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

Optimal. Leaf size=79 \[ -\frac {2 \cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {16 \cos (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \]

[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)

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4393, 4388, 4389, 4376} \begin {gather*} \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)}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\csc (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx &=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*}

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Mathematica [A]
time = 0.14, size = 52, normalized size = 0.66 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 31.79, size = 481, normalized size = 6.09

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

Verification of antiderivative is not currently implemented for this CAS.

[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)+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)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*tan(1/2*a+1/2*x*b)^2-12*(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)*(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)^6-(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)^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)^2-12*(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)^3-tan(1/2*a+1/2*x*b))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 3.41, size = 103, normalized size = 1.30 \begin {gather*} -\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 3.36, size = 136, normalized size = 1.72 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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