### 3.193 $$\int \log (\sin (\sqrt{x})) \, dx$$

Optimal. Leaf size=79 $i \sqrt{x} \text{PolyLog}\left (2,e^{2 i \sqrt{x}}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 i \sqrt{x}}\right )+\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )$

[Out]

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

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Rubi [A]  time = 0.102098, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 1.143, Rules used = {2548, 12, 3748, 3717, 2190, 2531, 2282, 6589} $i \sqrt{x} \text{PolyLog}\left (2,e^{2 i \sqrt{x}}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 i \sqrt{x}}\right )+\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[Sin[Sqrt[x]]],x]

[Out]

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

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3748

Int[((a_.) + Cot[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cot[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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 \log \left (\sin \left (\sqrt{x}\right )\right ) \, dx &=x \log \left (\sin \left (\sqrt{x}\right )\right )-\int \frac{1}{2} \sqrt{x} \cot \left (\sqrt{x}\right ) \, dx\\ &=x \log \left (\sin \left (\sqrt{x}\right )\right )-\frac{1}{2} \int \sqrt{x} \cot \left (\sqrt{x}\right ) \, dx\\ &=x \log \left (\sin \left (\sqrt{x}\right )\right )-\operatorname{Subst}\left (\int x^2 \cot (x) \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{3} i x^{3/2}+x \log \left (\sin \left (\sqrt{x}\right )\right )+2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} x^2}{1-e^{2 i x}} \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )+2 \operatorname{Subst}\left (\int x \log \left (1-e^{2 i x}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )+i \sqrt{x} \text{Li}_2\left (e^{2 i \sqrt{x}}\right )-i \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )+i \sqrt{x} \text{Li}_2\left (e^{2 i \sqrt{x}}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sqrt{x}}\right )\\ &=\frac{1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )+i \sqrt{x} \text{Li}_2\left (e^{2 i \sqrt{x}}\right )-\frac{1}{2} \text{Li}_3\left (e^{2 i \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0324164, size = 88, normalized size = 1.11 $-i \sqrt{x} \text{PolyLog}\left (2,e^{-2 i \sqrt{x}}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \sqrt{x}}\right )-\frac{1}{3} i x^{3/2}-x \log \left (1-e^{-2 i \sqrt{x}}\right )+x \log \left (\sin \left (\sqrt{x}\right )\right )+\frac{i \pi ^3}{24}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[Sin[Sqrt[x]]],x]

[Out]

(I/24)*Pi^3 - (I/3)*x^(3/2) - x*Log[1 - E^((-2*I)*Sqrt[x])] + x*Log[Sin[Sqrt[x]]] - I*Sqrt[x]*PolyLog[2, E^((-
2*I)*Sqrt[x])] - PolyLog[3, E^((-2*I)*Sqrt[x])]/2

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int \ln \left ( \sin \left ( \sqrt{x} \right ) \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(ln(sin(x^(1/2))),x)

[Out]

int(ln(sin(x^(1/2))),x)

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Maxima [B]  time = 1.06539, size = 188, normalized size = 2.38 \begin{align*} -i \, x \arctan \left (\sin \left (\sqrt{x}\right ), \cos \left (\sqrt{x}\right ) + 1\right ) + i \, x \arctan \left (\sin \left (\sqrt{x}\right ), -\cos \left (\sqrt{x}\right ) + 1\right ) - \frac{1}{2} \, x \log \left (\cos \left (\sqrt{x}\right )^{2} + \sin \left (\sqrt{x}\right )^{2} + 2 \, \cos \left (\sqrt{x}\right ) + 1\right ) - \frac{1}{2} \, x \log \left (\cos \left (\sqrt{x}\right )^{2} + \sin \left (\sqrt{x}\right )^{2} - 2 \, \cos \left (\sqrt{x}\right ) + 1\right ) + x \log \left (\sin \left (\sqrt{x}\right )\right ) + \frac{1}{3} i \, x^{\frac{3}{2}} + 2 i \, \sqrt{x}{\rm Li}_2\left (-e^{\left (i \, \sqrt{x}\right )}\right ) + 2 i \, \sqrt{x}{\rm Li}_2\left (e^{\left (i \, \sqrt{x}\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (i \, \sqrt{x}\right )}) - 2 \,{\rm Li}_{3}(e^{\left (i \, \sqrt{x}\right )}) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x^(1/2))),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x^(1/2))),x, algorithm="fricas")

[Out]

integral(log(sin(sqrt(x))), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(sin(x**(1/2))),x)

[Out]

Integral(log(sin(sqrt(x))), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(sin(x^(1/2))),x, algorithm="giac")

[Out]

integrate(log(sin(sqrt(x))), x)