\(\int \frac {-2-238 x+15 e^3 x+(-238 x+15 e^3 x) \log (x)}{-240 x+15 e^3 x} \, dx\) [6605]

Optimal result

Integrand size = 37, antiderivative size = 24 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {1}{2} \left (\frac {4 (-1+x)}{15 \left (-16+e^3\right )}+2 x\right ) \log (x) \]

[Out]

1/2*ln(x)*(2*x+(-1+x)/(15/4*exp(3)-60))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6, 12, 14, 45, 2332} \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {\left (238-15 e^3\right ) x \log (x)}{15 \left (16-e^3\right )}+\frac {2 \log (x)}{15 \left (16-e^3\right )} \]

[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 45

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 2332

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {\left (-2+\left (-238+15 e^3\right ) x\right ) \log (x)}{15 \left (-16+e^3\right )} \]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {15 x \,{\mathrm e}^{3} \ln \left (x \right )-238 x \ln \left (x \right )-2 \ln \left (x \right )}{15 \,{\mathrm e}^{3}-240}\)	\(26\)
norman	\(-\frac {2 \ln \left (x \right )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \left (x \right )}{15 \,{\mathrm e}^{3}-240}\)	\(29\)
risch	\(-\frac {2 \ln \left (x \right )}{15 \left ({\mathrm e}^{3}-16\right )}+\frac {\left (15 \,{\mathrm e}^{3}-238\right ) x \ln \left (x \right )}{15 \,{\mathrm e}^{3}-240}\)	\(29\)
parts	\(\frac {\left (15 \,{\mathrm e}^{3}-238\right ) \left (x \ln \left (x \right )-x \right )}{15 \,{\mathrm e}^{3}-240}+\frac {15 x \,{\mathrm e}^{3}-238 x -2 \ln \left (x \right )}{15 \,{\mathrm e}^{3}-240}\)	\(45\)
default	\(\frac {{\mathrm e}^{3} \left (x \ln \left (x \right )-x \right )}{{\mathrm e}^{3}-16}+\frac {{\mathrm e}^{3} x}{{\mathrm e}^{3}-16}-\frac {238 \left (x \ln \left (x \right )-x \right )}{15 \left ({\mathrm e}^{3}-16\right )}-\frac {238 x}{15 \left ({\mathrm e}^{3}-16\right )}-\frac {2 \ln \left (x \right )}{15 \left ({\mathrm e}^{3}-16\right )}\)	\(64\)

[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]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {{\left (15 \, x e^{3} - 238 \, x - 2\right )} \log \left (x\right )}{15 \, {\left (e^{3} - 16\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {\left (- 238 x + 15 x e^{3}\right ) \log {\left (x \right )}}{-240 + 15 e^{3}} - \frac {2 \log {\left (x \right )}}{-240 + 15 e^{3}} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.58 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx={\left (\frac {x \log \left (x\right )}{e^{3} - 16} - \frac {x}{e^{3} - 16}\right )} e^{3} + \frac {x e^{3}}{e^{3} - 16} - \frac {238 \, x \log \left (x\right )}{15 \, {\left (e^{3} - 16\right )}} - \frac {2 \, \log \left (x e^{3} - 16 \, x\right )}{15 \, {\left (e^{3} - 16\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=\frac {15 \, x e^{3} \log \left (x\right ) - 238 \, x \log \left (x\right ) - 2 \, \log \left (x\right )}{15 \, {\left (e^{3} - 16\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 11.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2-238 x+15 e^3 x+\left (-238 x+15 e^3 x\right ) \log (x)}{-240 x+15 e^3 x} \, dx=-\frac {\ln \left (x\right )\,\left (238\,x-15\,x\,{\mathrm {e}}^3+2\right )}{15\,\left ({\mathrm {e}}^3-16\right )} \]

[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))