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

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

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Rubi [A]  time = 0.10, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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]

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*}

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Mathematica [A]  time = 0.22, 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]

-1/500*(-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])/(Cos[x] + 3*Sin[x])^2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {5-\tan (x)-6 \tan ^2(x)}{(1+3 \tan (x))^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 1.09, size = 77, normalized size = 1.83 \[ -\frac {9 \, {\left (134 \, x - 1\right )} \tan \relax (x)^{2} + 56 \, {\left (9 \, \tan \relax (x)^{2} + 6 \, \tan \relax (x) + 1\right )} \log \left (\frac {9 \, \tan \relax (x)^{2} + 6 \, \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) + 12 \, {\left (67 \, x + 72\right )} \tan \relax (x) + 134 \, x + 639}{500 \, {\left (9 \, \tan \relax (x)^{2} + 6 \, \tan \relax (x) + 1\right )}} \]

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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giac [A]  time = 0.64, size = 39, normalized size = 0.93 \[ -\frac {67}{250} \, x - \frac {87 \, \tan \relax (x) + 64}{50 \, {\left (3 \, \tan \relax (x) + 1\right )}^{2}} + \frac {14}{125} \, \log \left (\tan \relax (x)^{2} + 1\right ) - \frac {28}{125} \, \log \left ({\left | 3 \, \tan \relax (x) + 1 \right |}\right ) \]

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))

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maple [A]  time = 0.10, size = 45, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {14 \ln \left (1+\tan ^{2}\relax (x )\right )}{125}-\frac {67 \arctan \left (\tan \relax (x )\right )}{250}-\frac {7}{10 \left (1+3 \tan \relax (x )\right )^{2}}-\frac {29}{50 \left (1+3 \tan \relax (x )\right )}-\frac {28 \ln \left (1+3 \tan \relax (x )\right )}{125}\)	\(45\)
default	\(\frac {14 \ln \left (1+\tan ^{2}\relax (x )\right )}{125}-\frac {67 \arctan \left (\tan \relax (x )\right )}{250}-\frac {7}{10 \left (1+3 \tan \relax (x )\right )^{2}}-\frac {29}{50 \left (1+3 \tan \relax (x )\right )}-\frac {28 \ln \left (1+3 \tan \relax (x )\right )}{125}\)	\(45\)
risch	\(-\frac {67 x}{250}+\frac {28 i x}{125}+\frac {\left (-\frac {36}{24125}-\frac {621 i}{48250}\right ) \left (965 \,{\mathrm e}^{2 i x}-324+768 i\right )}{\left (5 \,{\mathrm e}^{2 i x}-4+3 i\right )^{2}}-\frac {28 \ln \left ({\mathrm e}^{2 i x}-\frac {4}{5}+\frac {3 i}{5}\right )}{125}\)	\(49\)
norman	\(\frac {\frac {297 \tan \relax (x )}{50}+\frac {288 \left (\tan ^{2}\relax (x )\right )}{25}-\frac {67 x}{250}-\frac {201 x \tan \relax (x )}{125}-\frac {603 x \left (\tan ^{2}\relax (x )\right )}{250}}{\left (1+3 \tan \relax (x )\right )^{2}}-\frac {28 \ln \left (1+3 \tan \relax (x )\right )}{125}+\frac {14 \ln \left (1+\tan ^{2}\relax (x )\right )}{125}\)	\(55\)

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,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.00, size = 44, normalized size = 1.05 \[ -\frac {67}{250} \, x - \frac {87 \, \tan \relax (x) + 64}{50 \, {\left (9 \, \tan \relax (x)^{2} + 6 \, \tan \relax (x) + 1\right )}} + \frac {14}{125} \, \log \left (\tan \relax (x)^{2} + 1\right ) - \frac {28}{125} \, \log \left (3 \, \tan \relax (x) + 1\right ) \]

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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mupad [B]  time = 0.28, size = 48, normalized size = 1.14 \[ -\frac {28\,\ln \left (\mathrm {tan}\relax (x)+\frac {1}{3}\right )}{125}-\frac {\frac {29\,\mathrm {tan}\relax (x)}{150}+\frac {32}{225}}{{\mathrm {tan}\relax (x)}^2+\frac {2\,\mathrm {tan}\relax (x)}{3}+\frac {1}{9}}+\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,\left (\frac {14}{125}+\frac {67}{500}{}\mathrm {i}\right )+\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,\left (\frac {14}{125}-\frac {67}{500}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(x) - 1i)*(14/125 + 67i/500) - (28*log(tan(x) + 1/3))/125 + log(tan(x) + 1i)*(14/125 - 67i/500) - ((29*
tan(x))/150 + 32/225)/((2*tan(x))/3 + tan(x)^2 + 1/9)

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sympy [B]  time = 0.56, size = 252, normalized size = 6.00 \[ - \frac {603 x \tan ^{2}{\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {402 x \tan {\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {67 x}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {504 \log {\left (\tan {\relax (x )} + \frac {1}{3} \right )} \tan ^{2}{\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {336 \log {\left (\tan {\relax (x )} + \frac {1}{3} \right )} \tan {\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {56 \log {\left (\tan {\relax (x )} + \frac {1}{3} \right )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} + \frac {252 \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} \tan ^{2}{\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} + \frac {168 \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} \tan {\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} + \frac {28 \log {\left (\tan ^{2}{\relax (x )} + 1 \right )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {435 \tan {\relax (x )}}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} - \frac {320}{2250 \tan ^{2}{\relax (x )} + 1500 \tan {\relax (x )} + 250} \]

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]

-603*x*tan(x)**2/(2250*tan(x)**2 + 1500*tan(x) + 250) - 402*x*tan(x)/(2250*tan(x)**2 + 1500*tan(x) + 250) - 67
*x/(2250*tan(x)**2 + 1500*tan(x) + 250) - 504*log(tan(x) + 1/3)*tan(x)**2/(2250*tan(x)**2 + 1500*tan(x) + 250)
- 336*log(tan(x) + 1/3)*tan(x)/(2250*tan(x)**2 + 1500*tan(x) + 250) - 56*log(tan(x) + 1/3)/(2250*tan(x)**2 +
1500*tan(x) + 250) + 252*log(tan(x)**2 + 1)*tan(x)**2/(2250*tan(x)**2 + 1500*tan(x) + 250) + 168*log(tan(x)**2
+ 1)*tan(x)/(2250*tan(x)**2 + 1500*tan(x) + 250) + 28*log(tan(x)**2 + 1)/(2250*tan(x)**2 + 1500*tan(x) + 250)
- 435*tan(x)/(2250*tan(x)**2 + 1500*tan(x) + 250) - 320/(2250*tan(x)**2 + 1500*tan(x) + 250)

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