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]

-161/334084*x+60/83521*ln(5*cos(d*x+c)+3*sin(d*x+c))/d-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))

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Rubi [A]  time = 0.12, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.66, 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]

-1/668168*((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]] + 19652/(5 + 3*Tan[c + d*x])^3 + 8670/(5 + 3*Tan[c + d*x])^2 + 3366/(5 + 3*Tan[c + d*x]))/d

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fricas [B]  time = 0.46, size = 157, normalized size = 1.78 \[ -\frac {27 \, {\left (161 \, d x - 305\right )} \tan \left (d x + c\right )^{3} + 27 \, {\left (805 \, d x - 964\right )} \tan \left (d x + c\right )^{2} + 20125 \, d x - 120 \, {\left (27 \, \tan \left (d x + c\right )^{3} + 135 \, \tan \left (d x + c\right )^{2} + 225 \, \tan \left (d x + c\right ) + 125\right )} \log \left (\frac {9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25}{\tan \left (d x + c\right )^{2} + 1}\right ) + 45 \, {\left (805 \, d x - 114\right )} \tan \left (d x + c\right ) + 35451}{334084 \, {\left (27 \, d \tan \left (d x + c\right )^{3} + 135 \, d \tan \left (d x + c\right )^{2} + 225 \, d \tan \left (d x + c\right ) + 125 \, d\right )}} \]

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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giac [A]  time = 0.73, size = 83, normalized size = 0.94 \[ -\frac {161 \, d x + 161 \, c + \frac {11880 \, \tan \left (d x + c\right )^{3} + 74547 \, \tan \left (d x + c\right )^{2} + 162495 \, \tan \left (d x + c\right ) + 128576}{{\left (3 \, \tan \left (d x + c\right ) + 5\right )}^{3}} + 120 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{334084 \, d} \]

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

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maple [A]  time = 0.17, size = 97, normalized size = 1.10 \[ -\frac {30 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{83521 d}-\frac {161 \arctan \left (\tan \left (d x +c \right )\right )}{334084 d}-\frac {1}{34 d \left (5+3 \tan \left (d x +c \right )\right )^{3}}-\frac {15}{1156 d \left (5+3 \tan \left (d x +c \right )\right )^{2}}-\frac {99}{19652 d \left (5+3 \tan \left (d x +c \right )\right )}+\frac {60 \ln \left (5+3 \tan \left (d x +c \right )\right )}{83521 d} \]

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 = 0.59, size = 93, normalized size = 1.06 \[ -\frac {161 \, d x + 161 \, c + \frac {17 \, {\left (891 \, \tan \left (d x + c\right )^{2} + 3735 \, \tan \left (d x + c\right ) + 4328\right )}}{27 \, \tan \left (d x + c\right )^{3} + 135 \, \tan \left (d x + c\right )^{2} + 225 \, \tan \left (d x + c\right ) + 125} + 120 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{334084 \, d} \]

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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mupad [B]  time = 4.07, size = 104, normalized size = 1.18 \[ \frac {60\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {5}{3}\right )}{83521\,d}-\frac {\frac {33\,{\mathrm {tan}\left (c+d\,x\right )}^2}{19652}+\frac {415\,\mathrm {tan}\left (c+d\,x\right )}{58956}+\frac {1082}{132651}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3+5\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {25\,\mathrm {tan}\left (c+d\,x\right )}{3}+\frac {125}{27}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {30}{83521}+\frac {161}{668168}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {30}{83521}-\frac {161}{668168}{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*tan(c + d*x) + 5)^4,x)

[Out]

(60*log(tan(c + d*x) + 5/3))/(83521*d) - (log(tan(c + d*x) + 1i)*(30/83521 + 161i/668168))/d - (log(tan(c + d*
x) - 1i)*(30/83521 - 161i/668168))/d - ((415*tan(c + d*x))/58956 + (33*tan(c + d*x)^2)/19652 + 1082/132651)/(d
*((25*tan(c + d*x))/3 + 5*tan(c + d*x)^2 + tan(c + d*x)^3 + 125/27))

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sympy [A]  time = 1.14, size = 790, normalized size = 8.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4347*d*x*tan(c + d*x)**3/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(
c + d*x) + 41760500*d) - 21735*d*x*tan(c + d*x)**2/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 7
5168900*d*tan(c + d*x) + 41760500*d) - 36225*d*x*tan(c + d*x)/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c +
d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 20125*d*x/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*
x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) + 6480*log(tan(c + d*x) + 5/3)*tan(c + d*x)**3/(9020268*d*tan(c
+ d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) + 32400*log(tan(c + d*x) + 5/3)
*tan(c + d*x)**2/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*
d) + 54000*log(tan(c + d*x) + 5/3)*tan(c + d*x)/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 7516
8900*d*tan(c + d*x) + 41760500*d) + 30000*log(tan(c + d*x) + 5/3)/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(
c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 3240*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**3/(9020268*d
*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 16200*log(tan(c + d*x)
**2 + 1)*tan(c + d*x)**2/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 4
1760500*d) - 27000*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(9020268*d*tan(c + d*x)**3 + 45101340*d*tan(c + d*x)*
*2 + 75168900*d*tan(c + d*x) + 41760500*d) - 15000*log(tan(c + d*x)**2 + 1)/(9020268*d*tan(c + d*x)**3 + 45101
340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 15147*tan(c + d*x)**2/(9020268*d*tan(c + d*x)*
*3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 63495*tan(c + d*x)/(9020268*d*tan(c
+ d*x)**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d) - 73576/(9020268*d*tan(c + d*x)
**3 + 45101340*d*tan(c + d*x)**2 + 75168900*d*tan(c + d*x) + 41760500*d), Ne(d, 0)), (x/(3*tan(c) + 5)**4, Tru
e))

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