3.47.11 \(\int \frac {15 x-25 x \log (x)+(-4+9 x+2 x^2+3 x^3) \log ^2(x)+(5 x-5 x \log (x)) \log (x^2)}{4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ x-\log (x)+\frac {1}{4} x \left (5+x+x^2-\frac {5 \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \]

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Rubi [A]  time = 0.39, antiderivative size = 46, normalized size of antiderivative = 1.59, number of steps used = 16, number of rules used = 8, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 6742, 6688, 2297, 2298, 2360, 2361, 6496} \begin {gather*} \frac {x^3}{4}+\frac {x^2}{4}-\frac {5 x \log \left (x^2\right )}{4 \log (x)}+\frac {9 x}{4}-\frac {15 x}{4 \log (x)}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*x - 25*x*Log[x] + (-4 + 9*x + 2*x^2 + 3*x^3)*Log[x]^2 + (5*x - 5*x*Log[x])*Log[x^2])/(4*x*Log[x]^2),x]

[Out]

(9*x)/4 + x^2/4 + x^3/4 - (15*x)/(4*Log[x]) - Log[x] - (5*x*Log[x^2])/(4*Log[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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 6496

Int[LogIntegral[(b_.)*(x_)]/(x_), x_Symbol] :> -Simp[b*x, x] + Simp[Log[b*x]*LogIntegral[b*x], x] /; FreeQ[b,
x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {15 x-25 x \log (x)+\left (-4+9 x+2 x^2+3 x^3\right ) \log ^2(x)+(5 x-5 x \log (x)) \log \left (x^2\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (\frac {15 x-25 x \log (x)-4 \log ^2(x)+9 x \log ^2(x)+2 x^2 \log ^2(x)+3 x^3 \log ^2(x)}{x \log ^2(x)}-\frac {5 (-1+\log (x)) \log \left (x^2\right )}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{4} \int \frac {15 x-25 x \log (x)-4 \log ^2(x)+9 x \log ^2(x)+2 x^2 \log ^2(x)+3 x^3 \log ^2(x)}{x \log ^2(x)} \, dx-\frac {5}{4} \int \frac {(-1+\log (x)) \log \left (x^2\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (9-\frac {4}{x}+2 x+3 x^2+\frac {15}{\log ^2(x)}-\frac {25}{\log (x)}\right ) \, dx-\frac {5}{4} \int \left (-\frac {\log \left (x^2\right )}{\log ^2(x)}+\frac {\log \left (x^2\right )}{\log (x)}\right ) \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\log (x)+\frac {5}{4} \int \frac {\log \left (x^2\right )}{\log ^2(x)} \, dx-\frac {5}{4} \int \frac {\log \left (x^2\right )}{\log (x)} \, dx+\frac {15}{4} \int \frac {1}{\log ^2(x)} \, dx-\frac {25}{4} \int \frac {1}{\log (x)} \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}-\frac {25 \text {li}(x)}{4}+\frac {5}{2} \int \frac {\text {li}(x)}{x} \, dx-\frac {5}{2} \int \left (-\frac {1}{\log (x)}+\frac {\text {li}(x)}{x}\right ) \, dx+\frac {15}{4} \int \frac {1}{\log (x)} \, dx\\ &=-\frac {x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}-\frac {5 \text {li}(x)}{2}+\frac {5}{2} \log (x) \text {li}(x)+\frac {5}{2} \int \frac {1}{\log (x)} \, dx-\frac {5}{2} \int \frac {\text {li}(x)}{x} \, dx\\ &=\frac {9 x}{4}+\frac {x^2}{4}+\frac {x^3}{4}-\frac {15 x}{4 \log (x)}-\log (x)-\frac {5 x \log \left (x^2\right )}{4 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.03 \begin {gather*} \frac {1}{4} \left (x \left (9+x+x^2\right )-4 \log (x)-\frac {5 x \left (3+\log \left (x^2\right )\right )}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*x - 25*x*Log[x] + (-4 + 9*x + 2*x^2 + 3*x^3)*Log[x]^2 + (5*x - 5*x*Log[x])*Log[x^2])/(4*x*Log[x]
^2),x]

[Out]

(x*(9 + x + x^2) - 4*Log[x] - (5*x*(3 + Log[x^2]))/Log[x])/4

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fricas [A]  time = 0.68, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\left (x^{3} + x^{2} - x\right )} \log \relax (x) - 4 \, \log \relax (x)^{2} - 15 \, x}{4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25*x*log(x)+15*x)/x/log(x)^2,x, algorit
hm="fricas")

[Out]

1/4*((x^3 + x^2 - x)*log(x) - 4*log(x)^2 - 15*x)/log(x)

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giac [A]  time = 0.20, size = 25, normalized size = 0.86 \begin {gather*} \frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {1}{4} \, x - \frac {15 \, x}{4 \, \log \relax (x)} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25*x*log(x)+15*x)/x/log(x)^2,x, algorit
hm="giac")

[Out]

1/4*x^3 + 1/4*x^2 - 1/4*x - 15/4*x/log(x) - log(x)

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maple [C]  time = 0.05, size = 74, normalized size = 2.55

method	result	size
risch	\(\frac {x^{3}}{4}+\frac {x^{2}}{4}-\frac {x}{4}-\ln \relax (x )+\frac {5 i x \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+6 i\right )}{8 \ln \relax (x )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-5*x*ln(x)+5*x)*ln(x^2)+(3*x^3+2*x^2+9*x-4)*ln(x)^2-25*x*ln(x)+15*x)/x/ln(x)^2,x,method=_RETURNVERBO
SE)

[Out]

1/4*x^3+1/4*x^2-1/4*x-ln(x)+5/8*I*x*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+
6*I)/ln(x)

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maxima [C]  time = 0.38, size = 69, normalized size = 2.38 \begin {gather*} \frac {1}{4} \, x^{3} + \frac {1}{4} \, x^{2} - \frac {5}{4} \, {\rm Ei}\left (\log \relax (x)\right ) \log \left (x^{2}\right ) + \frac {5}{4} \, \Gamma \left (-1, -\log \relax (x)\right ) \log \left (x^{2}\right ) + \frac {5}{2} \, {\rm Ei}\left (\log \relax (x)\right ) \log \relax (x) - \frac {5}{2} \, \Gamma \left (-1, -\log \relax (x)\right ) \log \relax (x) - \frac {1}{4} \, x - \frac {15}{4} \, {\rm Ei}\left (\log \relax (x)\right ) + \frac {15}{4} \, \Gamma \left (-1, -\log \relax (x)\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*log(x)+5*x)*log(x^2)+(3*x^3+2*x^2+9*x-4)*log(x)^2-25*x*log(x)+15*x)/x/log(x)^2,x, algorit
hm="maxima")

[Out]

1/4*x^3 + 1/4*x^2 - 5/4*Ei(log(x))*log(x^2) + 5/4*gamma(-1, -log(x))*log(x^2) + 5/2*Ei(log(x))*log(x) - 5/2*ga
mma(-1, -log(x))*log(x) - 1/4*x - 15/4*Ei(log(x)) + 15/4*gamma(-1, -log(x)) - log(x)

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mupad [B]  time = 3.31, size = 35, normalized size = 1.21 \begin {gather*} \frac {9\,x}{4}-\ln \relax (x)+\frac {x^2}{4}+\frac {x^3}{4}-\frac {\frac {15\,x}{4}+\frac {5\,x\,\ln \left (x^2\right )}{4}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x)/4 + (log(x)^2*(9*x + 2*x^2 + 3*x^3 - 4))/4 + (log(x^2)*(5*x - 5*x*log(x)))/4 - (25*x*log(x))/4)/(x
*log(x)^2),x)

[Out]

(9*x)/4 - log(x) + x^2/4 + x^3/4 - ((15*x)/4 + (5*x*log(x^2))/4)/log(x)

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sympy [A]  time = 0.24, size = 24, normalized size = 0.83 \begin {gather*} \frac {x^{3}}{4} + \frac {x^{2}}{4} - \frac {x}{4} - \frac {15 x}{4 \log {\relax (x )}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-5*x*ln(x)+5*x)*ln(x**2)+(3*x**3+2*x**2+9*x-4)*ln(x)**2-25*x*ln(x)+15*x)/x/ln(x)**2,x)

[Out]

x**3/4 + x**2/4 - x/4 - 15*x/(4*log(x)) - log(x)

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