3.8 \(\int (1+x \tan (x)+\tan ^2(x)) \, dx\)

Optimal. Leaf size=42 \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\tan (x) \]

[Out]

(I/2)*x^2 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] + Tan[x]

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Rubi [A]  time = 0.0548971, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3719, 2190, 2279, 2391, 3473, 8} \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\tan (x) \]

Antiderivative was successfully verified.

[In]

Int[1 + x*Tan[x] + Tan[x]^2,x]

[Out]

(I/2)*x^2 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] + Tan[x]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \left (1+x \tan (x)+\tan ^2(x)\right ) \, dx &=x+\int x \tan (x) \, dx+\int \tan ^2(x) \, dx\\ &=x+\frac{i x^2}{2}+\tan (x)-2 i \int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx-\int 1 \, dx\\ &=\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\tan (x)+\int \log \left (1+e^{2 i x}\right ) \, dx\\ &=\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\tan (x)-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\frac{1}{2} i \text{Li}_2\left (-e^{2 i x}\right )+\tan (x)\\ \end{align*}

Mathematica [A]  time = 0.0124761, size = 42, normalized size = 1. \[ \frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i x}\right )+\frac{i x^2}{2}-x \log \left (1+e^{2 i x}\right )+\tan (x) \]

Antiderivative was successfully verified.

[In]

Integrate[1 + x*Tan[x] + Tan[x]^2,x]

[Out]

(I/2)*x^2 - x*Log[1 + E^((2*I)*x)] + (I/2)*PolyLog[2, -E^((2*I)*x)] + Tan[x]

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Maple [A]  time = 0.036, size = 43, normalized size = 1. \begin{align*}{\frac{i}{2}}{x}^{2}-x\ln \left ( 1+{{\rm e}^{2\,ix}} \right ) +{\frac{i}{2}}{\it polylog} \left ( 2,-{{\rm e}^{2\,ix}} \right ) +{\frac{2\,i}{1+{{\rm e}^{2\,ix}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1+x*tan(x)+tan(x)^2,x)

[Out]

1/2*I*x^2-x*ln(1+exp(2*I*x))+1/2*I*polylog(2,-exp(2*I*x))+2*I/(1+exp(2*I*x))

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Maxima [B]  time = 1.67233, size = 196, normalized size = 4.67 \begin{align*} x + \frac{x^{2} -{\left (2 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) +{\left (x^{2} + 2 i \, x\right )} \cos \left (2 \, x\right ) +{\left (\cos \left (2 \, x\right ) + i \, \sin \left (2 \, x\right ) + 1\right )}{\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) -{\left (-i \, x \cos \left (2 \, x\right ) + x \sin \left (2 \, x\right ) - i \, x\right )} \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) -{\left (-i \, x^{2} + 2 \, x\right )} \sin \left (2 \, x\right ) + 2 i \, x + 4}{-2 i \, \cos \left (2 \, x\right ) + 2 \, \sin \left (2 \, x\right ) - 2 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1+x*tan(x)+tan(x)^2,x, algorithm="maxima")

[Out]

x + (x^2 - (2*x*cos(2*x) + 2*I*x*sin(2*x) + 2*x)*arctan2(sin(2*x), cos(2*x) + 1) + (x^2 + 2*I*x)*cos(2*x) + (c
os(2*x) + I*sin(2*x) + 1)*dilog(-e^(2*I*x)) - (-I*x*cos(2*x) + x*sin(2*x) - I*x)*log(cos(2*x)^2 + sin(2*x)^2 +
2*cos(2*x) + 1) - (-I*x^2 + 2*x)*sin(2*x) + 2*I*x + 4)/(-2*I*cos(2*x) + 2*sin(2*x) - 2*I)

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Fricas [B]  time = 2.52804, size = 277, normalized size = 6.6 \begin{align*} -\frac{1}{2} \, x \log \left (-\frac{2 \,{\left (i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) - \frac{1}{2} \, x \log \left (-\frac{2 \,{\left (-i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1}\right ) - \frac{1}{4} i \,{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) + \frac{1}{4} i \,{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (x\right ) - 1\right )}}{\tan \left (x\right )^{2} + 1} + 1\right ) + \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1+x*tan(x)+tan(x)^2,x, algorithm="fricas")

[Out]

-1/2*x*log(-2*(I*tan(x) - 1)/(tan(x)^2 + 1)) - 1/2*x*log(-2*(-I*tan(x) - 1)/(tan(x)^2 + 1)) - 1/4*I*dilog(2*(I
*tan(x) - 1)/(tan(x)^2 + 1) + 1) + 1/4*I*dilog(2*(-I*tan(x) - 1)/(tan(x)^2 + 1) + 1) + tan(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (x \tan{\left (x \right )} + \tan ^{2}{\left (x \right )} + 1\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1+x*tan(x)+tan(x)**2,x)

[Out]

Integral(x*tan(x) + tan(x)**2 + 1, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \tan \left (x\right ) + \tan \left (x\right )^{2} + 1\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1+x*tan(x)+tan(x)^2,x, algorithm="giac")

[Out]

integrate(x*tan(x) + tan(x)^2 + 1, x)