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

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

[Out]

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

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Rubi [A]  time = 0.0637289, 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} \[ -\frac{3}{34 d (3 \tan (c+d x)+5)}+\frac{15 \log (3 \sin (c+d x)+5 \cos (c+d x))}{578 d}+\frac{4 x}{289} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Tan[c + d*x])^(-2),x]

[Out]

(4*x)/289 + (15*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(578*d) - 3/(34*d*(5 + 3*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}{(5+3 \tan (c+d x))^2} \, dx &=-\frac{3}{34 d (5+3 \tan (c+d x))}+\frac{1}{34} \int \frac{5-3 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx\\ &=\frac{4 x}{289}-\frac{3}{34 d (5+3 \tan (c+d x))}+\frac{15}{578} \int \frac{3-5 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx\\ &=\frac{4 x}{289}+\frac{15 \log (5 \cos (c+d x)+3 \sin (c+d x))}{578 d}-\frac{3}{34 d (5+3 \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.252701, size = 67, normalized size = 1.34 \[ -\frac{\frac{102}{3 \tan (c+d x)+5}+(15+8 i) \log (-\tan (c+d x)+i)+(15-8 i) \log (\tan (c+d x)+i)-30 \log (3 \tan (c+d x)+5)}{1156 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 3*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[5 + 3*Tan[c + d*x]] + 102/(5 +
3*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{3}{34\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }}+{\frac{15\,\ln \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }{578\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5+3*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))-3/34/d/(5+3*tan(d*x+c))+15/578/d*ln(5+3*tan(d*x+c))

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Maxima [A]  time = 1.52548, size = 72, normalized size = 1.44 \begin{align*} \frac{16 \, d x + 16 \, c - \frac{102}{3 \, \tan \left (d x + c\right ) + 5} - 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 30 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{1156 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1156*(16*d*x + 16*c - 102/(3*tan(d*x + c) + 5) - 15*log(tan(d*x + c)^2 + 1) + 30*log(3*tan(d*x + c) + 5))/d

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Fricas [A]  time = 1.70216, size = 232, normalized size = 4.64 \begin{align*} \frac{80 \, d x + 15 \,{\left (3 \, \tan \left (d x + c\right ) + 5\right )} \log \left (\frac{9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25}{\tan \left (d x + c\right )^{2} + 1}\right ) + 3 \,{\left (16 \, d x + 15\right )} \tan \left (d x + c\right ) - 27}{1156 \,{\left (3 \, d \tan \left (d x + c\right ) + 5 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1156*(80*d*x + 15*(3*tan(d*x + c) + 5)*log((9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25)/(tan(d*x + c)^2 + 1)) +
3*(16*d*x + 15)*tan(d*x + c) - 27)/(3*d*tan(d*x + c) + 5*d)

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Sympy [A]  time = 0.877957, size = 190, normalized size = 3.8 \begin{align*} \begin{cases} \frac{48 d x \tan{\left (c + d x \right )}}{3468 d \tan{\left (c + d x \right )} + 5780 d} + \frac{80 d x}{3468 d \tan{\left (c + d x \right )} + 5780 d} + \frac{90 \log{\left (\tan{\left (c + d x \right )} + \frac{5}{3} \right )} \tan{\left (c + d x \right )}}{3468 d \tan{\left (c + d x \right )} + 5780 d} + \frac{150 \log{\left (\tan{\left (c + d x \right )} + \frac{5}{3} \right )}}{3468 d \tan{\left (c + d x \right )} + 5780 d} - \frac{45 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{3468 d \tan{\left (c + d x \right )} + 5780 d} - \frac{75 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{3468 d \tan{\left (c + d x \right )} + 5780 d} - \frac{102}{3468 d \tan{\left (c + d x \right )} + 5780 d} & \text{for}\: d \neq 0 \\\frac{x}{\left (3 \tan{\left (c \right )} + 5\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((48*d*x*tan(c + d*x)/(3468*d*tan(c + d*x) + 5780*d) + 80*d*x/(3468*d*tan(c + d*x) + 5780*d) + 90*log
(tan(c + d*x) + 5/3)*tan(c + d*x)/(3468*d*tan(c + d*x) + 5780*d) + 150*log(tan(c + d*x) + 5/3)/(3468*d*tan(c +
d*x) + 5780*d) - 45*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(3468*d*tan(c + d*x) + 5780*d) - 75*log(tan(c + d*x
)**2 + 1)/(3468*d*tan(c + d*x) + 5780*d) - 102/(3468*d*tan(c + d*x) + 5780*d), Ne(d, 0)), (x/(3*tan(c) + 5)**2
, True))

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Giac [A]  time = 1.31514, size = 86, normalized size = 1.72 \begin{align*} \frac{16 \, d x + 16 \, c - \frac{18 \,{\left (5 \, \tan \left (d x + c\right ) + 14\right )}}{3 \, \tan \left (d x + c\right ) + 5} - 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 30 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{1156 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1156*(16*d*x + 16*c - 18*(5*tan(d*x + c) + 14)/(3*tan(d*x + c) + 5) - 15*log(tan(d*x + c)^2 + 1) + 30*log(ab
s(3*tan(d*x + c) + 5)))/d