3.6 \(\int \frac {1}{(a \sin ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=61 \[ -\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {3 \sin (x) \tanh ^{-1}(\cos (x))}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}} \]

[Out]

-1/4*cot(x)/a/(a*sin(x)^2)^(3/2)-3/8*cot(x)/a^2/(a*sin(x)^2)^(1/2)-3/8*arctanh(cos(x))*sin(x)/a^2/(a*sin(x)^2)
^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3204, 3207, 3770} \[ -\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {3 \sin (x) \tanh ^{-1}(\cos (x))}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[x]/(4*a*(a*Sin[x]^2)^(3/2)) - (3*Cot[x])/(8*a^2*Sqrt[a*Sin[x]^2]) - (3*ArcTanh[Cos[x]]*Sin[x])/(8*a^2*Sqr
t[a*Sin[x]^2])

Rule 3204

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^(p + 1))/(b*f*(
2*p + 1)), x] + Dist[(2*(p + 1))/(b*(2*p + 1)), Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x]
&&  !IntegerQ[p] && LtQ[p, -1]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \sin ^2(x)\right )^{5/2}} \, dx &=-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}+\frac {3 \int \frac {1}{\left (a \sin ^2(x)\right )^{3/2}} \, dx}{4 a}\\ &=-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}-\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}+\frac {3 \int \frac {1}{\sqrt {a \sin ^2(x)}} \, dx}{8 a^2}\\ &=-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}-\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}+\frac {(3 \sin (x)) \int \csc (x) \, dx}{8 a^2 \sqrt {a \sin ^2(x)}}\\ &=-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}-\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {3 \tanh ^{-1}(\cos (x)) \sin (x)}{8 a^2 \sqrt {a \sin ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 77, normalized size = 1.26 \[ -\frac {\csc (x) \sqrt {a \sin ^2(x)} \left (\csc ^4\left (\frac {x}{2}\right )+6 \csc ^2\left (\frac {x}{2}\right )-\sec ^4\left (\frac {x}{2}\right )-6 \sec ^2\left (\frac {x}{2}\right )+24 \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )\right )}{64 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(Csc[x]*(6*Csc[x/2]^2 + Csc[x/2]^4 + 24*(Log[Cos[x/2]] - Log[Sin[x/2]]) - 6*Sec[x/2]^2 - Sec[x/2]^4)*Sqr
t[a*Sin[x]^2])/a^3

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fricas [A]  time = 0.45, size = 78, normalized size = 1.28 \[ \frac {\sqrt {-a \cos \relax (x)^{2} + a} {\left (6 \, \cos \relax (x)^{3} + 3 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \log \left (-\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1}\right ) - 10 \, \cos \relax (x)\right )}}{16 \, {\left (a^{3} \cos \relax (x)^{4} - 2 \, a^{3} \cos \relax (x)^{2} + a^{3}\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(-a*cos(x)^2 + a)*(6*cos(x)^3 + 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(-(cos(x) - 1)/(cos(x) + 1)) - 10*co
s(x))/((a^3*cos(x)^4 - 2*a^3*cos(x)^2 + a^3)*sin(x))

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giac [A]  time = 0.25, size = 80, normalized size = 1.31 \[ \frac {\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + 8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2} + \frac {12 \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right )} - \frac {18 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 8 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4}}}{64 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sgn(tan(1/2*x))*tan(1/2*x)^4 + 8*sgn(tan(1/2*x))*tan(1/2*x)^2 + 12*log(tan(1/2*x)^2)/sgn(tan(1/2*x)) - (
18*tan(1/2*x)^4 + 8*tan(1/2*x)^2 + 1)/(sgn(tan(1/2*x))*tan(1/2*x)^4))/a^(5/2)

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maple [A]  time = 1.45, size = 89, normalized size = 1.46 \[ -\frac {\sqrt {a \left (\cos ^{2}\relax (x )\right )}\, \left (3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\relax (x )\right )}+2 a}{\sin \relax (x )}\right ) a \left (\sin ^{4}\relax (x )\right )+3 \sqrt {a \left (\cos ^{2}\relax (x )\right )}\, \left (\sin ^{2}\relax (x )\right ) \sqrt {a}+2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\relax (x )\right )}\right )}{8 a^{\frac {7}{2}} \sin \relax (x )^{3} \cos \relax (x ) \sqrt {a \left (\sin ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a^(7/2)/sin(x)^3*(a*cos(x)^2)^(1/2)*(3*ln(2*(a^(1/2)*(a*cos(x)^2)^(1/2)+a)/sin(x))*a*sin(x)^4+3*(a*cos(x)
^2)^(1/2)*sin(x)^2*a^(1/2)+2*a^(1/2)*(a*cos(x)^2)^(1/2))/cos(x)/(a*sin(x)^2)^(1/2)

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maxima [B]  time = 0.87, size = 931, normalized size = 15.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*(2*(4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 8*(6*cos(4*x) - 4*cos(2*x) + 1)*
cos(6*x) - 16*cos(6*x)^2 + 12*(4*cos(2*x) - 1)*cos(4*x) - 36*cos(4*x)^2 - 16*cos(2*x)^2 + 4*(2*sin(6*x) - 3*si
n(4*x) + 2*sin(2*x))*sin(8*x) - sin(8*x)^2 + 16*(3*sin(4*x) - 2*sin(2*x))*sin(6*x) - 16*sin(6*x)^2 - 36*sin(4*
x)^2 + 48*sin(4*x)*sin(2*x) - 16*sin(2*x)^2 + 8*cos(2*x) - 1)*arctan2(sin(x), cos(x) + 1) - 3*(2*(4*cos(6*x) -
6*cos(4*x) + 4*cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 8*(6*cos(4*x) - 4*cos(2*x) + 1)*cos(6*x) - 16*cos(6*x)^2
+ 12*(4*cos(2*x) - 1)*cos(4*x) - 36*cos(4*x)^2 - 16*cos(2*x)^2 + 4*(2*sin(6*x) - 3*sin(4*x) + 2*sin(2*x))*sin
(8*x) - sin(8*x)^2 + 16*(3*sin(4*x) - 2*sin(2*x))*sin(6*x) - 16*sin(6*x)^2 - 36*sin(4*x)^2 + 48*sin(4*x)*sin(2
*x) - 16*sin(2*x)^2 + 8*cos(2*x) - 1)*arctan2(sin(x), cos(x) - 1) + 2*(3*sin(7*x) - 11*sin(5*x) - 11*sin(3*x)
+ 3*sin(x))*cos(8*x) + 12*(2*sin(6*x) - 3*sin(4*x) + 2*sin(2*x))*cos(7*x) + 8*(11*sin(5*x) + 11*sin(3*x) - 3*s
in(x))*cos(6*x) + 44*(3*sin(4*x) - 2*sin(2*x))*cos(5*x) - 12*(11*sin(3*x) - 3*sin(x))*cos(4*x) - 2*(3*cos(7*x)
- 11*cos(5*x) - 11*cos(3*x) + 3*cos(x))*sin(8*x) - 6*(4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*sin(7*x) - 8*
(11*cos(5*x) + 11*cos(3*x) - 3*cos(x))*sin(6*x) - 22*(6*cos(4*x) - 4*cos(2*x) + 1)*sin(5*x) + 12*(11*cos(3*x)
- 3*cos(x))*sin(4*x) + 22*(4*cos(2*x) - 1)*sin(3*x) - 88*cos(3*x)*sin(2*x) + 24*cos(x)*sin(2*x) - 24*cos(2*x)*
sin(x) + 6*sin(x))*sqrt(-a)/(a^3*cos(8*x)^2 + 16*a^3*cos(6*x)^2 + 36*a^3*cos(4*x)^2 + 16*a^3*cos(2*x)^2 + a^3*
sin(8*x)^2 + 16*a^3*sin(6*x)^2 + 36*a^3*sin(4*x)^2 - 48*a^3*sin(4*x)*sin(2*x) + 16*a^3*sin(2*x)^2 - 8*a^3*cos(
2*x) + a^3 - 2*(4*a^3*cos(6*x) - 6*a^3*cos(4*x) + 4*a^3*cos(2*x) - a^3)*cos(8*x) - 8*(6*a^3*cos(4*x) - 4*a^3*c
os(2*x) + a^3)*cos(6*x) - 12*(4*a^3*cos(2*x) - a^3)*cos(4*x) - 4*(2*a^3*sin(6*x) - 3*a^3*sin(4*x) + 2*a^3*sin(
2*x))*sin(8*x) - 16*(3*a^3*sin(4*x) - 2*a^3*sin(2*x))*sin(6*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (a\,{\sin \relax (x)}^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*sin(x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sin ^{2}{\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)**2)**(5/2),x)

[Out]

Integral((a*sin(x)**2)**(-5/2), x)

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