3.857 \(\int \frac{\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx\)

Optimal. Leaf size=44 \[ \frac{1}{3 (2-\sin (x))}+\frac{1}{2} \log (1-\sin (x))-\frac{4}{9} \log (2-\sin (x))-\frac{1}{18} \log (\sin (x)+1) \]

[Out]

Log[1 - Sin[x]]/2 - (4*Log[2 - Sin[x]])/9 - Log[1 + Sin[x]]/18 + 1/(3*(2 - Sin[x]))

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Rubi [A]  time = 0.0556506, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {710, 801} \[ \frac{1}{3 (2-\sin (x))}+\frac{1}{2} \log (1-\sin (x))-\frac{4}{9} \log (2-\sin (x))-\frac{1}{18} \log (\sin (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(-5 + Cos[x]^2 + 4*Sin[x]),x]

[Out]

Log[1 - Sin[x]]/2 - (4*Log[2 - Sin[x]])/9 - Log[1 + Sin[x]]/18 + 1/(3*(2 - Sin[x]))

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\sec (x)}{-5+\cos ^2(x)+4 \sin (x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{(2-x)^2 \left (-1+x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac{1}{3 (2-\sin (x))}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{2+x}{(2-x) \left (-1+x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac{1}{3 (2-\sin (x))}+\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{4}{3 (-2+x)}+\frac{3}{2 (-1+x)}-\frac{1}{6 (1+x)}\right ) \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \log (1-\sin (x))-\frac{4}{9} \log (2-\sin (x))-\frac{1}{18} \log (1+\sin (x))+\frac{1}{3 (2-\sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.0769394, size = 38, normalized size = 0.86 \[ \frac{1}{18} \left (-\frac{6}{\sin (x)-2}+9 \log (1-\sin (x))-8 \log (2-\sin (x))-\log (\sin (x)+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(-5 + Cos[x]^2 + 4*Sin[x]),x]

[Out]

(9*Log[1 - Sin[x]] - 8*Log[2 - Sin[x]] - Log[1 + Sin[x]] - 6/(-2 + Sin[x]))/18

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Maple [A]  time = 0.071, size = 31, normalized size = 0.7 \begin{align*} -{\frac{1}{3\,\sin \left ( x \right ) -6}}-{\frac{4\,\ln \left ( \sin \left ( x \right ) -2 \right ) }{9}}-{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{18}}+{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(-5+cos(x)^2+4*sin(x)),x)

[Out]

-1/3/(sin(x)-2)-4/9*ln(sin(x)-2)-1/18*ln(1+sin(x))+1/2*ln(sin(x)-1)

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Maxima [A]  time = 0.962276, size = 41, normalized size = 0.93 \begin{align*} -\frac{1}{3 \,{\left (\sin \left (x\right ) - 2\right )}} - \frac{1}{18} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac{1}{2} \, \log \left (\sin \left (x\right ) - 1\right ) - \frac{4}{9} \, \log \left (\sin \left (x\right ) - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="maxima")

[Out]

-1/3/(sin(x) - 2) - 1/18*log(sin(x) + 1) + 1/2*log(sin(x) - 1) - 4/9*log(sin(x) - 2)

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Fricas [A]  time = 1.95965, size = 171, normalized size = 3.89 \begin{align*} -\frac{{\left (\sin \left (x\right ) - 2\right )} \log \left (\sin \left (x\right ) + 1\right ) + 8 \,{\left (\sin \left (x\right ) - 2\right )} \log \left (-\frac{1}{2} \, \sin \left (x\right ) + 1\right ) - 9 \,{\left (\sin \left (x\right ) - 2\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 6}{18 \,{\left (\sin \left (x\right ) - 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="fricas")

[Out]

-1/18*((sin(x) - 2)*log(sin(x) + 1) + 8*(sin(x) - 2)*log(-1/2*sin(x) + 1) - 9*(sin(x) - 2)*log(-sin(x) + 1) +
6)/(sin(x) - 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\left (x \right )}}{4 \sin{\left (x \right )} + \cos ^{2}{\left (x \right )} - 5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-5+cos(x)**2+4*sin(x)),x)

[Out]

Integral(sec(x)/(4*sin(x) + cos(x)**2 - 5), x)

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Giac [A]  time = 1.06596, size = 46, normalized size = 1.05 \begin{align*} -\frac{1}{3 \,{\left (\sin \left (x\right ) - 2\right )}} - \frac{1}{18} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{4}{9} \, \log \left (-\sin \left (x\right ) + 2\right ) + \frac{1}{2} \, \log \left (-\sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(-5+cos(x)^2+4*sin(x)),x, algorithm="giac")

[Out]

-1/3/(sin(x) - 2) - 1/18*log(sin(x) + 1) - 4/9*log(-sin(x) + 2) + 1/2*log(-sin(x) + 1)