\(\int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\) [103]

Optimal result

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

[Out]

-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)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4393, 4388, 4389, 4376} \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {8 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]

[In]

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

[Out]

(-2*Cos[a + b*x])/(5*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/(15*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a +
b*x])/(15*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4376

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 4388

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] + Dist[(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] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4389

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] + Dist[(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] && In
tegerQ[2*p]

Rule 4393

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[a + b*x]*(g*S
in[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]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{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}{5} \int \frac {\sin (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}{15} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (27 \csc (a+b x)+3 \csc ^3(a+b x)-5 \sec (a+b x) \tan (a+b x)\right )}{60 b} \]

[In]

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

[Out]

-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

Maple [C] (verified)

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

Time = 16.42 (sec) , antiderivative size = 481, normalized size of antiderivative = 6.09

method	result	size
default	\(-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (24 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-12 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+12 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{4}-12 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}+\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\right )}{80 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\)	\(481\)

[In]

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

[Out]

-1/80/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/tan(1/2*a+1/2*x*b)^3*(24*(tan(1/2*a+1/2*x*b)*(tan
(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))
^(1/2)*EllipticE((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2-12*(tan(1/2*a+1/2*x*b)*(tan(1/
2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1
/2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^2+(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/
2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^6+12*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^
4-(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^4-12*(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+
1/2*x*b))^(1/2)*tan(1/2*a+1/2*x*b)^2-(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^2+
(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2))/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 40 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{60 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 23.76 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.72 \[ \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx=\frac {8\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,2{}\mathrm {i}-{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}\,2{}\mathrm {i}-2{}\mathrm {i}\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2} \]

[In]

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

[Out]

(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*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)