3.366 \(\int \frac {1}{2 a+2 a \cos (d+e x)+2 c \sin (d+e x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {\log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{2 c e} \]

[Out]

1/2*ln(a+c*tan(1/2*e*x+1/2*d))/c/e

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3124, 31} \[ \frac {\log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{2 c e} \]

Antiderivative was successfully verified.

[In]

Int[(2*a + 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^(-1),x]

[Out]

Log[a + c*Tan[(d + e*x)/2]]/(2*c*e)

Rule 31

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

Rule 3124

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}{2 a+2 a \cos (d+e x)+2 c \sin (d+e x)} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{4 a+4 c x} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{e}\\ &=\frac {\log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{2 c e}\\ \end {align*}

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Mathematica [B]  time = 0.05, size = 57, normalized size = 2.28 \[ \frac {1}{2} \left (\frac {\log \left (a \cos \left (\frac {1}{2} (d+e x)\right )+c \sin \left (\frac {1}{2} (d+e x)\right )\right )}{c e}-\frac {\log \left (\cos \left (\frac {1}{2} (d+e x)\right )\right )}{c e}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2*a + 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^(-1),x]

[Out]

(-(Log[Cos[(d + e*x)/2]]/(c*e)) + Log[a*Cos[(d + e*x)/2] + c*Sin[(d + e*x)/2]]/(c*e))/2

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fricas [B]  time = 1.34, size = 60, normalized size = 2.40 \[ \frac {\log \left (a c \sin \left (e x + d\right ) + \frac {1}{2} \, a^{2} + \frac {1}{2} \, c^{2} + \frac {1}{2} \, {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right ) - \log \left (\frac {1}{2} \, \cos \left (e x + d\right ) + \frac {1}{2}\right )}{4 \, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d)),x, algorithm="fricas")

[Out]

1/4*(log(a*c*sin(e*x + d) + 1/2*a^2 + 1/2*c^2 + 1/2*(a^2 - c^2)*cos(e*x + d)) - log(1/2*cos(e*x + d) + 1/2))/(
c*e)

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giac [A]  time = 0.13, size = 23, normalized size = 0.92 \[ \frac {e^{\left (-1\right )} \log \left ({\left | c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a \right |}\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d)),x, algorithm="giac")

[Out]

1/2*e^(-1)*log(abs(c*tan(1/2*x*e + 1/2*d) + a))/c

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maple [A]  time = 0.38, size = 23, normalized size = 0.92 \[ \frac {\ln \left (a +c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{2 c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d)),x)

[Out]

1/2*ln(a+c*tan(1/2*d+1/2*e*x))/c/e

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maxima [A]  time = 0.31, size = 29, normalized size = 1.16 \[ \frac {\log \left (a + \frac {c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{2 \, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d)),x, algorithm="maxima")

[Out]

1/2*log(a + c*sin(e*x + d)/(cos(e*x + d) + 1))/(c*e)

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mupad [B]  time = 2.82, size = 22, normalized size = 0.88 \[ \frac {\ln \left (a+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )}{2\,c\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*a + 2*a*cos(d + e*x) + 2*c*sin(d + e*x)),x)

[Out]

log(a + c*tan(d/2 + (e*x)/2))/(2*c*e)

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sympy [A]  time = 1.10, size = 63, normalized size = 2.52 \[ \begin {cases} \frac {x}{2 a \cos {\relax (d )} + 2 a} & \text {for}\: c = 0 \wedge e = 0 \\\frac {\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{2 a e} & \text {for}\: c = 0 \\\frac {x}{2 a \cos {\relax (d )} + 2 a + 2 c \sin {\relax (d )}} & \text {for}\: e = 0 \\\frac {\log {\left (\frac {a}{c} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{2 c e} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d)),x)

[Out]

Piecewise((x/(2*a*cos(d) + 2*a), Eq(c, 0) & Eq(e, 0)), (tan(d/2 + e*x/2)/(2*a*e), Eq(c, 0)), (x/(2*a*cos(d) +
2*a + 2*c*sin(d)), Eq(e, 0)), (log(a/c + tan(d/2 + e*x/2))/(2*c*e), True))

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