3.502 \(\int \frac{1}{(5+3 \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=88 \[ -\frac{99}{19652 d (3 \tan (c+d x)+5)}-\frac{15}{1156 d (3 \tan (c+d x)+5)^2}-\frac{1}{34 d (3 \tan (c+d x)+5)^3}+\frac{60 \log (3 \sin (c+d x)+5 \cos (c+d x))}{83521 d}-\frac{161 x}{334084} \]

[Out]

(-161*x)/334084 + (60*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(83521*d) - 1/(34*d*(5 + 3*Tan[c + d*x])^3) - 15/(
1156*d*(5 + 3*Tan[c + d*x])^2) - 99/(19652*d*(5 + 3*Tan[c + d*x]))

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Rubi [A]  time = 0.117856, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ -\frac{99}{19652 d (3 \tan (c+d x)+5)}-\frac{15}{1156 d (3 \tan (c+d x)+5)^2}-\frac{1}{34 d (3 \tan (c+d x)+5)^3}+\frac{60 \log (3 \sin (c+d x)+5 \cos (c+d x))}{83521 d}-\frac{161 x}{334084} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Tan[c + d*x])^(-4),x]

[Out]

(-161*x)/334084 + (60*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(83521*d) - 1/(34*d*(5 + 3*Tan[c + d*x])^3) - 15/(
1156*d*(5 + 3*Tan[c + d*x])^2) - 99/(19652*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))^4} \, dx &=-\frac{1}{34 d (5+3 \tan (c+d x))^3}+\frac{1}{34} \int \frac{5-3 \tan (c+d x)}{(5+3 \tan (c+d x))^3} \, dx\\ &=-\frac{1}{34 d (5+3 \tan (c+d x))^3}-\frac{15}{1156 d (5+3 \tan (c+d x))^2}+\frac{\int \frac{16-30 \tan (c+d x)}{(5+3 \tan (c+d x))^2} \, dx}{1156}\\ &=-\frac{1}{34 d (5+3 \tan (c+d x))^3}-\frac{15}{1156 d (5+3 \tan (c+d x))^2}-\frac{99}{19652 d (5+3 \tan (c+d x))}+\frac{\int \frac{-10-198 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx}{39304}\\ &=-\frac{161 x}{334084}-\frac{1}{34 d (5+3 \tan (c+d x))^3}-\frac{15}{1156 d (5+3 \tan (c+d x))^2}-\frac{99}{19652 d (5+3 \tan (c+d x))}+\frac{60 \int \frac{3-5 \tan (c+d x)}{5+3 \tan (c+d x)} \, dx}{83521}\\ &=-\frac{161 x}{334084}+\frac{60 \log (5 \cos (c+d x)+3 \sin (c+d x))}{83521 d}-\frac{1}{34 d (5+3 \tan (c+d x))^3}-\frac{15}{1156 d (5+3 \tan (c+d x))^2}-\frac{99}{19652 d (5+3 \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.670275, size = 95, normalized size = 1.08 \[ -\frac{\frac{3366}{3 \tan (c+d x)+5}+\frac{8670}{(3 \tan (c+d x)+5)^2}+\frac{19652}{(3 \tan (c+d x)+5)^3}+(240-161 i) \log (-\tan (c+d x)+i)+(240+161 i) \log (\tan (c+d x)+i)-480 \log (3 \tan (c+d x)+5)}{668168 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 3*Tan[c + d*x])^(-4),x]

[Out]

-((240 - 161*I)*Log[I - Tan[c + d*x]] + (240 + 161*I)*Log[I + Tan[c + d*x]] - 480*Log[5 + 3*Tan[c + d*x]] + 19
652/(5 + 3*Tan[c + d*x])^3 + 8670/(5 + 3*Tan[c + d*x])^2 + 3366/(5 + 3*Tan[c + d*x]))/(668168*d)

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Maple [A]  time = 0.021, size = 97, normalized size = 1.1 \begin{align*} -{\frac{30\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{83521\,d}}-{\frac{161\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{334084\,d}}-{\frac{1}{34\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{15}{1156\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{99}{19652\,d \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }}+{\frac{60\,\ln \left ( 5+3\,\tan \left ( dx+c \right ) \right ) }{83521\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5+3*tan(d*x+c))^4,x)

[Out]

-30/83521/d*ln(1+tan(d*x+c)^2)-161/334084/d*arctan(tan(d*x+c))-1/34/d/(5+3*tan(d*x+c))^3-15/1156/d/(5+3*tan(d*
x+c))^2-99/19652/d/(5+3*tan(d*x+c))+60/83521/d*ln(5+3*tan(d*x+c))

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Maxima [A]  time = 1.56489, size = 126, normalized size = 1.43 \begin{align*} -\frac{161 \, d x + 161 \, c + \frac{17 \,{\left (891 \, \tan \left (d x + c\right )^{2} + 3735 \, \tan \left (d x + c\right ) + 4328\right )}}{27 \, \tan \left (d x + c\right )^{3} + 135 \, \tan \left (d x + c\right )^{2} + 225 \, \tan \left (d x + c\right ) + 125} + 120 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{334084 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/334084*(161*d*x + 161*c + 17*(891*tan(d*x + c)^2 + 3735*tan(d*x + c) + 4328)/(27*tan(d*x + c)^3 + 135*tan(d
*x + c)^2 + 225*tan(d*x + c) + 125) + 120*log(tan(d*x + c)^2 + 1) - 240*log(3*tan(d*x + c) + 5))/d

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Fricas [B]  time = 1.76429, size = 474, normalized size = 5.39 \begin{align*} -\frac{27 \,{\left (161 \, d x - 305\right )} \tan \left (d x + c\right )^{3} + 27 \,{\left (805 \, d x - 964\right )} \tan \left (d x + c\right )^{2} + 20125 \, d x - 120 \,{\left (27 \, \tan \left (d x + c\right )^{3} + 135 \, \tan \left (d x + c\right )^{2} + 225 \, \tan \left (d x + c\right ) + 125\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 ) + 45 \,{\left (805 \, d x - 114\right )} \tan \left (d x + c\right ) + 35451}{334084 \,{\left (27 \, d \tan \left (d x + c\right )^{3} + 135 \, d \tan \left (d x + c\right )^{2} + 225 \, d \tan \left (d x + c\right ) + 125 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/334084*(27*(161*d*x - 305)*tan(d*x + c)^3 + 27*(805*d*x - 964)*tan(d*x + c)^2 + 20125*d*x - 120*(27*tan(d*x
+ c)^3 + 135*tan(d*x + c)^2 + 225*tan(d*x + c) + 125)*log((9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25)/(tan(d*x
+ c)^2 + 1)) + 45*(805*d*x - 114)*tan(d*x + c) + 35451)/(27*d*tan(d*x + c)^3 + 135*d*tan(d*x + c)^2 + 225*d*ta
n(d*x + c) + 125*d)

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Sympy [A]  time = 1.41232, size = 790, normalized size = 8.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*tan(d*x+c))**4,x)

[Out]

Piecewise((-21735*d*x*tan(c + d*x)**3/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*
tan(c + d*x) + 208802500*d) - 108675*d*x*tan(c + d*x)**2/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x
)**2 + 375844500*d*tan(c + d*x) + 208802500*d) - 181125*d*x*tan(c + d*x)/(45101340*d*tan(c + d*x)**3 + 2255067
00*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) - 100625*d*x/(45101340*d*tan(c + d*x)**3 + 2255
06700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) + 32400*log(tan(c + d*x) + 5/3)*tan(c + d*x)
**3/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) + 1620
00*log(tan(c + d*x) + 5/3)*tan(c + d*x)**2/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 3758445
00*d*tan(c + d*x) + 208802500*d) + 270000*log(tan(c + d*x) + 5/3)*tan(c + d*x)/(45101340*d*tan(c + d*x)**3 + 2
25506700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) + 150000*log(tan(c + d*x) + 5/3)/(4510134
0*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) - 16200*log(tan(c
+ d*x)**2 + 1)*tan(c + d*x)**3/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(c +
d*x) + 208802500*d) - 81000*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(45101340*d*tan(c + d*x)**3 + 225506700*
d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) - 135000*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(45
101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208802500*d) - 75000*log(t
an(c + d*x)**2 + 1)/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(c + d*x) + 208
802500*d) + 15147*tan(c + d*x)**3/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 375844500*d*tan(
c + d*x) + 208802500*d) - 191250*tan(c + d*x)/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 3758
44500*d*tan(c + d*x) + 208802500*d) - 297755/(45101340*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 37584
4500*d*tan(c + d*x) + 208802500*d), Ne(d, 0)), (x/(3*tan(c) + 5)**4, True))

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Giac [A]  time = 1.23258, size = 112, normalized size = 1.27 \begin{align*} -\frac{161 \, d x + 161 \, c + \frac{11880 \, \tan \left (d x + c\right )^{3} + 74547 \, \tan \left (d x + c\right )^{2} + 162495 \, \tan \left (d x + c\right ) + 128576}{{\left (3 \, \tan \left (d x + c\right ) + 5\right )}^{3}} + 120 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{334084 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/334084*(161*d*x + 161*c + (11880*tan(d*x + c)^3 + 74547*tan(d*x + c)^2 + 162495*tan(d*x + c) + 128576)/(3*t
an(d*x + c) + 5)^3 + 120*log(tan(d*x + c)^2 + 1) - 240*log(abs(3*tan(d*x + c) + 5)))/d