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

Optimal. Leaf size=69 \[ -\frac{15}{578 d (3 \tan (c+d x)+5)}-\frac{3}{68 d (3 \tan (c+d x)+5)^2}+\frac{99 \log (3 \sin (c+d x)+5 \cos (c+d x))}{19652 d}-\frac{5 x}{19652} \]

[Out]

(-5*x)/19652 + (99*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(19652*d) - 3/(68*d*(5 + 3*Tan[c + d*x])^2) - 15/(578
*d*(5 + 3*Tan[c + d*x]))

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Rubi [A]  time = 0.0906826, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ -\frac{15}{578 d (3 \tan (c+d x)+5)}-\frac{3}{68 d (3 \tan (c+d x)+5)^2}+\frac{99 \log (3 \sin (c+d x)+5 \cos (c+d x))}{19652 d}-\frac{5 x}{19652} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Tan[c + d*x])^(-3),x]

[Out]

(-5*x)/19652 + (99*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(19652*d) - 3/(68*d*(5 + 3*Tan[c + d*x])^2) - 15/(578
*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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -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))^3} \, dx &=-\frac{3}{68 d (5+3 \tan (c+d x))^2}+\frac{1}{34} \int \frac{5-3 \tan (c+d x)}{(5+3 \tan (c+d x))^2} \, dx\\ &=-\frac{3}{68 d (5+3 \tan (c+d x))^2}-\frac{15}{578 d (5+3 \tan (c+d x))}+\frac{\int \frac{16-30 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx}{1156}\\ &=-\frac{5 x}{19652}-\frac{3}{68 d (5+3 \tan (c+d x))^2}-\frac{15}{578 d (5+3 \tan (c+d x))}+\frac{99 \int \frac{3-5 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx}{19652}\\ &=-\frac{5 x}{19652}+\frac{99 \log (5 \cos (c+d x)+3 \sin (c+d x))}{19652 d}-\frac{3}{68 d (5+3 \tan (c+d x))^2}-\frac{15}{578 d (5+3 \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.658262, size = 86, normalized size = 1.25 \[ \frac{\left (\frac{1}{39304}+\frac{i}{39304}\right ) \left ((-47+52 i) \log (-\tan (c+d x)+i)-(52-47 i) \log (\tan (c+d x)+i)+(3-3 i) \left (33 \log (3 \tan (c+d x)+5)-\frac{17 (30 \tan (c+d x)+67)}{(3 \tan (c+d x)+5)^2}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/39304 + I/39304)*((-47 + 52*I)*Log[I - Tan[c + d*x]] - (52 - 47*I)*Log[I + Tan[c + d*x]] + (3 - 3*I)*(33*L
og[5 + 3*Tan[c + d*x]] - (17*(67 + 30*Tan[c + d*x]))/(5 + 3*Tan[c + d*x])^2)))/d

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Maple [A]  time = 0.02, size = 80, normalized size = 1.2 \begin{align*} -{\frac{99\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{39304\,d}}-{\frac{5\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{19652\,d}}-{\frac{3}{68\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15}{578\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }}+{\frac{99\,\ln \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }{19652\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-99/39304/d*ln(1+tan(d*x+c)^2)-5/19652/d*arctan(tan(d*x+c))-3/68/d/(5+3*tan(d*x+c))^2-15/578/d/(5+3*tan(d*x+c)
)+99/19652/d*ln(5+3*tan(d*x+c))

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Maxima [A]  time = 1.51513, size = 99, normalized size = 1.43 \begin{align*} -\frac{10 \, d x + 10 \, c + \frac{102 \,{\left (30 \, \tan \left (d x + c\right ) + 67\right )}}{9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25} + 99 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 198 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{39304 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(10*d*x + 10*c + 102*(30*tan(d*x + c) + 67)/(9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25) + 99*log(tan(d*
x + c)^2 + 1) - 198*log(3*tan(d*x + c) + 5))/d

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Fricas [A]  time = 1.67222, size = 343, normalized size = 4.97 \begin{align*} -\frac{18 \,{\left (5 \, d x - 87\right )} \tan \left (d x + c\right )^{2} + 250 \, d x - 99 \,{\left (9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25\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 ) + 60 \,{\left (5 \, d x - 36\right )} \tan \left (d x + c\right ) + 2484}{39304 \,{\left (9 \, d \tan \left (d x + c\right )^{2} + 30 \, d \tan \left (d x + c\right ) + 25 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(18*(5*d*x - 87)*tan(d*x + c)^2 + 250*d*x - 99*(9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25)*log((9*tan(d
*x + c)^2 + 30*tan(d*x + c) + 25)/(tan(d*x + c)^2 + 1)) + 60*(5*d*x - 36)*tan(d*x + c) + 2484)/(9*d*tan(d*x +
c)^2 + 30*d*tan(d*x + c) + 25*d)

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Sympy [A]  time = 1.18905, size = 442, normalized size = 6.41 \begin{align*} \begin{cases} - \frac{90 d x \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{300 d x \tan{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{250 d x}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} + \frac{1782 \log{\left (\tan{\left (c + d x \right )} + \frac{5}{3} \right )} \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} + \frac{5940 \log{\left (\tan{\left (c + d x \right )} + \frac{5}{3} \right )} \tan{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} + \frac{4950 \log{\left (\tan{\left (c + d x \right )} + \frac{5}{3} \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{891 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{2970 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{2475 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{3060 \tan{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} - \frac{6834}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 982600 d} & \text{for}\: d \neq 0 \\\frac{x}{\left (3 \tan{\left (c \right )} + 5\right )^{3}} & \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))**3,x)

[Out]

Piecewise((-90*d*x*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 300*d*x*ta
n(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 250*d*x/(353736*d*tan(c + d*x)**2
+ 1179120*d*tan(c + d*x) + 982600*d) + 1782*log(tan(c + d*x) + 5/3)*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2
+ 1179120*d*tan(c + d*x) + 982600*d) + 5940*log(tan(c + d*x) + 5/3)*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1
179120*d*tan(c + d*x) + 982600*d) + 4950*log(tan(c + d*x) + 5/3)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c +
d*x) + 982600*d) - 891*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c +
d*x) + 982600*d) - 2970*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d
*x) + 982600*d) - 2475*log(tan(c + d*x)**2 + 1)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d)
- 3060*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 6834/(353736*d*tan(c + d
*x)**2 + 1179120*d*tan(c + d*x) + 982600*d), Ne(d, 0)), (x/(3*tan(c) + 5)**3, True))

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Giac [A]  time = 1.27404, size = 100, normalized size = 1.45 \begin{align*} -\frac{10 \, d x + 10 \, c + \frac{3 \,{\left (891 \, \tan \left (d x + c\right )^{2} + 3990 \, \tan \left (d x + c\right ) + 4753\right )}}{{\left (3 \, \tan \left (d x + c\right ) + 5\right )}^{2}} + 99 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 198 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{39304 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(10*d*x + 10*c + 3*(891*tan(d*x + c)^2 + 3990*tan(d*x + c) + 4753)/(3*tan(d*x + c) + 5)^2 + 99*log(ta
n(d*x + c)^2 + 1) - 198*log(abs(3*tan(d*x + c) + 5)))/d