b*x]^(5/2))/(6*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 + (S 
in[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 - (Cos[a + b*x]*Sin[2*a + 2* 
b*x]^(3/2))/(4*b)))/6)
 

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 4794

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]
 

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 = 100.34 (sec) , antiderivative size = 973, normalized size of antiderivative = 5.93

method	result	size
default	\(\text {Expression too large to display}\)	\(973\)

Input:

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

32/5*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(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)+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) 
*EllipticE((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- 
(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)+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)^3-tan(1/2*a+1/2*b*x))^(1/2)*tan(1/2*a+1 
/2*b*x)^6-4*(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)+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)*EllipticE((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+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)+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)^2-4*(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)^4+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)+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)*EllipticE((tan(1/2*a+ 
1/2*b*x)+1)^(1/2),1/2*2^(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)+1)^(1/2)*(-2*tan(1/2*...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.71 \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) + 10 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 10 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 5 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{16 \, b} \] Input:

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")
 

Output:

1/16*(8*sqrt(2)*(4*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a))*sin 
(b*x + a) + 10*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)) - 10*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))) + 5 
*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*c 
os(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1))/b
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")
 

Output:

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(7/2), x)
 

Giac [F]

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

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2),x, algorithm="giac")
 

Output:

integrate(csc(b*x + a)^3*sin(2*b*x + 2*a)^(7/2), x)
 

Mupad [F(-1)]

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

int(sin(2*a + 2*b*x)^(7/2)/sin(a + b*x)^3,x)
 

Output:

int(sin(2*a + 2*b*x)^(7/2)/sin(a + b*x)^3, x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int(csc(b*x+a)^3*sin(2*b*x+2*a)^(7/2),x)
 

Output:

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