3.263 \(\int \frac {\cos ^5(x)}{a-a \sin ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {\sin (x)}{a}-\frac {\sin ^3(x)}{3 a} \]

[Out]

sin(x)/a-1/3*sin(x)^3/a

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Rubi [A]  time = 0.05, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3175, 2633} \[ \frac {\sin (x)}{a}-\frac {\sin ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^5/(a - a*Sin[x]^2),x]

[Out]

Sin[x]/a - Sin[x]^3/(3*a)

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^5(x)}{a-a \sin ^2(x)} \, dx &=\frac {\int \cos ^3(x) \, dx}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (x)\right )}{a}\\ &=\frac {\sin (x)}{a}-\frac {\sin ^3(x)}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.06 \[ \frac {\frac {3 \sin (x)}{4}+\frac {1}{12} \sin (3 x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*Sin[x])/4 + Sin[3*x]/12)/a

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fricas [A]  time = 0.44, size = 13, normalized size = 0.72 \[ \frac {{\left (\cos \relax (x)^{2} + 2\right )} \sin \relax (x)}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(cos(x)^2 + 2)*sin(x)/a

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giac [A]  time = 0.14, size = 14, normalized size = 0.78 \[ -\frac {\sin \relax (x)^{3} - 3 \, \sin \relax (x)}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(sin(x)^3 - 3*sin(x))/a

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maple [A]  time = 0.15, size = 14, normalized size = 0.78 \[ \frac {-\frac {\left (\sin ^{3}\relax (x )\right )}{3}+\sin \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*sin(x)^3+sin(x))

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maxima [A]  time = 0.32, size = 14, normalized size = 0.78 \[ -\frac {\sin \relax (x)^{3} - 3 \, \sin \relax (x)}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(sin(x)^3 - 3*sin(x))/a

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mupad [B]  time = 0.06, size = 16, normalized size = 0.89 \[ \frac {3\,\sin \relax (x)-{\sin \relax (x)}^3}{3\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^5/(a - a*sin(x)^2),x)

[Out]

(3*sin(x) - sin(x)^3)/(3*a)

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sympy [B]  time = 7.95, size = 124, normalized size = 6.89 \[ \frac {6 \tan ^{5}{\left (\frac {x}{2} \right )}}{3 a \tan ^{6}{\left (\frac {x}{2} \right )} + 9 a \tan ^{4}{\left (\frac {x}{2} \right )} + 9 a \tan ^{2}{\left (\frac {x}{2} \right )} + 3 a} + \frac {4 \tan ^{3}{\left (\frac {x}{2} \right )}}{3 a \tan ^{6}{\left (\frac {x}{2} \right )} + 9 a \tan ^{4}{\left (\frac {x}{2} \right )} + 9 a \tan ^{2}{\left (\frac {x}{2} \right )} + 3 a} + \frac {6 \tan {\left (\frac {x}{2} \right )}}{3 a \tan ^{6}{\left (\frac {x}{2} \right )} + 9 a \tan ^{4}{\left (\frac {x}{2} \right )} + 9 a \tan ^{2}{\left (\frac {x}{2} \right )} + 3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**5/(a-a*sin(x)**2),x)

[Out]

6*tan(x/2)**5/(3*a*tan(x/2)**6 + 9*a*tan(x/2)**4 + 9*a*tan(x/2)**2 + 3*a) + 4*tan(x/2)**3/(3*a*tan(x/2)**6 + 9
*a*tan(x/2)**4 + 9*a*tan(x/2)**2 + 3*a) + 6*tan(x/2)/(3*a*tan(x/2)**6 + 9*a*tan(x/2)**4 + 9*a*tan(x/2)**2 + 3*
a)

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