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

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

[Out]

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

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Rubi [A]  time = 0.0915674, 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} \[ -\frac{15}{578 d (5 \tan (c+d x)+3)}-\frac{5}{68 d (5 \tan (c+d x)+3)^2}+\frac{5 \log (5 \sin (c+d x)+3 \cos (c+d x))}{19652 d}-\frac{99 x}{19652} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Tan[c + d*x])^(-3),x]

[Out]

(-99*x)/19652 + (5*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(19652*d) - 5/(68*d*(3 + 5*Tan[c + d*x])^2) - 15/(578
*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 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}{(3+5 \tan (c+d x))^3} \, dx &=-\frac{5}{68 d (3+5 \tan (c+d x))^2}+\frac{1}{34} \int \frac{3-5 \tan (c+d x)}{(3+5 \tan (c+d x))^2} \, dx\\ &=-\frac{5}{68 d (3+5 \tan (c+d x))^2}-\frac{15}{578 d (3+5 \tan (c+d x))}+\frac{\int \frac{-16-30 \tan (c+d x)}{3+5 \tan (c+d x)} \, dx}{1156}\\ &=-\frac{99 x}{19652}-\frac{5}{68 d (3+5 \tan (c+d x))^2}-\frac{15}{578 d (3+5 \tan (c+d x))}+\frac{5 \int \frac{5-3 \tan (c+d x)}{3+5 \tan (c+d x)} \, dx}{19652}\\ &=-\frac{99 x}{19652}+\frac{5 \log (3 \cos (c+d x)+5 \sin (c+d x))}{19652 d}-\frac{5}{68 d (3+5 \tan (c+d x))^2}-\frac{15}{578 d (3+5 \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.561182, size = 84, normalized size = 1.22 \[ \frac{\left (\frac{1}{39304}+\frac{i}{39304}\right ) \left ((47+52 i) \log (-\tan (c+d x)+i)-(52+47 i) \log (\tan (c+d x)+i)+(5-5 i) \left (\log (5 \tan (c+d x)+3)-\frac{85 (6 \tan (c+d x)+7)}{(5 \tan (c+d x)+3)^2}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*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]] + (5 - 5*I)*(Log[3
+ 5*Tan[c + d*x]] - (85*(7 + 6*Tan[c + d*x]))/(3 + 5*Tan[c + d*x])^2)))/d

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Maple [A]  time = 0.019, size = 80, normalized size = 1.2 \begin{align*} -{\frac{5\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{39304\,d}}-{\frac{99\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{19652\,d}}-{\frac{5}{68\,d \left ( 3+5\,\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15}{578\,d \left ( 3+5\,\tan \left ( dx+c \right ) \right ) }}+{\frac{5\,\ln \left ( 3+5\,\tan \left ( dx+c \right ) \right ) }{19652\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.62769, size = 99, normalized size = 1.43 \begin{align*} -\frac{198 \, d x + 198 \, c + \frac{850 \,{\left (6 \, \tan \left (d x + c\right ) + 7\right )}}{25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9} + 5 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 10 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{39304 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(198*d*x + 198*c + 850*(6*tan(d*x + c) + 7)/(25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9) + 5*log(tan(d*x
+ c)^2 + 1) - 10*log(5*tan(d*x + c) + 3))/d

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Fricas [A]  time = 1.89132, size = 347, normalized size = 5.03 \begin{align*} -\frac{50 \,{\left (99 \, d x - 25\right )} \tan \left (d x + c\right )^{2} + 1782 \, d x - 5 \,{\left (25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9\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 ) + 180 \,{\left (33 \, d x + 20\right )} \tan \left (d x + c\right ) + 5500}{39304 \,{\left (25 \, d \tan \left (d x + c\right )^{2} + 30 \, d \tan \left (d x + c\right ) + 9 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(50*(99*d*x - 25)*tan(d*x + c)^2 + 1782*d*x - 5*(25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9)*log((25*tan
(d*x + c)^2 + 30*tan(d*x + c) + 9)/(tan(d*x + c)^2 + 1)) + 180*(33*d*x + 20)*tan(d*x + c) + 5500)/(25*d*tan(d*
x + c)^2 + 30*d*tan(d*x + c) + 9*d)

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Sympy [A]  time = 1.14318, size = 442, normalized size = 6.41 \begin{align*} \begin{cases} - \frac{4950 d x \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{5940 d x \tan{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{1782 d x}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} + \frac{250 \log{\left (\tan{\left (c + d x \right )} + \frac{3}{5} \right )} \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} + \frac{300 \log{\left (\tan{\left (c + d x \right )} + \frac{3}{5} \right )} \tan{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} + \frac{90 \log{\left (\tan{\left (c + d x \right )} + \frac{3}{5} \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{125 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{150 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{45 \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{5100 \tan{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} - \frac{5950}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan{\left (c + d x \right )} + 353736 d} & \text{for}\: d \neq 0 \\\frac{x}{\left (5 \tan{\left (c \right )} + 3\right )^{3}} & \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))**3,x)

[Out]

Piecewise((-4950*d*x*tan(c + d*x)**2/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 5940*d*x
*tan(c + d*x)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 1782*d*x/(982600*d*tan(c + d*x)
**2 + 1179120*d*tan(c + d*x) + 353736*d) + 250*log(tan(c + d*x) + 3/5)*tan(c + d*x)**2/(982600*d*tan(c + d*x)*
*2 + 1179120*d*tan(c + d*x) + 353736*d) + 300*log(tan(c + d*x) + 3/5)*tan(c + d*x)/(982600*d*tan(c + d*x)**2 +
1179120*d*tan(c + d*x) + 353736*d) + 90*log(tan(c + d*x) + 3/5)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c +
d*x) + 353736*d) - 125*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c +
d*x) + 353736*d) - 150*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*
x) + 353736*d) - 45*log(tan(c + d*x)**2 + 1)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) -
5100*tan(c + d*x)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 5950/(982600*d*tan(c + d*x)
**2 + 1179120*d*tan(c + d*x) + 353736*d), Ne(d, 0)), (x/(5*tan(c) + 3)**3, True))

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Giac [A]  time = 1.25842, size = 100, normalized size = 1.45 \begin{align*} -\frac{198 \, d x + 198 \, c + \frac{5 \,{\left (75 \, \tan \left (d x + c\right )^{2} + 1110 \, \tan \left (d x + c\right ) + 1217\right )}}{{\left (5 \, \tan \left (d x + c\right ) + 3\right )}^{2}} + 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 )}{39304 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/39304*(198*d*x + 198*c + 5*(75*tan(d*x + c)^2 + 1110*tan(d*x + c) + 1217)/(5*tan(d*x + c) + 3)^2 + 5*log(ta
n(d*x + c)^2 + 1) - 10*log(abs(5*tan(d*x + c) + 3)))/d