3.276 \(\int \frac {\cos ^3(x)}{(a-a \sin ^2(x))^2} \, dx\)

Optimal. Leaf size=7 \[ \frac {\tanh ^{-1}(\sin (x))}{a^2} \]

[Out]

arctanh(sin(x))/a^2

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Rubi [A]  time = 0.04, antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3175, 3770} \[ \frac {\tanh ^{-1}(\sin (x))}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3/(a - a*Sin[x]^2)^2,x]

[Out]

ArcTanh[Sin[x]]/a^2

Rule 3175

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

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 {\cos ^3(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx &=\frac {\int \sec (x) \, dx}{a^2}\\ &=\frac {\tanh ^{-1}(\sin (x))}{a^2}\\ \end {align*}

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Mathematica [B]  time = 0.00, size = 37, normalized size = 5.29 \[ \frac {\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^3/(a - a*Sin[x]^2)^2,x]

[Out]

(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]])/a^2

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fricas [B]  time = 0.42, size = 20, normalized size = 2.86 \[ \frac {\log \left (\sin \relax (x) + 1\right ) - \log \left (-\sin \relax (x) + 1\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a-a*sin(x)^2)^2,x, algorithm="fricas")

[Out]

1/2*(log(sin(x) + 1) - log(-sin(x) + 1))/a^2

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giac [B]  time = 0.13, size = 23, normalized size = 3.29 \[ \frac {\log \left (\sin \relax (x) + 1\right )}{2 \, a^{2}} - \frac {\log \left (-\sin \relax (x) + 1\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a-a*sin(x)^2)^2,x, algorithm="giac")

[Out]

1/2*log(sin(x) + 1)/a^2 - 1/2*log(-sin(x) + 1)/a^2

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maple [A]  time = 0.18, size = 8, normalized size = 1.14 \[ \frac {\arctanh \left (\sin \relax (x )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a-a*sin(x)^2)^2,x)

[Out]

arctanh(sin(x))/a^2

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maxima [B]  time = 0.33, size = 21, normalized size = 3.00 \[ \frac {\log \left (\sin \relax (x) + 1\right )}{2 \, a^{2}} - \frac {\log \left (\sin \relax (x) - 1\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a-a*sin(x)^2)^2,x, algorithm="maxima")

[Out]

1/2*log(sin(x) + 1)/a^2 - 1/2*log(sin(x) - 1)/a^2

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mupad [B]  time = 0.06, size = 7, normalized size = 1.00 \[ \frac {\mathrm {atanh}\left (\sin \relax (x)\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a - a*sin(x)^2)^2,x)

[Out]

atanh(sin(x))/a^2

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sympy [B]  time = 7.46, size = 22, normalized size = 3.14 \[ - \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} - 1 \right )}}{a^{2}} + \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**3/(a-a*sin(x)**2)**2,x)

[Out]

-log(tan(x/2) - 1)/a**2 + log(tan(x/2) + 1)/a**2

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