3.48.55 \(\int \frac {-3 x^2-9 x^2 \log (\frac {75}{x})+4 x^3 \log ^2(\frac {75}{x})}{5 \log ^2(\frac {75}{x})} \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{5} x^3 \left (x-\frac {3}{\log \left (\frac {75}{x}\right )}\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6688, 14, 2306, 2310, 2178} \begin {gather*} \frac {x^4}{5}-\frac {3 x^3}{5 \log \left (\frac {75}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^2 - 9*x^2*Log[75/x] + 4*x^3*Log[75/x]^2)/(5*Log[75/x]^2),x]

[Out]

x^4/5 - (3*x^3)/(5*Log[75/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2310

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-3 x^2-9 x^2 \log \left (\frac {75}{x}\right )+4 x^3 \log ^2\left (\frac {75}{x}\right )}{\log ^2\left (\frac {75}{x}\right )} \, dx\\ &=\frac {1}{5} \int x^2 \left (4 x-\frac {3}{\log ^2\left (\frac {75}{x}\right )}-\frac {9}{\log \left (\frac {75}{x}\right )}\right ) \, dx\\ &=\frac {1}{5} \int \left (4 x^3-\frac {3 x^2}{\log ^2\left (\frac {75}{x}\right )}-\frac {9 x^2}{\log \left (\frac {75}{x}\right )}\right ) \, dx\\ &=\frac {x^4}{5}-\frac {3}{5} \int \frac {x^2}{\log ^2\left (\frac {75}{x}\right )} \, dx-\frac {9}{5} \int \frac {x^2}{\log \left (\frac {75}{x}\right )} \, dx\\ &=\frac {x^4}{5}-\frac {3 x^3}{5 \log \left (\frac {75}{x}\right )}+\frac {9}{5} \int \frac {x^2}{\log \left (\frac {75}{x}\right )} \, dx+759375 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{x} \, dx,x,\log \left (\frac {75}{x}\right )\right )\\ &=\frac {x^4}{5}+759375 \text {Ei}\left (-3 \log \left (\frac {75}{x}\right )\right )-\frac {3 x^3}{5 \log \left (\frac {75}{x}\right )}-759375 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{x} \, dx,x,\log \left (\frac {75}{x}\right )\right )\\ &=\frac {x^4}{5}-\frac {3 x^3}{5 \log \left (\frac {75}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 23, normalized size = 1.21 \begin {gather*} \frac {x^4}{5}-\frac {3 x^3}{5 \log \left (\frac {75}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 - 9*x^2*Log[75/x] + 4*x^3*Log[75/x]^2)/(5*Log[75/x]^2),x]

[Out]

x^4/5 - (3*x^3)/(5*Log[75/x])

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fricas [A]  time = 0.65, size = 26, normalized size = 1.37 \begin {gather*} \frac {x^{4} \log \left (\frac {75}{x}\right ) - 3 \, x^{3}}{5 \, \log \left (\frac {75}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3*log(75/x)^2-9*x^2*log(75/x)-3*x^2)/log(75/x)^2,x, algorithm="fricas")

[Out]

1/5*(x^4*log(75/x) - 3*x^3)/log(75/x)

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giac [A]  time = 0.22, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, x^{4} - \frac {3 \, x^{3}}{5 \, \log \left (\frac {75}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3*log(75/x)^2-9*x^2*log(75/x)-3*x^2)/log(75/x)^2,x, algorithm="giac")

[Out]

1/5*x^4 - 3/5*x^3/log(75/x)

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maple [A]  time = 0.13, size = 20, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {x^{4}}{5}-\frac {3 x^{3}}{5 \ln \left (\frac {75}{x}\right )}\)	\(20\)
default	\(\frac {x^{4}}{5}-\frac {3 x^{3}}{5 \ln \left (\frac {75}{x}\right )}\)	\(20\)
risch	\(\frac {x^{4}}{5}-\frac {3 x^{3}}{5 \ln \left (\frac {75}{x}\right )}\)	\(20\)
norman	\(\frac {-\frac {3 x^{3}}{5}+\frac {x^{4} \ln \left (\frac {75}{x}\right )}{5}}{\ln \left (\frac {75}{x}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*x^3*ln(75/x)^2-9*x^2*ln(75/x)-3*x^2)/ln(75/x)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*x^4-3/5*x^3/ln(75/x)

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maxima [B]  time = 0.48, size = 39, normalized size = 2.05 \begin {gather*} \frac {x^{4} {\left (2 \, \log \relax (5) + \log \relax (3)\right )} - x^{4} \log \relax (x) - 3 \, x^{3}}{5 \, {\left (2 \, \log \relax (5) + \log \relax (3) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3*log(75/x)^2-9*x^2*log(75/x)-3*x^2)/log(75/x)^2,x, algorithm="maxima")

[Out]

1/5*(x^4*(2*log(5) + log(3)) - x^4*log(x) - 3*x^3)/(2*log(5) + log(3) - log(x))

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mupad [B]  time = 3.28, size = 19, normalized size = 1.00 \begin {gather*} \frac {x^4}{5}-\frac {3\,x^3}{5\,\ln \left (\frac {75}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((3*x^2)/5 - (4*x^3*log(75/x)^2)/5 + (9*x^2*log(75/x))/5)/log(75/x)^2,x)

[Out]

x^4/5 - (3*x^3)/(5*log(75/x))

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sympy [A]  time = 0.11, size = 15, normalized size = 0.79 \begin {gather*} \frac {x^{4}}{5} - \frac {3 x^{3}}{5 \log {\left (\frac {75}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x**3*ln(75/x)**2-9*x**2*ln(75/x)-3*x**2)/ln(75/x)**2,x)

[Out]

x**4/5 - 3*x**3/(5*log(75/x))

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