3.5.95 \(\int \frac {1}{3+5 \tan (c+d x)} \, dx\) [495]

Optimal. Leaf size=31 \[ \frac {3 x}{34}+\frac {5 \log (3 \cos (c+d x)+5 \sin (c+d x))}{34 d} \]

[Out]

3/34*x+5/34*ln(3*cos(d*x+c)+5*sin(d*x+c))/d

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Rubi [A]
time = 0.03, antiderivative size = 31, 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 = {3565, 3611} \begin {gather*} \frac {5 \log (5 \sin (c+d x)+3 \cos (c+d x))}{34 d}+\frac {3 x}{34} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Tan[c + d*x])^(-1),x]

[Out]

(3*x)/34 + (5*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(34*d)

Rule 3565

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[a*(x/(a^2 + b^2)), x] + Dist[b/(a^2 + b^2),
Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{3+5 \tan (c+d x)} \, dx &=\frac {3 x}{34}+\frac {5}{34} \int \frac {5-3 \tan (c+d x)}{3+5 \tan (c+d x)} \, dx\\ &=\frac {3 x}{34}+\frac {5 \log (3 \cos (c+d x)+5 \sin (c+d x))}{34 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.04, size = 65, normalized size = 2.10 \begin {gather*} -\frac {\left (\frac {5}{68}+\frac {3 i}{68}\right ) \log (i-\tan (c+d x))}{d}-\frac {\left (\frac {5}{68}-\frac {3 i}{68}\right ) \log (i+\tan (c+d x))}{d}+\frac {5 \log (3+5 \tan (c+d x))}{34 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-5/68 - (3*I)/68)*Log[I - Tan[c + d*x]])/d - ((5/68 - (3*I)/68)*Log[I + Tan[c + d*x]])/d + (5*Log[3 + 5*Tan[
c + d*x]])/(34*d)

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Maple [A]
time = 0.08, size = 41, normalized size = 1.32

method	result	size
risch	\(\frac {3 x}{34}-\frac {5 i x}{34}-\frac {5 i c}{17 d}+\frac {5 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {8}{17}+\frac {15 i}{17}\right )}{34 d}\)	\(35\)
norman	\(\frac {3 x}{34}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{34 d}-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{68 d}\)	\(37\)
derivativedivides	\(\frac {-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{68}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{34}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{34}}{d}\)	\(41\)
default	\(\frac {-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{68}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{34}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{34}}{d}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-5/68*ln(1+tan(d*x+c)^2)+3/34*arctan(tan(d*x+c))+5/34*ln(3+5*tan(d*x+c)))

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Maxima [A]
time = 0.49, size = 39, normalized size = 1.26 \begin {gather*} \frac {6 \, d x + 6 \, c - 5 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 10 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{68 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/68*(6*d*x + 6*c - 5*log(tan(d*x + c)^2 + 1) + 10*log(5*tan(d*x + c) + 3))/d

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Fricas [A]
time = 1.38, size = 46, normalized size = 1.48 \begin {gather*} \frac {6 \, d x + 5 \, \log \left (\frac {25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9}{\tan \left (d x + c\right )^{2} + 1}\right )}{68 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/68*(6*d*x + 5*log((25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9)/(tan(d*x + c)^2 + 1)))/d

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Sympy [A]
time = 0.19, size = 46, normalized size = 1.48 \begin {gather*} \begin {cases} \frac {3 x}{34} + \frac {5 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )}}{34 d} - \frac {5 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{68 d} & \text {for}\: d \neq 0 \\\frac {x}{5 \tan {\left (c \right )} + 3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*x/34 + 5*log(5*tan(c + d*x) + 3)/(34*d) - 5*log(tan(c + d*x)**2 + 1)/(68*d), Ne(d, 0)), (x/(5*tan
(c) + 3), True))

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Giac [A]
time = 0.48, size = 40, normalized size = 1.29 \begin {gather*} \frac {6 \, d x + 6 \, c - 5 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 10 \, \log \left ({\left | 5 \, \tan \left (d x + c\right ) + 3 \right |}\right )}{68 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/68*(6*d*x + 6*c - 5*log(tan(d*x + c)^2 + 1) + 10*log(abs(5*tan(d*x + c) + 3)))/d

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Mupad [B]
time = 4.11, size = 49, normalized size = 1.58 \begin {gather*} \frac {5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {3}{5}\right )}{34\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {5}{68}-\frac {3}{68}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {5}{68}+\frac {3}{68}{}\mathrm {i}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*tan(c + d*x) + 3),x)

[Out]

(5*log(tan(c + d*x) + 3/5))/(34*d) - (log(tan(c + d*x) + 1i)*(5/68 - 3i/68))/d - (log(tan(c + d*x) - 1i)*(5/68
+ 3i/68))/d

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