3.496 \(\int \frac{1}{(3+5 \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{5}{34 d (5 \tan (c+d x)+3)}+\frac{15 \log (5 \sin (c+d x)+3 \cos (c+d x))}{578 d}-\frac{4 x}{289} \]

[Out]

(-4*x)/289 + (15*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(578*d) - 5/(34*d*(3 + 5*Tan[c + d*x]))

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Rubi [A]  time = 0.064283, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3483, 3531, 3530} \[ -\frac{5}{34 d (5 \tan (c+d x)+3)}+\frac{15 \log (5 \sin (c+d x)+3 \cos (c+d x))}{578 d}-\frac{4 x}{289} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Tan[c + d*x])^(-2),x]

[Out]

(-4*x)/289 + (15*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(578*d) - 5/(34*d*(3 + 5*Tan[c + d*x]))

Rule 3483

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[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] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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))^2} \, dx &=-\frac{5}{34 d (3+5 \tan (c+d x))}+\frac{1}{34} \int \frac{3-5 \tan (c+d x)}{3+5 \tan (c+d x)} \, dx\\ &=-\frac{4 x}{289}-\frac{5}{34 d (3+5 \tan (c+d x))}+\frac{15}{578} \int \frac{5-3 \tan (c+d x)}{3+5 \tan (c+d x)} \, dx\\ &=-\frac{4 x}{289}+\frac{15 \log (3 \cos (c+d x)+5 \sin (c+d x))}{578 d}-\frac{5}{34 d (3+5 \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.249933, size = 67, normalized size = 1.34 \[ -\frac{\frac{170}{5 \tan (c+d x)+3}+(15-8 i) \log (-\tan (c+d x)+i)+(15+8 i) \log (\tan (c+d x)+i)-30 \log (5 \tan (c+d x)+3)}{1156 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*Tan[c + d*x])^(-2),x]

[Out]

-((15 - 8*I)*Log[I - Tan[c + d*x]] + (15 + 8*I)*Log[I + Tan[c + d*x]] - 30*Log[3 + 5*Tan[c + d*x]] + 170/(3 +
5*Tan[c + d*x]))/(1156*d)

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Maple [A]  time = 0.017, size = 63, normalized size = 1.3 \begin{align*} -{\frac{15\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{1156\,d}}-{\frac{4\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{289\,d}}-{\frac{5}{34\,d \left ( 3+5\,\tan \left ( dx+c \right ) \right ) }}+{\frac{15\,\ln \left ( 3+5\,\tan \left ( dx+c \right ) \right ) }{578\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+5*tan(d*x+c))^2,x)

[Out]

-15/1156/d*ln(1+tan(d*x+c)^2)-4/289/d*arctan(tan(d*x+c))-5/34/d/(3+5*tan(d*x+c))+15/578/d*ln(3+5*tan(d*x+c))

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Maxima [A]  time = 1.65626, size = 72, normalized size = 1.44 \begin{align*} -\frac{16 \, d x + 16 \, c + \frac{170}{5 \, \tan \left (d x + c\right ) + 3} + 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 30 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{1156 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1156*(16*d*x + 16*c + 170/(5*tan(d*x + c) + 3) + 15*log(tan(d*x + c)^2 + 1) - 30*log(5*tan(d*x + c) + 3))/d

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Fricas [A]  time = 1.68205, size = 235, normalized size = 4.7 \begin{align*} -\frac{48 \, d x - 15 \,{\left (5 \, \tan \left (d x + c\right ) + 3\right )} \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 ) + 5 \,{\left (16 \, d x - 15\right )} \tan \left (d x + c\right ) + 125}{1156 \,{\left (5 \, d \tan \left (d x + c\right ) + 3 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/1156*(48*d*x - 15*(5*tan(d*x + c) + 3)*log((25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9)/(tan(d*x + c)^2 + 1))
+ 5*(16*d*x - 15)*tan(d*x + c) + 125)/(5*d*tan(d*x + c) + 3*d)

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Sympy [A]  time = 0.812839, size = 190, normalized size = 3.8 \begin{align*} \begin{cases} - \frac{80 d x \tan{\left (c + d x \right )}}{5780 d \tan{\left (c + d x \right )} + 3468 d} - \frac{48 d x}{5780 d \tan{\left (c + d x \right )} + 3468 d} + \frac{150 \log{\left (\tan{\left (c + d x \right )} + \frac{3}{5} \right )} \tan{\left (c + d x \right )}}{5780 d \tan{\left (c + d x \right )} + 3468 d} + \frac{90 \log{\left (\tan{\left (c + d x \right )} + \frac{3}{5} \right )}}{5780 d \tan{\left (c + d x \right )} + 3468 d} - \frac{75 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{5780 d \tan{\left (c + d x \right )} + 3468 d} - \frac{45 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{5780 d \tan{\left (c + d x \right )} + 3468 d} - \frac{170}{5780 d \tan{\left (c + d x \right )} + 3468 d} & \text{for}\: d \neq 0 \\\frac{x}{\left (5 \tan{\left (c \right )} + 3\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*tan(d*x+c))**2,x)

[Out]

Piecewise((-80*d*x*tan(c + d*x)/(5780*d*tan(c + d*x) + 3468*d) - 48*d*x/(5780*d*tan(c + d*x) + 3468*d) + 150*l
og(tan(c + d*x) + 3/5)*tan(c + d*x)/(5780*d*tan(c + d*x) + 3468*d) + 90*log(tan(c + d*x) + 3/5)/(5780*d*tan(c
+ d*x) + 3468*d) - 75*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(5780*d*tan(c + d*x) + 3468*d) - 45*log(tan(c + d*
x)**2 + 1)/(5780*d*tan(c + d*x) + 3468*d) - 170/(5780*d*tan(c + d*x) + 3468*d), Ne(d, 0)), (x/(5*tan(c) + 3)**
2, True))

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Giac [A]  time = 1.32015, size = 86, normalized size = 1.72 \begin{align*} -\frac{16 \, d x + 16 \, c + \frac{10 \,{\left (15 \, \tan \left (d x + c\right ) + 26\right )}}{5 \, \tan \left (d x + c\right ) + 3} + 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 30 \, \log \left ({\left | 5 \, \tan \left (d x + c\right ) + 3 \right |}\right )}{1156 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/1156*(16*d*x + 16*c + 10*(15*tan(d*x + c) + 26)/(5*tan(d*x + c) + 3) + 15*log(tan(d*x + c)^2 + 1) - 30*log(
abs(5*tan(d*x + c) + 3)))/d