3.382 \(\int \frac{5-\tan (x)-6 \tan ^2(x)}{(1+3 \tan (x))^3} \, dx\)

Optimal. Leaf size=42 \[ -\frac{67 x}{250}-\frac{29}{50 (3 \tan (x)+1)}-\frac{7}{10 (3 \tan (x)+1)^2}-\frac{28}{125} \log (3 \sin (x)+\cos (x)) \]

[Out]

(-67*x)/250 - (28*Log[Cos[x] + 3*Sin[x]])/125 - 7/(10*(1 + 3*Tan[x])^2) - 29/(50*(1 + 3*Tan[x]))

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Rubi [A]  time = 0.0958575, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3628, 3529, 3531, 3530} \[ -\frac{67 x}{250}-\frac{29}{50 (3 \tan (x)+1)}-\frac{7}{10 (3 \tan (x)+1)^2}-\frac{28}{125} \log (3 \sin (x)+\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(5 - Tan[x] - 6*Tan[x]^2)/(1 + 3*Tan[x])^3,x]

[Out]

(-67*x)/250 - (28*Log[Cos[x] + 3*Sin[x]])/125 - 7/(10*(1 + 3*Tan[x])^2) - 29/(50*(1 + 3*Tan[x]))

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

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{5-\tan (x)-6 \tan ^2(x)}{(1+3 \tan (x))^3} \, dx &=-\frac{7}{10 (1+3 \tan (x))^2}+\frac{1}{10} \int \frac{8-34 \tan (x)}{(1+3 \tan (x))^2} \, dx\\ &=-\frac{7}{10 (1+3 \tan (x))^2}-\frac{29}{50 (1+3 \tan (x))}+\frac{1}{100} \int \frac{-94-58 \tan (x)}{1+3 \tan (x)} \, dx\\ &=-\frac{67 x}{250}-\frac{7}{10 (1+3 \tan (x))^2}-\frac{29}{50 (1+3 \tan (x))}-\frac{28}{125} \int \frac{3-\tan (x)}{1+3 \tan (x)} \, dx\\ &=-\frac{67 x}{250}-\frac{28}{125} \log (\cos (x)+3 \sin (x))-\frac{7}{10 (1+3 \tan (x))^2}-\frac{29}{50 (1+3 \tan (x))}\\ \end{align*}

Mathematica [A]  time = 0.21688, size = 70, normalized size = 1.67 \[ -\frac{670 x+560 \log (3 \sin (x)+\cos (x))-4 \cos (2 x) (134 x+112 \log (3 \sin (x)+\cos (x))-405)+6 \sin (2 x) (67 x+56 \log (3 \sin (x)+\cos (x))-90)-1305}{500 (3 \sin (x)+\cos (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - Tan[x] - 6*Tan[x]^2)/(1 + 3*Tan[x])^3,x]

[Out]

-(-1305 + 670*x + 560*Log[Cos[x] + 3*Sin[x]] - 4*Cos[2*x]*(-405 + 134*x + 112*Log[Cos[x] + 3*Sin[x]]) + 6*(-90
+ 67*x + 56*Log[Cos[x] + 3*Sin[x]])*Sin[2*x])/(500*(Cos[x] + 3*Sin[x])^2)

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Maple [A]  time = 0.025, size = 43, normalized size = 1. \begin{align*}{\frac{14\,\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }{125}}-{\frac{7}{10\, \left ( 1+3\,\tan \left ( x \right ) \right ) ^{2}}}-{\frac{29}{50+150\,\tan \left ( x \right ) }}-{\frac{28\,\ln \left ( 1+3\,\tan \left ( x \right ) \right ) }{125}}-{\frac{67\,x}{250}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-tan(x)-6*tan(x)^2)/(1+3*tan(x))^3,x)

[Out]

14/125*ln(tan(x)^2+1)-7/10/(1+3*tan(x))^2-29/50/(1+3*tan(x))-28/125*ln(1+3*tan(x))-67/250*x

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Maxima [A]  time = 1.41712, size = 59, normalized size = 1.4 \begin{align*} -\frac{67}{250} \, x - \frac{87 \, \tan \left (x\right ) + 64}{50 \,{\left (9 \, \tan \left (x\right )^{2} + 6 \, \tan \left (x\right ) + 1\right )}} + \frac{14}{125} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac{28}{125} \, \log \left (3 \, \tan \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-tan(x)-6*tan(x)^2)/(1+3*tan(x))^3,x, algorithm="maxima")

[Out]

-67/250*x - 1/50*(87*tan(x) + 64)/(9*tan(x)^2 + 6*tan(x) + 1) + 14/125*log(tan(x)^2 + 1) - 28/125*log(3*tan(x)
+ 1)

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Fricas [B]  time = 2.24171, size = 243, normalized size = 5.79 \begin{align*} -\frac{9 \,{\left (134 \, x - 1\right )} \tan \left (x\right )^{2} + 56 \,{\left (9 \, \tan \left (x\right )^{2} + 6 \, \tan \left (x\right ) + 1\right )} \log \left (\frac{9 \, \tan \left (x\right )^{2} + 6 \, \tan \left (x\right ) + 1}{\tan \left (x\right )^{2} + 1}\right ) + 12 \,{\left (67 \, x + 72\right )} \tan \left (x\right ) + 134 \, x + 639}{500 \,{\left (9 \, \tan \left (x\right )^{2} + 6 \, \tan \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-tan(x)-6*tan(x)^2)/(1+3*tan(x))^3,x, algorithm="fricas")

[Out]

-1/500*(9*(134*x - 1)*tan(x)^2 + 56*(9*tan(x)^2 + 6*tan(x) + 1)*log((9*tan(x)^2 + 6*tan(x) + 1)/(tan(x)^2 + 1)
) + 12*(67*x + 72)*tan(x) + 134*x + 639)/(9*tan(x)^2 + 6*tan(x) + 1)

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Sympy [B]  time = 0.539959, size = 253, normalized size = 6.02 \begin{align*} - \frac{1206 x \tan ^{2}{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{804 x \tan{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{134 x}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{1008 \log{\left (\tan{\left (x \right )} + \frac{1}{3} \right )} \tan ^{2}{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{672 \log{\left (\tan{\left (x \right )} + \frac{1}{3} \right )} \tan{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{112 \log{\left (\tan{\left (x \right )} + \frac{1}{3} \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} + \frac{504 \log{\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan ^{2}{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} + \frac{336 \log{\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} + \frac{56 \log{\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} + \frac{1305 \tan ^{2}{\left (x \right )}}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} - \frac{495}{4500 \tan ^{2}{\left (x \right )} + 3000 \tan{\left (x \right )} + 500} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-tan(x)-6*tan(x)**2)/(1+3*tan(x))**3,x)

[Out]

-1206*x*tan(x)**2/(4500*tan(x)**2 + 3000*tan(x) + 500) - 804*x*tan(x)/(4500*tan(x)**2 + 3000*tan(x) + 500) - 1
34*x/(4500*tan(x)**2 + 3000*tan(x) + 500) - 1008*log(tan(x) + 1/3)*tan(x)**2/(4500*tan(x)**2 + 3000*tan(x) + 5
00) - 672*log(tan(x) + 1/3)*tan(x)/(4500*tan(x)**2 + 3000*tan(x) + 500) - 112*log(tan(x) + 1/3)/(4500*tan(x)**
2 + 3000*tan(x) + 500) + 504*log(tan(x)**2 + 1)*tan(x)**2/(4500*tan(x)**2 + 3000*tan(x) + 500) + 336*log(tan(x
)**2 + 1)*tan(x)/(4500*tan(x)**2 + 3000*tan(x) + 500) + 56*log(tan(x)**2 + 1)/(4500*tan(x)**2 + 3000*tan(x) +
500) + 1305*tan(x)**2/(4500*tan(x)**2 + 3000*tan(x) + 500) - 495/(4500*tan(x)**2 + 3000*tan(x) + 500)

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Giac [A]  time = 1.12964, size = 53, normalized size = 1.26 \begin{align*} -\frac{67}{250} \, x - \frac{87 \, \tan \left (x\right ) + 64}{50 \,{\left (3 \, \tan \left (x\right ) + 1\right )}^{2}} + \frac{14}{125} \, \log \left (\tan \left (x\right )^{2} + 1\right ) - \frac{28}{125} \, \log \left ({\left | 3 \, \tan \left (x\right ) + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-tan(x)-6*tan(x)^2)/(1+3*tan(x))^3,x, algorithm="giac")

[Out]

-67/250*x - 1/50*(87*tan(x) + 64)/(3*tan(x) + 1)^2 + 14/125*log(tan(x)^2 + 1) - 28/125*log(abs(3*tan(x) + 1))