3.3.43 \(\int \frac {1}{3-5 \sin (x)} \, dx\) [243]

Optimal. Leaf size=43 \[ -\frac {1}{4} \log \left (\cos \left (\frac {x}{2}\right )-3 \sin \left (\frac {x}{2}\right )\right )+\frac {1}{4} \log \left (3 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right ) \]

[Out]

-1/4*ln(cos(1/2*x)-3*sin(1/2*x))+1/4*ln(3*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}{4} \log \left (3 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\frac {1}{4} \log \left (\cos \left (\frac {x}{2}\right )-3 \sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 5*Sin[x])^(-1),x]

[Out]

-1/4*Log[Cos[x/2] - 3*Sin[x/2]] + Log[3*Cos[x/2] - Sin[x/2]]/4

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}{3-5 \sin (x)} \, dx &=2 \text {Subst}\left (\int \frac {1}{3-10 x+3 x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {3}{4} \text {Subst}\left (\int \frac {1}{-9+3 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{-1+3 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\frac {1}{4} \log \left (1-3 \tan \left (\frac {x}{2}\right )\right )+\frac {1}{4} \log \left (3-\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}{4} \log \left (\cos \left (\frac {x}{2}\right )-3 \sin \left (\frac {x}{2}\right )\right )+\frac {1}{4} \log \left (3 \cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 5*Sin[x])^(-1),x]

[Out]

-1/4*Log[Cos[x/2] - 3*Sin[x/2]] + Log[3*Cos[x/2] - Sin[x/2]]/4

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Mathics [A]
time = 1.97, size = 21, normalized size = 0.49 \begin {gather*} -\frac {\text {Log}\left [-1+3 \text {Tan}\left [\frac {x}{2}\right ]\right ]}{4}+\frac {\text {Log}\left [-3+\text {Tan}\left [\frac {x}{2}\right ]\right ]}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(3 - 5*Sin[x]),x]')

[Out]

-Log[-1 + 3 Tan[x / 2]] / 4 + Log[-3 + Tan[x / 2]] / 4

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3-5*sin(x)),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(tan(1/2*x)-3)-1/4*ln(3*tan(1/2*x)-1)

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Maxima [A]
time = 0.25, size = 30, normalized size = 0.70 \begin {gather*} -\frac {1}{4} \, \log \left (\frac {3 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \frac {1}{4} \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-5*sin(x)),x, algorithm="maxima")

[Out]

-1/4*log(3*sin(x)/(cos(x) + 1) - 1) + 1/4*log(sin(x)/(cos(x) + 1) - 3)

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Fricas [A]
time = 0.33, size = 27, normalized size = 0.63 \begin {gather*} \frac {1}{8} \, \log \left (4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right ) + 5\right ) - \frac {1}{8} \, \log \left (-4 \, \cos \left (x\right ) - 3 \, \sin \left (x\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-5*sin(x)),x, algorithm="fricas")

[Out]

1/8*log(4*cos(x) - 3*sin(x) + 5) - 1/8*log(-4*cos(x) - 3*sin(x) + 5)

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Sympy [A]
time = 0.11, size = 20, normalized size = 0.47 \begin {gather*} \frac {\log {\left (\tan {\left (\frac {x}{2} \right )} - 3 \right )}}{4} - \frac {\log {\left (3 \tan {\left (\frac {x}{2} \right )} - 1 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-5*sin(x)),x)

[Out]

log(tan(x/2) - 3)/4 - log(3*tan(x/2) - 1)/4

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Giac [A]
time = 0.00, size = 30, normalized size = 0.70 \begin {gather*} 2 \left (\frac {\ln \left |\tan \left (\frac {x}{2}\right )-3\right |}{8}-\frac {\ln \left |3 \tan \left (\frac {x}{2}\right )-1\right |}{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-5*sin(x)),x)

[Out]

-1/4*log(abs(3*tan(1/2*x) - 1)) + 1/4*log(abs(tan(1/2*x) - 3))

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Mupad [B]
time = 0.39, size = 11, normalized size = 0.26 \begin {gather*} -\frac {\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {x}{2}\right )}{4}-\frac {5}{4}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(5*sin(x) - 3),x)

[Out]

-atanh((3*tan(x/2))/4 - 5/4)/2

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