3.1.7 \(\int \frac {3}{5+4 \sin (x)} \, dx\) [7]

Optimal. Leaf size=14 \[ x+2 \tan ^{-1}\left (\frac {\cos (x)}{2+\sin (x)}\right ) \]

[Out]

x+2*arctan(cos(x)/(2+sin(x)))

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Rubi [A]
time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2736} \begin {gather*} x+2 \tan ^{-1}\left (\frac {\cos (x)}{\sin (x)+2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[3/(5 + 4*Sin[x]),x]

[Out]

x + 2*ArcTan[Cos[x]/(2 + Sin[x])]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2736

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(14)=28\).
time = 0.01, size = 79, normalized size = 5.64 \begin {gather*} 3 \left (-\frac {1}{3} \tan ^{-1}\left (\frac {2 \cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+2 \sin \left (\frac {x}{2}\right )}\right )+\frac {1}{3} \tan ^{-1}\left (\frac {\cos \left (\frac {x}{2}\right )+2 \sin \left (\frac {x}{2}\right )}{2 \cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[3/(5 + 4*Sin[x]),x]

[Out]

3*(-1/3*ArcTan[(2*Cos[x/2] + Sin[x/2])/(Cos[x/2] + 2*Sin[x/2])] + ArcTan[(Cos[x/2] + 2*Sin[x/2])/(2*Cos[x/2] +
Sin[x/2])]/3)

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.17, size = 24, normalized size = 1.71 \begin {gather*} 2 \text {Pi} \text {Floor}\left [-\frac {1}{2}+\frac {x}{2 \text {Pi}}\right ]+2 \text {ArcTan}\left [\frac {4}{3}+\frac {5 \text {Tan}\left [\frac {x}{2}\right ]}{3}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[3/(5 + 4*Sin[x]),x]')

[Out]

2 Pi Floor[-1 / 2 + x / (2 Pi)] + 2 ArcTan[4 / 3 + 5 Tan[x / 2] / 3]

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Maple [A]
time = 0.03, size = 12, normalized size = 0.86

method	result	size
default	\(2 \arctan \left (\frac {5 \tan \left (\frac {x}{2}\right )}{3}+\frac {4}{3}\right )\)	\(12\)
risch	\(i \ln \left ({\mathrm e}^{i x}+2 i\right )-i \ln \left ({\mathrm e}^{i x}+\frac {i}{2}\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3/(5+4*sin(x)),x,method=_RETURNVERBOSE)

[Out]

2*arctan(5/3*tan(1/2*x)+4/3)

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Maxima [A]
time = 0.36, size = 15, normalized size = 1.07 \begin {gather*} 2 \, \arctan \left (\frac {5 \, \sin \left (x\right )}{3 \, {\left (\cos \left (x\right ) + 1\right )}} + \frac {4}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/(5+4*sin(x)),x, algorithm="maxima")

[Out]

2*arctan(5/3*sin(x)/(cos(x) + 1) + 4/3)

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Fricas [A]
time = 0.33, size = 13, normalized size = 0.93 \begin {gather*} \arctan \left (\frac {5 \, \sin \left (x\right ) + 4}{3 \, \cos \left (x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/(5+4*sin(x)),x, algorithm="fricas")

[Out]

arctan(1/3*(5*sin(x) + 4)/cos(x))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\)
time = 0.13, size = 27, normalized size = 1.93 \begin {gather*} 2 \operatorname {atan}{\left (\frac {5 \tan {\left (\frac {x}{2} \right )}}{3} + \frac {4}{3} \right )} + 2 \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/(5+4*sin(x)),x)

[Out]

2*atan(5*tan(x/2)/3 + 4/3) + 2*pi*floor((x/2 - pi/2)/pi)

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Giac [A]
time = 0.00, size = 32, normalized size = 2.29 \begin {gather*} \frac {2}{3}\cdot 3 \left (\arctan \left (\frac {-2 \cos x-\sin x-2}{\cos x-2 \sin x-4}\right )+\frac {x}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/(5+4*sin(x)),x)

[Out]

x + 2*arctan(-(2*cos(x) + sin(x) + 2)/(cos(x) - 2*sin(x) - 4))

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Mupad [B]
time = 0.22, size = 20, normalized size = 1.43 \begin {gather*} x+2\,\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {4}{3}\right )-2\,\mathrm {atan}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3/(4*sin(x) + 5),x)

[Out]

x + 2*atan((5*tan(x/2))/3 + 4/3) - 2*atan(tan(x/2))

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