3.380 \(\int \frac {1}{1+\tan ^3(x)} \, dx\)

Optimal. Leaf size=37 \[ \frac {x}{2}-\frac {1}{3} \log \left (\tan ^2(x)-\tan (x)+1\right )+\frac {1}{6} \log (\tan (x)+1)-\frac {1}{2} \log (\cos (x)) \]

[Out]

1/2*x-1/2*ln(cos(x))+1/6*ln(1+tan(x))-1/3*ln(1-tan(x)+tan(x)^2)

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Rubi [A]  time = 0.06, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3661, 2074, 635, 203, 260, 628} \[ \frac {x}{2}-\frac {1}{3} \log \left (\tan ^2(x)-\tan (x)+1\right )+\frac {1}{6} \log (\tan (x)+1)-\frac {1}{2} \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[x]^3)^(-1),x]

[Out]

x/2 - Log[Cos[x]]/2 + Log[1 + Tan[x]]/6 - Log[1 - Tan[x] + Tan[x]^2]/3

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {1}{1+\tan ^3(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (1+x^3\right )} \, dx,x,\tan (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{6 (1+x)}+\frac {1+x}{2 \left (1+x^2\right )}+\frac {1-2 x}{3 \left (1-x+x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=\frac {1}{6} \log (1+\tan (x))+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1-2 x}{1-x+x^2} \, dx,x,\tan (x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {1}{6} \log (1+\tan (x))-\frac {1}{3} \log \left (1-\tan (x)+\tan ^2(x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {x}{2}-\frac {1}{2} \log (\cos (x))+\frac {1}{6} \log (1+\tan (x))-\frac {1}{3} \log \left (1-\tan (x)+\tan ^2(x)\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 1.54 \[ -\frac {1}{3} \log \left (\tan ^2(x)-\tan (x)+1\right )+\left (\frac {1}{4}-\frac {i}{4}\right ) \log (-\tan (x)+i)+\left (\frac {1}{4}+\frac {i}{4}\right ) \log (\tan (x)+i)+\frac {1}{6} \log (\tan (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[x]^3)^(-1),x]

[Out]

(1/4 - I/4)*Log[I - Tan[x]] + (1/4 + I/4)*Log[I + Tan[x]] + Log[1 + Tan[x]]/6 - Log[1 - Tan[x] + Tan[x]^2]/3

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fricas [A]  time = 0.54, size = 48, normalized size = 1.30 \[ \frac {1}{2} \, x + \frac {1}{12} \, \log \left (\frac {\tan \relax (x)^{2} + 2 \, \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) - \frac {1}{3} \, \log \left (\frac {\tan \relax (x)^{2} - \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tan(x)^3),x, algorithm="fricas")

[Out]

1/2*x + 1/12*log((tan(x)^2 + 2*tan(x) + 1)/(tan(x)^2 + 1)) - 1/3*log((tan(x)^2 - tan(x) + 1)/(tan(x)^2 + 1))

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giac [A]  time = 0.59, size = 34, normalized size = 0.92 \[ \frac {1}{2} \, x - \frac {1}{3} \, \log \left (\tan \relax (x)^{2} - \tan \relax (x) + 1\right ) + \frac {1}{4} \, \log \left (\tan \relax (x)^{2} + 1\right ) + \frac {1}{6} \, \log \left ({\left | \tan \relax (x) + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tan(x)^3),x, algorithm="giac")

[Out]

1/2*x - 1/3*log(tan(x)^2 - tan(x) + 1) + 1/4*log(tan(x)^2 + 1) + 1/6*log(abs(tan(x) + 1))

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maple [A]  time = 0.10, size = 34, normalized size = 0.92 \[ -\frac {\ln \left (1-\tan \relax (x )+\tan ^{2}\relax (x )\right )}{3}+\frac {\ln \left (1+\tan \relax (x )\right )}{6}+\frac {\ln \left (1+\tan ^{2}\relax (x )\right )}{4}+\frac {x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+tan(x)^3),x)

[Out]

-1/3*ln(1-tan(x)+tan(x)^2)+1/6*ln(1+tan(x))+1/4*ln(1+tan(x)^2)+1/2*x

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maxima [A]  time = 0.75, size = 33, normalized size = 0.89 \[ \frac {1}{2} \, x - \frac {1}{3} \, \log \left (\tan \relax (x)^{2} - \tan \relax (x) + 1\right ) + \frac {1}{4} \, \log \left (\tan \relax (x)^{2} + 1\right ) + \frac {1}{6} \, \log \left (\tan \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tan(x)^3),x, algorithm="maxima")

[Out]

1/2*x - 1/3*log(tan(x)^2 - tan(x) + 1) + 1/4*log(tan(x)^2 + 1) + 1/6*log(tan(x) + 1)

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mupad [B]  time = 11.62, size = 41, normalized size = 1.11 \[ \frac {\ln \left (\mathrm {tan}\relax (x)+1\right )}{6}-\frac {\ln \left ({\mathrm {tan}\relax (x)}^2-\mathrm {tan}\relax (x)+1\right )}{3}+\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x)^3 + 1),x)

[Out]

log(tan(x) + 1)/6 - log(tan(x)^2 - tan(x) + 1)/3 + log(tan(x) - 1i)*(1/4 - 1i/4) + log(tan(x) + 1i)*(1/4 + 1i/
4)

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sympy [A]  time = 0.18, size = 34, normalized size = 0.92 \[ \frac {x}{2} + \frac {\log {\left (\tan {\relax (x )} + 1 \right )}}{6} + \frac {\log {\left (\tan ^{2}{\relax (x )} + 1 \right )}}{4} - \frac {\log {\left (\tan ^{2}{\relax (x )} - \tan {\relax (x )} + 1 \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tan(x)**3),x)

[Out]

x/2 + log(tan(x) + 1)/6 + log(tan(x)**2 + 1)/4 - log(tan(x)**2 - tan(x) + 1)/3

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