3.67 \(\int \frac{1}{(a \csc ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{63 x \csc ^2(x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{63 \cot (x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{\sin ^7(x) \cos (x)}{10 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \sin ^5(x) \cos (x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \sin ^3(x) \cos (x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \sin (x) \cos (x)}{128 a^2 \sqrt{a \csc ^4(x)}} \]

[Out]

(-63*Cot[x])/(256*a^2*Sqrt[a*Csc[x]^4]) + (63*x*Csc[x]^2)/(256*a^2*Sqrt[a*Csc[x]^4]) - (21*Cos[x]*Sin[x])/(128
*a^2*Sqrt[a*Csc[x]^4]) - (21*Cos[x]*Sin[x]^3)/(160*a^2*Sqrt[a*Csc[x]^4]) - (9*Cos[x]*Sin[x]^5)/(80*a^2*Sqrt[a*
Csc[x]^4]) - (Cos[x]*Sin[x]^7)/(10*a^2*Sqrt[a*Csc[x]^4])

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Rubi [A]  time = 0.0446798, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4123, 2635, 8} \[ \frac{63 x \csc ^2(x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{63 \cot (x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{\sin ^7(x) \cos (x)}{10 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \sin ^5(x) \cos (x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \sin ^3(x) \cos (x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \sin (x) \cos (x)}{128 a^2 \sqrt{a \csc ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Csc[x]^4)^(-5/2),x]

[Out]

(-63*Cot[x])/(256*a^2*Sqrt[a*Csc[x]^4]) + (63*x*Csc[x]^2)/(256*a^2*Sqrt[a*Csc[x]^4]) - (21*Cos[x]*Sin[x])/(128
*a^2*Sqrt[a*Csc[x]^4]) - (21*Cos[x]*Sin[x]^3)/(160*a^2*Sqrt[a*Csc[x]^4]) - (9*Cos[x]*Sin[x]^5)/(80*a^2*Sqrt[a*
Csc[x]^4]) - (Cos[x]*Sin[x]^7)/(10*a^2*Sqrt[a*Csc[x]^4])

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{\left (a \csc ^4(x)\right )^{5/2}} \, dx &=\frac{\csc ^2(x) \int \sin ^{10}(x) \, dx}{a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}+\frac{\left (9 \csc ^2(x)\right ) \int \sin ^8(x) \, dx}{10 a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{9 \cos (x) \sin ^5(x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}+\frac{\left (63 \csc ^2(x)\right ) \int \sin ^6(x) \, dx}{80 a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{21 \cos (x) \sin ^3(x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \cos (x) \sin ^5(x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}+\frac{\left (21 \csc ^2(x)\right ) \int \sin ^4(x) \, dx}{32 a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{21 \cos (x) \sin (x)}{128 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \cos (x) \sin ^3(x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \cos (x) \sin ^5(x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}+\frac{\left (63 \csc ^2(x)\right ) \int \sin ^2(x) \, dx}{128 a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{63 \cot (x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \cos (x) \sin (x)}{128 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \cos (x) \sin ^3(x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \cos (x) \sin ^5(x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}+\frac{\left (63 \csc ^2(x)\right ) \int 1 \, dx}{256 a^2 \sqrt{a \csc ^4(x)}}\\ &=-\frac{63 \cot (x)}{256 a^2 \sqrt{a \csc ^4(x)}}+\frac{63 x \csc ^2(x)}{256 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \cos (x) \sin (x)}{128 a^2 \sqrt{a \csc ^4(x)}}-\frac{21 \cos (x) \sin ^3(x)}{160 a^2 \sqrt{a \csc ^4(x)}}-\frac{9 \cos (x) \sin ^5(x)}{80 a^2 \sqrt{a \csc ^4(x)}}-\frac{\cos (x) \sin ^7(x)}{10 a^2 \sqrt{a \csc ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0948023, size = 55, normalized size = 0.42 \[ \frac{\sin ^2(x) (2520 x-2100 \sin (2 x)+600 \sin (4 x)-150 \sin (6 x)+25 \sin (8 x)-2 \sin (10 x)) \sqrt{a \csc ^4(x)}}{10240 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Csc[x]^4)^(-5/2),x]

[Out]

(Sqrt[a*Csc[x]^4]*Sin[x]^2*(2520*x - 2100*Sin[2*x] + 600*Sin[4*x] - 150*Sin[6*x] + 25*Sin[8*x] - 2*Sin[10*x]))
/(10240*a^3)

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Maple [A]  time = 0.245, size = 57, normalized size = 0.4 \begin{align*} -{\frac{128\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{9}-656\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{7}+1368\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{5}-1490\,\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{3}+965\,\cos \left ( x \right ) \sin \left ( x \right ) -315\,x}{1280\, \left ( \sin \left ( x \right ) \right ) ^{10}} \left ({\frac{a}{ \left ( \sin \left ( x \right ) \right ) ^{4}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*csc(x)^4)^(5/2),x)

[Out]

-1/1280*(128*sin(x)*cos(x)^9-656*sin(x)*cos(x)^7+1368*sin(x)*cos(x)^5-1490*sin(x)*cos(x)^3+965*cos(x)*sin(x)-3
15*x)/(a/sin(x)^4)^(5/2)/sin(x)^10

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Maxima [A]  time = 1.53539, size = 119, normalized size = 0.9 \begin{align*} -\frac{965 \, \tan \left (x\right )^{9} + 2370 \, \tan \left (x\right )^{7} + 2688 \, \tan \left (x\right )^{5} + 1470 \, \tan \left (x\right )^{3} + 315 \, \tan \left (x\right )}{1280 \,{\left (a^{\frac{5}{2}} \tan \left (x\right )^{10} + 5 \, a^{\frac{5}{2}} \tan \left (x\right )^{8} + 10 \, a^{\frac{5}{2}} \tan \left (x\right )^{6} + 10 \, a^{\frac{5}{2}} \tan \left (x\right )^{4} + 5 \, a^{\frac{5}{2}} \tan \left (x\right )^{2} + a^{\frac{5}{2}}\right )}} + \frac{63 \, x}{256 \, a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)^4)^(5/2),x, algorithm="maxima")

[Out]

-1/1280*(965*tan(x)^9 + 2370*tan(x)^7 + 2688*tan(x)^5 + 1470*tan(x)^3 + 315*tan(x))/(a^(5/2)*tan(x)^10 + 5*a^(
5/2)*tan(x)^8 + 10*a^(5/2)*tan(x)^6 + 10*a^(5/2)*tan(x)^4 + 5*a^(5/2)*tan(x)^2 + a^(5/2)) + 63/256*x/a^(5/2)

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Fricas [A]  time = 0.523596, size = 236, normalized size = 1.79 \begin{align*} -\frac{{\left (315 \, x \cos \left (x\right )^{2} -{\left (128 \, \cos \left (x\right )^{11} - 784 \, \cos \left (x\right )^{9} + 2024 \, \cos \left (x\right )^{7} - 2858 \, \cos \left (x\right )^{5} + 2455 \, \cos \left (x\right )^{3} - 965 \, \cos \left (x\right )\right )} \sin \left (x\right ) - 315 \, x\right )} \sqrt{\frac{a}{\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1}}}{1280 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)^4)^(5/2),x, algorithm="fricas")

[Out]

-1/1280*(315*x*cos(x)^2 - (128*cos(x)^11 - 784*cos(x)^9 + 2024*cos(x)^7 - 2858*cos(x)^5 + 2455*cos(x)^3 - 965*
cos(x))*sin(x) - 315*x)*sqrt(a/(cos(x)^4 - 2*cos(x)^2 + 1))/a^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \csc ^{4}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)**4)**(5/2),x)

[Out]

Integral((a*csc(x)**4)**(-5/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*csc(x)^4)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError