3.40.58 \(\int \frac {750 \log (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}})-125 \log ^2(\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}})}{13 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {125 \log ^2\left (x \left (x+\frac {x}{e^5}\right )^2\right )}{13 x} \]

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Rubi [A]  time = 0.08, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 4, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 14, 2305, 2304} \begin {gather*} \frac {125 \left (2 \left (5-\log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{13 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(750*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10] - 125*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10]^2)/(13*x^2),x]

[Out]

(125*(2*(5 - Log[1 + E^5]) - Log[x^3])^2)/(13*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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 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] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{13} \int \frac {750 \log \left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )-125 \log ^2\left (\frac {x^3+2 e^5 x^3+e^{10} x^3}{e^{10}}\right )}{x^2} \, dx\\ &=\frac {1}{13} \int \left (-\frac {125 \left (10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{x^2}+\frac {750 \left (-10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )+\log \left (x^3\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {125}{13} \int \frac {\left (10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{x^2} \, dx\right )+\frac {750}{13} \int \frac {-10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )+\log \left (x^3\right )}{x^2} \, dx\\ &=-\frac {2250}{13 x}+\frac {750 \left (2 \left (5-\log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )}{13 x}+\frac {125 \left (2 \left (5-\log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{13 x}+\frac {750}{13} \int \frac {10 \left (1-\frac {1}{5} \log \left (1+e^5\right )\right )-\log \left (x^3\right )}{x^2} \, dx\\ &=\frac {125 \left (2 \left (5-\log \left (1+e^5\right )\right )-\log \left (x^3\right )\right )^2}{13 x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.05, size = 69, normalized size = 3.29 \begin {gather*} -\frac {125}{13} \left (-\frac {100}{x}+\frac {40 \log \left (1+e^5\right )}{x}-\frac {4 \log ^2\left (1+e^5\right )}{x}+\frac {20 \log \left (x^3\right )}{x}-\frac {4 \log \left (1+e^5\right ) \log \left (x^3\right )}{x}-\frac {\log ^2\left (x^3\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(750*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10] - 125*Log[(x^3 + 2*E^5*x^3 + E^10*x^3)/E^10]^2)/(13*x^2)
,x]

[Out]

(-125*(-100/x + (40*Log[1 + E^5])/x - (4*Log[1 + E^5]^2)/x + (20*Log[x^3])/x - (4*Log[1 + E^5]*Log[x^3])/x - L
og[x^3]^2/x))/13

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fricas [A]  time = 0.87, size = 28, normalized size = 1.33 \begin {gather*} \frac {125 \, \log \left ({\left (x^{3} e^{10} + 2 \, x^{3} e^{5} + x^{3}\right )} e^{\left (-10\right )}\right )^{2}}{13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/e
xp(5)^2))/x^2,x, algorithm="fricas")

[Out]

125/13*log((x^3*e^10 + 2*x^3*e^5 + x^3)*e^(-10))^2/x

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giac [B]  time = 0.14, size = 53, normalized size = 2.52 \begin {gather*} \frac {125 \, {\left (\log \left (x^{3} e^{15} + 2 \, x^{3} e^{10} + x^{3} e^{5}\right )^{2} - 30 \, \log \left (x^{3} e^{15} + 2 \, x^{3} e^{10} + x^{3} e^{5}\right ) + 225\right )}}{13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/e
xp(5)^2))/x^2,x, algorithm="giac")

[Out]

125/13*(log(x^3*e^15 + 2*x^3*e^10 + x^3*e^5)^2 - 30*log(x^3*e^15 + 2*x^3*e^10 + x^3*e^5) + 225)/x

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maple [A]  time = 0.06, size = 29, normalized size = 1.38

method	result	size
risch	\(\frac {125 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )^{2}}{13 x}\)	\(29\)
norman	\(\frac {125 \ln \left (\left (x^{3} {\mathrm e}^{10}+2 x^{3} {\mathrm e}^{5}+x^{3}\right ) {\mathrm e}^{-10}\right )^{2}}{13 x}\)	\(33\)
default	\(\frac {250 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right ) \ln \left (x^{3}\right )}{13 x}-\frac {2500 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )}{13 x}-\frac {250 \ln \left (x^{3}\right )}{x}+\frac {10250}{13 x}+\frac {\frac {2250}{13}+\frac {125 \ln \left (x^{3}\right )^{2}}{13}+\frac {750 \ln \left (x^{3}\right )}{13}}{x}+\frac {125 \ln \left ({\mathrm e}^{10}+2 \,{\mathrm e}^{5}+1\right )^{2}}{13 x}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/13*(-125*ln((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*ln((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)
)/x^2,x,method=_RETURNVERBOSE)

[Out]

125/13*ln((x^3*exp(10)+2*x^3*exp(5)+x^3)*exp(-10))^2/x

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maxima [A]  time = 0.39, size = 25, normalized size = 1.19 \begin {gather*} \frac {125 \, \log \left (2 \, x^{3} e^{\left (-5\right )} + x^{3} e^{\left (-10\right )} + x^{3}\right )^{2}}{13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-125*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/exp(5)^2)^2+750*log((x^3*exp(5)^2+2*x^3*exp(5)+x^3)/e
xp(5)^2))/x^2,x, algorithm="maxima")

[Out]

125/13*log(2*x^3*e^(-5) + x^3*e^(-10) + x^3)^2/x

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mupad [B]  time = 2.80, size = 22, normalized size = 1.05 \begin {gather*} \frac {125\,{\ln \left (x^3\,{\mathrm {e}}^{-10}\,\left (2\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+1\right )\right )}^2}{13\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((750*log(exp(-10)*(2*x^3*exp(5) + x^3*exp(10) + x^3)))/13 - (125*log(exp(-10)*(2*x^3*exp(5) + x^3*exp(10)
+ x^3))^2)/13)/x^2,x)

[Out]

(125*log(x^3*exp(-10)*(2*exp(5) + exp(10) + 1))^2)/(13*x)

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sympy [A]  time = 0.13, size = 29, normalized size = 1.38 \begin {gather*} \frac {125 \log {\left (\frac {x^{3} + 2 x^{3} e^{5} + x^{3} e^{10}}{e^{10}} \right )}^{2}}{13 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-125*ln((x**3*exp(5)**2+2*x**3*exp(5)+x**3)/exp(5)**2)**2+750*ln((x**3*exp(5)**2+2*x**3*exp(5)
+x**3)/exp(5)**2))/x**2,x)

[Out]

125*log((x**3 + 2*x**3*exp(5) + x**3*exp(10))*exp(-10))**2/(13*x)

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