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

Optimal. Leaf size=22 \[ \frac {\tanh ^{-1}(\sin (x))}{2 a^2}+\frac {\tan (x) \sec (x)}{2 a^2} \]

[Out]

1/2*arctanh(sin(x))/a^2+1/2*sec(x)*tan(x)/a^2

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3175, 3768, 3770} \[ \frac {\tanh ^{-1}(\sin (x))}{2 a^2}+\frac {\tan (x) \sec (x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(a - a*Sin[x]^2)^2,x]

[Out]

ArcTanh[Sin[x]]/(2*a^2) + (Sec[x]*Tan[x])/(2*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 3768

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

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

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Mathematica [B]  time = 0.01, size = 45, normalized size = 2.05 \[ \frac {\tan (x) \sec (x)-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 37, normalized size = 1.68 \[ \frac {\cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) - \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) + 2 \, \sin \relax (x)}{4 \, a^{2} \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.15, size = 38, normalized size = 1.73 \[ \frac {\log \left (\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {\log \left (-\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {\sin \relax (x)}{2 \, {\left (\sin \relax (x)^{2} - 1\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 44, normalized size = 2.00 \[ -\frac {1}{4 a^{2} \left (-1+\sin \relax (x )\right )}-\frac {\ln \left (-1+\sin \relax (x )\right )}{4 a^{2}}-\frac {1}{4 a^{2} \left (1+\sin \relax (x )\right )}+\frac {\ln \left (1+\sin \relax (x )\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.37, size = 41, normalized size = 1.86 \[ -\frac {\sin \relax (x)}{2 \, {\left (a^{2} \sin \relax (x)^{2} - a^{2}\right )}} + \frac {\log \left (\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {\log \left (\sin \relax (x) - 1\right )}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 30, normalized size = 1.36 \[ \frac {\mathrm {atanh}\left (\sin \relax (x)\right )}{2\,a^2}-\frac {\sin \relax (x)}{2\,\left (a^2\,{\sin \relax (x)}^2-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a - a*sin(x)^2)^2,x)

[Out]

atanh(sin(x))/(2*a^2) - sin(x)/(2*(a^2*sin(x)^2 - a^2))

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sympy [B]  time = 1.06, size = 117, normalized size = 5.32 \[ - \frac {\log {\left (\sin {\relax (x )} - 1 \right )} \sin ^{2}{\relax (x )}}{4 a^{2} \sin ^{2}{\relax (x )} - 4 a^{2}} + \frac {\log {\left (\sin {\relax (x )} - 1 \right )}}{4 a^{2} \sin ^{2}{\relax (x )} - 4 a^{2}} + \frac {\log {\left (\sin {\relax (x )} + 1 \right )} \sin ^{2}{\relax (x )}}{4 a^{2} \sin ^{2}{\relax (x )} - 4 a^{2}} - \frac {\log {\left (\sin {\relax (x )} + 1 \right )}}{4 a^{2} \sin ^{2}{\relax (x )} - 4 a^{2}} - \frac {2 \sin {\relax (x )}}{4 a^{2} \sin ^{2}{\relax (x )} - 4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sin(x) - 1)*sin(x)**2/(4*a**2*sin(x)**2 - 4*a**2) + log(sin(x) - 1)/(4*a**2*sin(x)**2 - 4*a**2) + log(sin
(x) + 1)*sin(x)**2/(4*a**2*sin(x)**2 - 4*a**2) - log(sin(x) + 1)/(4*a**2*sin(x)**2 - 4*a**2) - 2*sin(x)/(4*a**
2*sin(x)**2 - 4*a**2)

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