3.2.93 \(\int \log (\sin (\sqrt {x})) \, dx\) [193]

Optimal. Leaf size=79 \[ \frac {1}{3} i x^{3/2}-x \log \left (1-e^{2 i \sqrt {x}}\right )+x \log \left (\sin \left (\sqrt {x}\right )\right )+i \sqrt {x} \text {Li}_2\left (e^{2 i \sqrt {x}}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 i \sqrt {x}}\right ) \]

[Out]

1/3*I*x^(3/2)-x*ln(1-exp(2*I*x^(1/2)))+x*ln(sin(x^(1/2)))-1/2*polylog(3,exp(2*I*x^(1/2)))+I*polylog(2,exp(2*I*
x^(1/2)))*x^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 = {2628, 12, 3833, 3798, 2221, 2611, 2320, 6724} \begin {gather*} 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 ) \end {gather*}

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 12

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 2628

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

Rule 3798

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 3833

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 6724

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 )-\text {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 \text {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 \text {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 \text {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} \text {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*}

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Mathematica [A]
time = 0.03, size = 88, normalized size = 1.11 \begin {gather*} \frac {i \pi ^3}{24}-\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 {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \ln \left (\sin \left (\sqrt {x}\right )\right )\, dx\]

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (49) = 98\).
time = 0.29, size = 139, normalized size = 1.76 \begin {gather*} -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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \log {\left (\sin {\left (\sqrt {x} \right )} \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \ln \left (\sin \left (\sqrt {x}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(log(sin(x^(1/2))), x)

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