3.60 \(\int \sqrt{x} \tan (\sqrt{x}) \, dx\)

Optimal. Leaf size=70 \[ 2 i \sqrt{x} \text{PolyLog}\left (2,-e^{2 i \sqrt{x}}\right )-\text{PolyLog}\left (3,-e^{2 i \sqrt{x}}\right )+\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right ) \]

[Out]

((2*I)/3)*x^(3/2) - 2*x*Log[1 + E^((2*I)*Sqrt[x])] + (2*I)*Sqrt[x]*PolyLog[2, -E^((2*I)*Sqrt[x])] - PolyLog[3,
-E^((2*I)*Sqrt[x])]

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Rubi [A]  time = 0.0912281, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3747, 3719, 2190, 2531, 2282, 6589} \[ 2 i \sqrt{x} \text{PolyLog}\left (2,-e^{2 i \sqrt{x}}\right )-\text{PolyLog}\left (3,-e^{2 i \sqrt{x}}\right )+\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Tan[Sqrt[x]],x]

[Out]

((2*I)/3)*x^(3/2) - 2*x*Log[1 + E^((2*I)*Sqrt[x])] + (2*I)*Sqrt[x]*PolyLog[2, -E^((2*I)*Sqrt[x])] - PolyLog[3,
-E^((2*I)*Sqrt[x])]

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \sqrt{x} \tan \left (\sqrt{x}\right ) \, dx &=2 \operatorname{Subst}\left (\int x^2 \tan (x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} i x^{3/2}-4 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right )+4 \operatorname{Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right )+2 i \sqrt{x} \text{Li}_2\left (-e^{2 i \sqrt{x}}\right )-2 i \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right )+2 i \sqrt{x} \text{Li}_2\left (-e^{2 i \sqrt{x}}\right )-\operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sqrt{x}}\right )\\ &=\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right )+2 i \sqrt{x} \text{Li}_2\left (-e^{2 i \sqrt{x}}\right )-\text{Li}_3\left (-e^{2 i \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0205173, size = 70, normalized size = 1. \[ 2 i \sqrt{x} \text{PolyLog}\left (2,-e^{2 i \sqrt{x}}\right )-\text{PolyLog}\left (3,-e^{2 i \sqrt{x}}\right )+\frac{2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Tan[Sqrt[x]],x]

[Out]

((2*I)/3)*x^(3/2) - 2*x*Log[1 + E^((2*I)*Sqrt[x])] + (2*I)*Sqrt[x]*PolyLog[2, -E^((2*I)*Sqrt[x])] - PolyLog[3,
-E^((2*I)*Sqrt[x])]

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int \sqrt{x}\tan \left ( \sqrt{x} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.52134, size = 108, normalized size = 1.54 \begin{align*} -2 i \, x \arctan \left (\sin \left (2 \, \sqrt{x}\right ), \cos \left (2 \, \sqrt{x}\right ) + 1\right ) - x \log \left (\cos \left (2 \, \sqrt{x}\right )^{2} + \sin \left (2 \, \sqrt{x}\right )^{2} + 2 \, \cos \left (2 \, \sqrt{x}\right ) + 1\right ) + \frac{2}{3} i \, x^{\frac{3}{2}} + 2 i \, \sqrt{x}{\rm Li}_2\left (-e^{\left (2 i \, \sqrt{x}\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 i \, \sqrt{x}\right )}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*x*arctan2(sin(2*sqrt(x)), cos(2*sqrt(x)) + 1) - x*log(cos(2*sqrt(x))^2 + sin(2*sqrt(x))^2 + 2*cos(2*sqrt(
x)) + 1) + 2/3*I*x^(3/2) + 2*I*sqrt(x)*dilog(-e^(2*I*sqrt(x))) - polylog(3, -e^(2*I*sqrt(x)))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{x} \tan \left (\sqrt{x}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x)*tan(sqrt(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x)*tan(sqrt(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x)*tan(sqrt(x)), x)