3.67.5 \(\int \frac {-2-238 x+15 e^3 x+(-238 x+15 e^3 x) \log (x)}{-240 x+15 e^3 x} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{2} \left (\frac {4 (-1+x)}{15 \left (-16+e^3\right )}+2 x\right ) \log (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.62, number of steps used = 8, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6, 12, 14, 43, 2295} \begin {gather*} \frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}+\frac {2 \log (x)}{15 \left (16-e^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 238*x + 15*E^3*x + (-238*x + 15*E^3*x)*Log[x])/(-240*x + 15*E^3*x),x]

[Out]

(2*Log[x])/(15*(16 - E^3)) + ((238 - 15*E^3)*x*Log[x])/(15*(16 - E^3))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{\left (-240+15 e^3\right ) x} \, dx\\ &=\int \frac {-2+\left (-238+15 e^3\right ) x+\left (-238 x+15 e^3 x\right ) \log (x)}{\left (-240+15 e^3\right ) x} \, dx\\ &=-\frac {\int \frac {-2+\left (-238+15 e^3\right ) x+\left (-238 x+15 e^3 x\right ) \log (x)}{x} \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\int \left (\frac {-2-\left (238-15 e^3\right ) x}{x}+\left (-238+15 e^3\right ) \log (x)\right ) \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\int \frac {-2-\left (238-15 e^3\right ) x}{x} \, dx}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) \int \log (x) \, dx}{15 \left (16-e^3\right )}\\ &=-\frac {\left (238-15 e^3\right ) x}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}-\frac {\int \left (-238+15 e^3-\frac {2}{x}\right ) \, dx}{15 \left (16-e^3\right )}\\ &=\frac {2 \log (x)}{15 \left (16-e^3\right )}+\frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.08 \begin {gather*} \frac {2 \log (x)+\left (238-15 e^3\right ) x \log (x)}{240-15 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 238*x + 15*E^3*x + (-238*x + 15*E^3*x)*Log[x])/(-240*x + 15*E^3*x),x]

[Out]

(2*Log[x] + (238 - 15*E^3)*x*Log[x])/(240 - 15*E^3)

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fricas [A]  time = 0.50, size = 20, normalized size = 0.83 \begin {gather*} \frac {{\left (15 \, x e^{3} - 238 \, x - 2\right )} \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x*exp(3)-238*x)*log(x)+15*x*exp(3)-238*x-2)/(15*x*exp(3)-240*x),x, algorithm="fricas")

[Out]

1/15*(15*x*e^3 - 238*x - 2)*log(x)/(e^3 - 16)

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giac [A]  time = 0.20, size = 25, normalized size = 1.04 \begin {gather*} \frac {15 \, x e^{3} \log \relax (x) - 238 \, x \log \relax (x) - 2 \, \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x*exp(3)-238*x)*log(x)+15*x*exp(3)-238*x-2)/(15*x*exp(3)-240*x),x, algorithm="giac")

[Out]

1/15*(15*x*e^3*log(x) - 238*x*log(x) - 2*log(x))/(e^3 - 16)

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maple [A]  time = 0.06, size = 29, normalized size = 1.21

method	result	size
norman	\(-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \relax (x )}{15 \,{\mathrm e}^{3}-240}\)	\(29\)
risch	\(-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \relax (x )}{15 \,{\mathrm e}^{3}-240}\)	\(29\)
default	\(\frac {{\mathrm e}^{3} \left (x \ln \relax (x )-x \right )}{{\mathrm e}^{3}-16}-\frac {238 \left (x \ln \relax (x )-x \right )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {{\mathrm e}^{3} x}{{\mathrm e}^{3}-16}-\frac {238 x}{15 \left ({\mathrm e}^{3}-16\right )}-\frac {2 \ln \relax (x )}{15 \left ({\mathrm e}^{3}-16\right )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x*exp(3)-238*x)*ln(x)+15*x*exp(3)-238*x-2)/(15*x*exp(3)-240*x),x,method=_RETURNVERBOSE)

[Out]

-2/15/(exp(3)-16)*ln(x)+1/15*(15*exp(3)-238)/(exp(3)-16)*x*ln(x)

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maxima [B]  time = 0.42, size = 62, normalized size = 2.58 \begin {gather*} {\left (\frac {x \log \relax (x)}{e^{3} - 16} - \frac {x}{e^{3} - 16}\right )} e^{3} + \frac {x e^{3}}{e^{3} - 16} - \frac {238 \, x \log \relax (x)}{15 \, {\left (e^{3} - 16\right )}} - \frac {2 \, \log \left (x e^{3} - 16 \, x\right )}{15 \, {\left (e^{3} - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x*exp(3)-238*x)*log(x)+15*x*exp(3)-238*x-2)/(15*x*exp(3)-240*x),x, algorithm="maxima")

[Out]

(x*log(x)/(e^3 - 16) - x/(e^3 - 16))*e^3 + x*e^3/(e^3 - 16) - 238/15*x*log(x)/(e^3 - 16) - 2/15*log(x*e^3 - 16
*x)/(e^3 - 16)

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mupad [B]  time = 4.32, size = 22, normalized size = 0.92 \begin {gather*} -\frac {\ln \relax (x)\,\left (238\,x-15\,x\,{\mathrm {e}}^3+2\right )}{15\,\left ({\mathrm {e}}^3-16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((238*x - 15*x*exp(3) + log(x)*(238*x - 15*x*exp(3)) + 2)/(240*x - 15*x*exp(3)),x)

[Out]

-(log(x)*(238*x - 15*x*exp(3) + 2))/(15*(exp(3) - 16))

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sympy [A]  time = 0.14, size = 31, normalized size = 1.29 \begin {gather*} \frac {\left (- 238 x + 15 x e^{3}\right ) \log {\relax (x )}}{-240 + 15 e^{3}} - \frac {2 \log {\relax (x )}}{-240 + 15 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x*exp(3)-238*x)*ln(x)+15*x*exp(3)-238*x-2)/(15*x*exp(3)-240*x),x)

[Out]

(-238*x + 15*x*exp(3))*log(x)/(-240 + 15*exp(3)) - 2*log(x)/(-240 + 15*exp(3))

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