\(\int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\) [530]

Optimal result

Integrand size = 22, antiderivative size = 55 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {4 \csc (a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b}-\frac {\csc ^3(a+b x) \sqrt {\sin (2 a+2 b x)}}{5 b} \] Output:

-4/5*csc(b*x+a)*sin(2*b*x+2*a)^(1/2)/b-1/5*csc(b*x+a)^3*sin(2*b*x+2*a)^(1/ 
2)/b
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.64 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\csc (a+b x) \left (4+\csc ^2(a+b x)\right ) \sqrt {\sin (2 (a+b x))}}{5 b} \] Input:

Integrate[Csc[a + b*x]^3/Sqrt[Sin[2*a + 2*b*x]],x]
 

Output:

-1/5*(Csc[a + b*x]*(4 + Csc[a + b*x]^2)*Sqrt[Sin[2*(a + b*x)]])/b
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4788, 3042, 4780}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sqrt {\sin (2 a+2 b x)}}dx\)

\(\Big \downarrow \) 4788

\(\displaystyle \frac {4}{5} \int \frac {\csc (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} \int \frac {1}{\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \csc ^3(a+b x)}{5 b}\)

\(\Big \downarrow \) 4780

\(\displaystyle -\frac {\sqrt {\sin (2 a+2 b x)} \csc ^3(a+b x)}{5 b}-\frac {4 \sqrt {\sin (2 a+2 b x)} \csc (a+b x)}{5 b}\)

Input:

Int[Csc[a + b*x]^3/Sqrt[Sin[2*a + 2*b*x]],x]
 

Output:

(-4*Csc[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(5*b) - (Csc[a + b*x]^3*Sqrt[Sin[ 
2*a + 2*b*x]])/(5*b)
 

Defintions of rubi rules used

rule 3042

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

rule 4780

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)/(b*g*m) 
), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b 
, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
 

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 [C] (verified)

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

Time = 10.22 (sec) , antiderivative size = 482, normalized size of antiderivative = 8.76

method	result	size
default	\(\frac {\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 (16 \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 )+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 {EllipticE}\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}-8 \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 )+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 ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6}+\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 )}\, \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+8 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+\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 )}\, \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-8 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \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 )}\right )}{20 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, b}\)	\(482\)

Input:

int(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/20*(-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*(16*(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 
)*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 
-8*(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)*Elli 
pticF((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)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*tan(1/2*a+1/2*b*x)^6+(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)^4+8*(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+(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)^2-8*(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)^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)/b
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.38 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {\sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 4 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{5 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \] Input:

integrate(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")
 

Output:

-1/5*(sqrt(2)*(4*cos(b*x + a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 4*( 
cos(b*x + a)^2 - 1)*sin(b*x + a))/((b*cos(b*x + a)^2 - b)*sin(b*x + a))
 

Sympy [F(-1)]

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

integrate(csc(b*x+a)**3/sin(2*b*x+2*a)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 24.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.69 \[ \int \frac {\csc ^3(a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx=-\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (-{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{5\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3} \] Input:

int(1/(sin(a + b*x)^3*sin(2*a + 2*b*x)^(1/2)),x)
 

Output:

-(8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)* 
1i)/2)^(1/2)*(exp(a*4i + b*x*4i)*1i - exp(a*2i + b*x*2i)*3i + 1i))/(5*b*(e 
xp(a*2i + b*x*2i) - 1)^3)
 

Reduce [F]

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

int(csc(b*x+a)^3/sin(2*b*x+2*a)^(1/2),x)
 

Output:

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