3.319 \(\int \frac {\cos (x)}{(a+b \sin ^2(x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\sin (x)}{2 a \left (a+b \sin ^2(x)\right )} \]

[Out]

1/2*sin(x)/a/(a+b*sin(x)^2)+1/2*arctan(sin(x)*b^(1/2)/a^(1/2))/a^(3/2)/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3190, 199, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\sin (x)}{2 a \left (a+b \sin ^2(x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(a + b*Sin[x]^2)^2,x]

[Out]

ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]) + Sin[x]/(2*a*(a + b*Sin[x]^2))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\sin (x)}{2 a \left (a+b \sin ^2(x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]) + Sin[x]/(2*a*(a + b*Sin[x]^2))

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fricas [A]  time = 0.44, size = 165, normalized size = 3.44 \[ \left [-\frac {2 \, a b \sin \relax (x) + {\left (b \cos \relax (x)^{2} - a - b\right )} \sqrt {-a b} \log \left (-\frac {b \cos \relax (x)^{2} + 2 \, \sqrt {-a b} \sin \relax (x) + a - b}{b \cos \relax (x)^{2} - a - b}\right )}{4 \, {\left (a^{2} b^{2} \cos \relax (x)^{2} - a^{3} b - a^{2} b^{2}\right )}}, -\frac {a b \sin \relax (x) - {\left (b \cos \relax (x)^{2} - a - b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \sin \relax (x)}{a}\right )}{2 \, {\left (a^{2} b^{2} \cos \relax (x)^{2} - a^{3} b - a^{2} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*b*sin(x) + (b*cos(x)^2 - a - b)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sqrt(-a*b)*sin(x) + a - b)/(b*cos(x
)^2 - a - b)))/(a^2*b^2*cos(x)^2 - a^3*b - a^2*b^2), -1/2*(a*b*sin(x) - (b*cos(x)^2 - a - b)*sqrt(a*b)*arctan(
sqrt(a*b)*sin(x)/a))/(a^2*b^2*cos(x)^2 - a^3*b - a^2*b^2)]

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giac [A]  time = 0.14, size = 38, normalized size = 0.79 \[ \frac {\arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} + \frac {\sin \relax (x)}{2 \, {\left (b \sin \relax (x)^{2} + a\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*a) + 1/2*sin(x)/((b*sin(x)^2 + a)*a)

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maple [A]  time = 0.11, size = 39, normalized size = 0.81 \[ \frac {\sin \relax (x )}{2 a \left (a +b \left (\sin ^{2}\relax (x )\right )\right )}+\frac {\arctan \left (\frac {\sin \relax (x ) b}{\sqrt {a b}}\right )}{2 a \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sin(x)/a/(a+b*sin(x)^2)+1/2/a/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))

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maxima [A]  time = 0.70, size = 38, normalized size = 0.79 \[ \frac {\sin \relax (x)}{2 \, {\left (a b \sin \relax (x)^{2} + a^{2}\right )}} + \frac {\arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sin(x)/(a*b*sin(x)^2 + a^2) + 1/2*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*a)

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mupad [B]  time = 15.38, size = 36, normalized size = 0.75 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \relax (x)}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {b}}+\frac {\sin \relax (x)}{2\,a\,\left (b\,{\sin \relax (x)}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a + b*sin(x)^2)^2,x)

[Out]

atan((b^(1/2)*sin(x))/a^(1/2))/(2*a^(3/2)*b^(1/2)) + sin(x)/(2*a*(a + b*sin(x)^2))

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sympy [A]  time = 15.38, size = 340, normalized size = 7.08 \[ \begin {cases} \frac {\tilde {\infty }}{\sin ^{3}{\relax (x )}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {1}{3 b^{2} \sin ^{3}{\relax (x )}} & \text {for}\: a = 0 \\\frac {\sin {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \\\frac {2 i \sqrt {a} b \sqrt {\frac {1}{b}} \sin {\relax (x )}}{4 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} + 4 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \sin ^{2}{\relax (x )}} + \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sin {\relax (x )} \right )}}{4 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} + 4 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \sin ^{2}{\relax (x )}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sin {\relax (x )} \right )}}{4 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} + 4 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \sin ^{2}{\relax (x )}} + \frac {b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sin {\relax (x )} \right )} \sin ^{2}{\relax (x )}}{4 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} + 4 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \sin ^{2}{\relax (x )}} - \frac {b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sin {\relax (x )} \right )} \sin ^{2}{\relax (x )}}{4 i a^{\frac {5}{2}} b \sqrt {\frac {1}{b}} + 4 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} \sin ^{2}{\relax (x )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(a+b*sin(x)**2)**2,x)

[Out]

Piecewise((zoo/sin(x)**3, Eq(a, 0) & Eq(b, 0)), (-1/(3*b**2*sin(x)**3), Eq(a, 0)), (sin(x)/a**2, Eq(b, 0)), (2
*I*sqrt(a)*b*sqrt(1/b)*sin(x)/(4*I*a**(5/2)*b*sqrt(1/b) + 4*I*a**(3/2)*b**2*sqrt(1/b)*sin(x)**2) + a*log(-I*sq
rt(a)*sqrt(1/b) + sin(x))/(4*I*a**(5/2)*b*sqrt(1/b) + 4*I*a**(3/2)*b**2*sqrt(1/b)*sin(x)**2) - a*log(I*sqrt(a)
*sqrt(1/b) + sin(x))/(4*I*a**(5/2)*b*sqrt(1/b) + 4*I*a**(3/2)*b**2*sqrt(1/b)*sin(x)**2) + b*log(-I*sqrt(a)*sqr
t(1/b) + sin(x))*sin(x)**2/(4*I*a**(5/2)*b*sqrt(1/b) + 4*I*a**(3/2)*b**2*sqrt(1/b)*sin(x)**2) - b*log(I*sqrt(a
)*sqrt(1/b) + sin(x))*sin(x)**2/(4*I*a**(5/2)*b*sqrt(1/b) + 4*I*a**(3/2)*b**2*sqrt(1/b)*sin(x)**2), True))

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