\(\int \frac {1}{3+5 \cos (c+d x)} \, dx\) [34]

Optimal result

Integrand size = 12, antiderivative size = 65 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=-\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[Out]

-1/4*ln(2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d+1/4*ln(2*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2738, 212} \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[In]

Int[(3 + 5*Cos[c + d*x])^(-1),x]

[Out]

-1/4*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]/d + Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]/(4*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {1}{8-2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = -\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=-\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}+\frac {\log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[In]

Integrate[(3 + 5*Cos[c + d*x])^(-1),x]

[Out]

-1/4*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]/d + Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]/(4*d)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.51

method	result	size
parallelrisch	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{4 d}\)	\(33\)
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{4}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{4}}{d}\)	\(34\)
default	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{4}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{4}}{d}\)	\(34\)
norman	\(-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{4 d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{4 d}\)	\(36\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}-\frac {4 i}{5}\right )}{4 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}+\frac {4 i}{5}\right )}{4 d}\)	\(40\)

[In]

int(1/(3+5*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/4*(ln(tan(1/2*d*x+1/2*c)+2)-ln(tan(1/2*d*x+1/2*c)-2))/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\frac {\log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right )}{8 \, d} \]

[In]

integrate(1/(3+5*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/8*(log(3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2) - log(3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2))/d

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.63 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\begin {cases} - \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{4 d} + \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{4 d} & \text {for}\: d \neq 0 \\\frac {x}{5 \cos {\left (c \right )} + 3} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(3+5*cos(d*x+c)),x)

[Out]

Piecewise((-log(tan(c/2 + d*x/2) - 2)/(4*d) + log(tan(c/2 + d*x/2) + 2)/(4*d), Ne(d, 0)), (x/(5*cos(c) + 3), T
rue))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{4 \, d} \]

[In]

integrate(1/(3+5*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.52 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) - \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{4 \, d} \]

[In]

integrate(1/(3+5*cos(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.42 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.26 \[ \int \frac {1}{3+5 \cos (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{2\,d} \]

[In]

int(1/(5*cos(c + d*x) + 3),x)

[Out]

atanh(tan(c/2 + (d*x)/2)/2)/(2*d)