3.3.49 \(\int \frac {-15 x^2 \log (4)+45 x^2 \log (4) \log (x)+4 \log ^2(x)}{5 \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ -6+\frac {1}{5} (-2-x)+x+\frac {3 x^3 \log (4)}{\log (x)} \]

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Rubi [A]  time = 0.14, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {3 x^3 \log (4)}{\log (x)}+\frac {4 x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x^2*Log[4] + 45*x^2*Log[4]*Log[x] + 4*Log[x]^2)/(5*Log[x]^2),x]

[Out]

(4*x)/5 + (3*x^3*Log[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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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}{5} \int \frac {-15 x^2 \log (4)+45 x^2 \log (4) \log (x)+4 \log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (4-\frac {15 x^2 \log (4)}{\log ^2(x)}+\frac {45 x^2 \log (4)}{\log (x)}\right ) \, dx\\ &=\frac {4 x}{5}-(3 \log (4)) \int \frac {x^2}{\log ^2(x)} \, dx+(9 \log (4)) \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {4 x}{5}+\frac {3 x^3 \log (4)}{\log (x)}-(9 \log (4)) \int \frac {x^2}{\log (x)} \, dx+(9 \log (4)) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4 x}{5}+9 \text {Ei}(3 \log (x)) \log (4)+\frac {3 x^3 \log (4)}{\log (x)}-(9 \log (4)) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {4 x}{5}+\frac {3 x^3 \log (4)}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.74 \begin {gather*} \frac {4 x}{5}+\frac {3 x^3 \log (4)}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x^2*Log[4] + 45*x^2*Log[4]*Log[x] + 4*Log[x]^2)/(5*Log[x]^2),x]

[Out]

(4*x)/5 + (3*x^3*Log[4])/Log[x]

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fricas [A]  time = 0.69, size = 19, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (15 \, x^{3} \log \relax (2) + 2 \, x \log \relax (x)\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*log(x)^2+90*x^2*log(2)*log(x)-30*x^2*log(2))/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 15, normalized size = 0.65 \begin {gather*} \frac {6 \, x^{3} \log \relax (2)}{\log \relax (x)} + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*log(x)^2+90*x^2*log(2)*log(x)-30*x^2*log(2))/log(x)^2,x, algorithm="giac")

[Out]

6*x^3*log(2)/log(x) + 4/5*x

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maple [A]  time = 0.08, size = 16, normalized size = 0.70

method	result	size
risch	\(\frac {4 x}{5}+\frac {6 x^{3} \ln \relax (2)}{\ln \relax (x )}\)	\(16\)
norman	\(\frac {\frac {4 x \ln \relax (x )}{5}+6 x^{3} \ln \relax (2)}{\ln \relax (x )}\)	\(19\)
default	\(-18 \ln \relax (2) \expIntegralEi \left (1, -3 \ln \relax (x )\right )-6 \ln \relax (2) \left (-\frac {x^{3}}{\ln \relax (x )}-3 \expIntegralEi \left (1, -3 \ln \relax (x )\right )\right )+\frac {4 x}{5}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*ln(x)^2+90*x^2*ln(2)*ln(x)-30*x^2*ln(2))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

4/5*x+6*x^3*ln(2)/ln(x)

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maxima [C]  time = 0.45, size = 23, normalized size = 1.00 \begin {gather*} 18 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) \log \relax (2) - 18 \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \log \relax (2) + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*log(x)^2+90*x^2*log(2)*log(x)-30*x^2*log(2))/log(x)^2,x, algorithm="maxima")

[Out]

18*Ei(3*log(x))*log(2) - 18*gamma(-1, -3*log(x))*log(2) + 4/5*x

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mupad [B]  time = 0.32, size = 15, normalized size = 0.65 \begin {gather*} \frac {4\,x}{5}+\frac {6\,x^3\,\ln \relax (2)}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*log(x)^2)/5 - 6*x^2*log(2) + 18*x^2*log(2)*log(x))/log(x)^2,x)

[Out]

(4*x)/5 + (6*x^3*log(2))/log(x)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.65 \begin {gather*} \frac {6 x^{3} \log {\relax (2 )}}{\log {\relax (x )}} + \frac {4 x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*ln(x)**2+90*x**2*ln(2)*ln(x)-30*x**2*ln(2))/ln(x)**2,x)

[Out]

6*x**3*log(2)/log(x) + 4*x/5

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