3.4.5 \(\int \frac {1}{4-5 \sin (x)} \, dx\) [305]

Optimal. Leaf size=43 \[ -\frac {1}{3} \log \left (\cos \left (\frac {x}{2}\right )-2 \sin \left (\frac {x}{2}\right )\right )+\frac {1}{3} \log \left (2 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right ) \]

[Out]

-1/3*ln(cos(1/2*x)-2*sin(1/2*x))+1/3*ln(2*cos(1/2*x)-sin(1/2*x))

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2739, 630, 31} \begin {gather*} \frac {1}{3} \log \left (2 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\frac {1}{3} \log \left (\cos \left (\frac {x}{2}\right )-2 \sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 5*Sin[x])^(-1),x]

[Out]

-1/3*Log[Cos[x/2] - 2*Sin[x/2]] + Log[2*Cos[x/2] - Sin[x/2]]/3

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{4-5 \sin (x)} \, dx &=2 \text {Subst}\left (\int \frac {1}{4-10 x+4 x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {4}{3} \text {Subst}\left (\int \frac {1}{-8+4 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {4}{3} \text {Subst}\left (\int \frac {1}{-2+4 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\frac {1}{3} \log \left (1-2 \tan \left (\frac {x}{2}\right )\right )+\frac {1}{3} \log \left (2-\tan \left (\frac {x}{2}\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 1.00 \begin {gather*} -\frac {1}{3} \log \left (\cos \left (\frac {x}{2}\right )-2 \sin \left (\frac {x}{2}\right )\right )+\frac {1}{3} \log \left (2 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 5*Sin[x])^(-1),x]

[Out]

-1/3*Log[Cos[x/2] - 2*Sin[x/2]] + Log[2*Cos[x/2] - Sin[x/2]]/3

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Maple [A]
time = 0.03, size = 22, normalized size = 0.51

method	result	size
default	\(-\frac {\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )}{3}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )-2\right )}{3}\)	\(22\)
norman	\(-\frac {\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )}{3}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )-2\right )}{3}\)	\(22\)
risch	\(\frac {\ln \left ({\mathrm e}^{i x}+\frac {3}{5}-\frac {4 i}{5}\right )}{3}-\frac {\ln \left ({\mathrm e}^{i x}-\frac {3}{5}-\frac {4 i}{5}\right )}{3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4-5*sin(x)),x,method=_RETURNVERBOSE)

[Out]

-1/3*ln(2*tan(1/2*x)-1)+1/3*ln(tan(1/2*x)-2)

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Maxima [A]
time = 2.09, size = 30, normalized size = 0.70 \begin {gather*} -\frac {1}{3} \, \log \left (\frac {2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \frac {1}{3} \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sin(x)),x, algorithm="maxima")

[Out]

-1/3*log(2*sin(x)/(cos(x) + 1) - 1) + 1/3*log(sin(x)/(cos(x) + 1) - 2)

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Fricas [A]
time = 0.63, size = 27, normalized size = 0.63 \begin {gather*} \frac {1}{6} \, \log \left (\frac {3}{2} \, \cos \left (x\right ) - 2 \, \sin \left (x\right ) + \frac {5}{2}\right ) - \frac {1}{6} \, \log \left (-\frac {3}{2} \, \cos \left (x\right ) - 2 \, \sin \left (x\right ) + \frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sin(x)),x, algorithm="fricas")

[Out]

1/6*log(3/2*cos(x) - 2*sin(x) + 5/2) - 1/6*log(-3/2*cos(x) - 2*sin(x) + 5/2)

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Sympy [A]
time = 0.10, size = 20, normalized size = 0.47 \begin {gather*} \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} - 2 \right )}}{3} - \frac {\log {\left (2 \tan {\left (\frac {x}{2} \right )} - 1 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sin(x)),x)

[Out]

log(tan(x/2) - 2)/3 - log(2*tan(x/2) - 1)/3

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Giac [A]
time = 1.06, size = 23, normalized size = 0.53 \begin {gather*} -\frac {1}{3} \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-5*sin(x)),x, algorithm="giac")

[Out]

-1/3*log(abs(2*tan(1/2*x) - 1)) + 1/3*log(abs(tan(1/2*x) - 2))

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Mupad [B]
time = 0.40, size = 11, normalized size = 0.26 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}-\frac {5}{3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(5*sin(x) - 4),x)

[Out]

-(2*atanh((4*tan(x/2))/3 - 5/3))/3

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