3.1.53 \(\int \frac {1}{(1-\cos ^2(x))^{3/2}} \, dx\) [53]

Optimal. Leaf size=32 \[ -\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}-\frac {\tanh ^{-1}(\cos (x)) \sin (x)}{2 \sqrt {\sin ^2(x)}} \]

[Out]

-1/2*cot(x)/(sin(x)^2)^(1/2)-1/2*arctanh(cos(x))*sin(x)/(sin(x)^2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3255, 3283, 3286, 3855} \begin {gather*} -\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}-\frac {\sin (x) \tanh ^{-1}(\cos (x))}{2 \sqrt {\sin ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[x]^2)^(-3/2),x]

[Out]

-1/2*Cot[x]/Sqrt[Sin[x]^2] - (ArcTanh[Cos[x]]*Sin[x])/(2*Sqrt[Sin[x]^2])

Rule 3255

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3283

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 3286

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 3855

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

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Mathematica [A]
time = 0.06, size = 51, normalized size = 1.59 \begin {gather*} -\frac {\left (\csc ^2\left (\frac {x}{2}\right )+4 \log \left (\cos \left (\frac {x}{2}\right )\right )-4 \log \left (\sin \left (\frac {x}{2}\right )\right )-\sec ^2\left (\frac {x}{2}\right )\right ) \sin (x)}{8 \sqrt {\sin ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[x]^2)^(-3/2),x]

[Out]

-1/8*((Csc[x/2]^2 + 4*Log[Cos[x/2]] - 4*Log[Sin[x/2]] - Sec[x/2]^2)*Sin[x])/Sqrt[Sin[x]^2]

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Maple [A]
time = 0.35, size = 37, normalized size = 1.16

method	result	size
default	\(-\frac {2 \left (\frac {\cos \left (x \right )}{2}+\frac {\left (\ln \left (\cos \left (x \right )+1\right )-\ln \left (-1+\cos \left (x \right )\right )\right ) \left (\sin ^{2}\left (x \right )\right )}{4}\right )}{\sin \left (x \right ) \sqrt {2-2 \cos \left (2 x \right )}}\)	\(37\)
risch	\(-\frac {i \left ({\mathrm e}^{2 i x}+1\right )}{\left ({\mathrm e}^{2 i x}-1\right ) \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(1/2*cos(x)+1/4*(ln(cos(x)+1)-ln(-1+cos(x)))*sin(x)^2)/sin(x)/(sin(x)^2)^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (24) = 48\).
time = 0.52, size = 300, normalized size = 9.38 \begin {gather*} \frac {4 \, {\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \cos \left (4 \, x\right ) - 4 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \cos \left (x\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) + 4 \, {\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - 8 \, \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - 8 \, \sin \left (2 \, x\right ) \sin \left (x\right ) + 4 \, \cos \left (x\right )}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/4*(4*(cos(3*x) + cos(x))*cos(4*x) - 4*(2*cos(2*x) - 1)*cos(3*x) - 8*cos(2*x)*cos(x) + (2*(2*cos(2*x) - 1)*co
s(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*log(cos
(x)^2 + sin(x)^2 + 2*cos(x) + 1) - (2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*s
in(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + 4*(sin(3*x) + sin(
x))*sin(4*x) - 8*sin(3*x)*sin(2*x) - 8*sin(2*x)*sin(x) + 4*cos(x))/(2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 -
4*cos(2*x)^2 - sin(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)

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Fricas [A]
time = 0.38, size = 44, normalized size = 1.38 \begin {gather*} -\frac {{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 2 \, \cos \left (x\right )}{4 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/4*((cos(x)^2 - 1)*log(1/2*cos(x) + 1/2) - (cos(x)^2 - 1)*log(-1/2*cos(x) + 1/2) - 2*cos(x))/(cos(x)^2 - 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (1 - \cos ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)**2)**(3/2),x)

[Out]

Integral((1 - cos(x)**2)**(-3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (24) = 48\).
time = 0.42, size = 78, normalized size = 2.44 \begin {gather*} \frac {\tan \left (\frac {1}{2} \, x\right )^{2}}{8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} + \frac {\log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{4 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} - \frac {2 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1}{8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="giac")

[Out]

1/8*tan(1/2*x)^2/sgn(tan(1/2*x)^3 + tan(1/2*x)) + 1/4*log(tan(1/2*x)^2)/sgn(tan(1/2*x)^3 + tan(1/2*x)) - 1/8*(
2*tan(1/2*x)^2 + 1)/(sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{{\left (1-{\cos \left (x\right )}^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - cos(x)^2)^(3/2),x)

[Out]

int(1/(1 - cos(x)^2)^(3/2), x)

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