\(\int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [513]

Optimal result

Integrand size = 20, antiderivative size = 79 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \] Output:

-2/5*cos(b*x+a)/b/sin(2*b*x+2*a)^(5/2)+8/15*sin(b*x+a)/b/sin(2*b*x+2*a)^(3 
/2)-16/15*cos(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (27 \csc (a+b x)+3 \csc ^3(a+b x)-5 \sec (a+b x) \tan (a+b x)\right )}{60 b} \] Input:

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

Output:

-1/60*(Sqrt[Sin[2*(a + b*x)]]*(27*Csc[a + b*x] + 3*Csc[a + b*x]^3 - 5*Sec[ 
a + b*x]*Tan[a + b*x]))/b
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4796, 3042, 4791, 3042, 4792, 3042, 4779}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^{5/2}}dx\)

\(\Big \downarrow \) 4796

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4791

\(\displaystyle 2 \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\)

\(\Big \downarrow \) 4792

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

\(\Big \downarrow \) 3042

\(\displaystyle 2 \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )\)

\(\Big \downarrow \) 4779

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

Input:

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

Output:

2*((4*(Sin[a + b*x]/(3*b*Sin[2*a + 2*b*x]^(3/2)) - (2*Cos[a + b*x])/(3*b*S 
qrt[Sin[2*a + 2*b*x]])))/5 - Cos[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)))
 

Defintions of rubi rules used

rule 3042

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

rule 4779

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( 
p_), x_Symbol] :> Simp[(-(e*Cos[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 4791

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[Cos[a + b*x]*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Simp 
[(2*p + 3)/(2*g*(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] &&  !Int 
egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 4792

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[(-Sin[a + b*x])*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + S 
imp[(2*p + 3)/(2*g*(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] && LtQ[p, -1] && IntegerQ[2*p]
 

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 = 81.95 (sec) , antiderivative size = 481, normalized size of antiderivative = 6.09

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 (24 \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}-12 \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}+12 \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}-12 \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 )}{80 b \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 )}}\)	\(481\)

Input:

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

Output:

-1/80/b*(-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*(24*(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 
)^2-12*(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)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*tan(1/2*a+1/2*b*x)^6-(t 
an(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+12* 
(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)^2-12*(ta 
n(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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{60 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:

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

Output:

-1/60*(sqrt(2)*(32*cos(b*x + a)^4 - 40*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + 
a)*sin(b*x + a)) + 32*(cos(b*x + a)^4 - cos(b*x + a)^2)*sin(b*x + a))/((b* 
cos(b*x + a)^4 - b*cos(b*x + a)^2)*sin(b*x + a))
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 22.91 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.72 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,2{}\mathrm {i}-{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}\,2{}\mathrm {i}-2{}\mathrm {i}\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \] Input:

int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^(5/2)),x)
 

Output:

(8*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1 
i)/2)^(1/2)*(exp(a*2i + b*x*2i)*2i + exp(a*4i + b*x*4i)*3i + exp(a*6i + b* 
x*6i)*2i - exp(a*8i + b*x*8i)*2i - 2i))/(15*b*(exp(a*2i + b*x*2i) - 1)^3*( 
exp(a*2i + b*x*2i) + 1)^2)
 

Reduce [F]

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

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

Output:

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