3.1.4 \(\int \frac {1}{3+3 \cos (x)+4 \sin (x)} \, dx\) [4]

Optimal. Leaf size=15 \[ \frac {1}{4} \log \left (3+4 \tan \left (\frac {x}{2}\right )\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3203, 31} \begin {gather*} \frac {1}{4} \log \left (4 \tan \left (\frac {x}{2}\right )+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 3*Cos[x] + 4*Sin[x])^(-1),x]

[Out]

Log[3 + 4*Tan[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 3203

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[2*(f/e), Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{3+3 \cos (x)+4 \sin (x)} \, dx &=2 \text {Subst}\left (\int \frac {1}{6+8 x} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=\frac {1}{4} \log \left (3+4 \tan \left (\frac {x}{2}\right )\right )\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(34\) vs. \(2(15)=30\).
time = 0.01, size = 34, normalized size = 2.27 \begin {gather*} -\frac {1}{4} \log \left (\cos \left (\frac {x}{2}\right )\right )+\frac {1}{4} \log \left (3 \cos \left (\frac {x}{2}\right )+4 \sin \left (\frac {x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 3*Cos[x] + 4*Sin[x])^(-1),x]

[Out]

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

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Mathics [A]
time = 1.75, size = 11, normalized size = 0.73 \begin {gather*} \frac {\text {Log}\left [3+4 \text {Tan}\left [\frac {x}{2}\right ]\right ]}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(3 + 3*Cos[x] + 4*Sin[x]),x]')

[Out]

Log[3 + 4 Tan[x / 2]] / 4

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Maple [A]
time = 0.08, size = 12, normalized size = 0.80

method	result	size
default	\(\frac {\ln \left (3+4 \tan \left (\frac {x}{2}\right )\right )}{4}\)	\(12\)
norman	\(\frac {\ln \left (6+8 \tan \left (\frac {x}{2}\right )\right )}{4}\)	\(12\)
risch	\(-\frac {\ln \left (1+{\mathrm e}^{i x}\right )}{4}+\frac {\ln \left ({\mathrm e}^{i x}-\frac {7}{25}+\frac {24 i}{25}\right )}{4}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+3*cos(x)+4*sin(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.24, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{4} \, \log \left (\frac {4 \, \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+3*cos(x)+4*sin(x)),x, algorithm="maxima")

[Out]

1/4*log(4*sin(x)/(cos(x) + 1) + 3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (11) = 22\).
time = 0.33, size = 23, normalized size = 1.53 \begin {gather*} -\frac {1}{8} \, \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + \frac {1}{8} \, \log \left (-\frac {7}{2} \, \cos \left (x\right ) + 12 \, \sin \left (x\right ) + \frac {25}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+3*cos(x)+4*sin(x)),x, algorithm="fricas")

[Out]

-1/8*log(1/2*cos(x) + 1/2) + 1/8*log(-7/2*cos(x) + 12*sin(x) + 25/2)

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Sympy [A]
time = 0.13, size = 10, normalized size = 0.67 \begin {gather*} \frac {\log {\left (4 \tan {\left (\frac {x}{2} \right )} + 3 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+3*cos(x)+4*sin(x)),x)

[Out]

log(4*tan(x/2) + 3)/4

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Giac [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {2}{8} \ln \left |4 \tan \left (\frac {x}{2}\right )+3\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+3*cos(x)+4*sin(x)),x)

[Out]

1/4*log(abs(4*tan(1/2*x) + 3))

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Mupad [B]
time = 0.08, size = 11, normalized size = 0.73 \begin {gather*} \frac {\ln \left (4\,\mathrm {tan}\left (\frac {x}{2}\right )+3\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*cos(x) + 4*sin(x) + 3),x)

[Out]

log(4*tan(x/2) + 3)/4

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