\(\int \frac {1}{12} (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)) \, dx\) [7777]

Optimal result

Integrand size = 29, antiderivative size = 33 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=e^x+\frac {5 x}{4}-x^2 \left (2-(-5+x)^2+\frac {1}{3} (-x+\log (x))\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2225, 2341} \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=x^4-\frac {29 x^3}{3}+23 x^2-\frac {1}{3} x^2 \log (x)+\frac {5 x}{4}+e^x \]

[In]

Int[(15 + 12*E^x + 548*x - 348*x^2 + 48*x^3 - 8*x*Log[x])/12,x]

[Out]

E^x + (5*x)/4 + 23*x^2 - (29*x^3)/3 + x^4 - (x^2*Log[x])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2341

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
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{12} \int \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx \\ & = \frac {5 x}{4}+\frac {137 x^2}{6}-\frac {29 x^3}{3}+x^4-\frac {2}{3} \int x \log (x) \, dx+\int e^x \, dx \\ & = e^x+\frac {5 x}{4}+23 x^2-\frac {29 x^3}{3}+x^4-\frac {1}{3} x^2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=e^x+\frac {5 x}{4}+23 x^2-\frac {29 x^3}{3}+x^4-\frac {1}{3} x^2 \log (x) \]

[In]

Integrate[(15 + 12*E^x + 548*x - 348*x^2 + 48*x^3 - 8*x*Log[x])/12,x]

[Out]

E^x + (5*x)/4 + 23*x^2 - (29*x^3)/3 + x^4 - (x^2*Log[x])/3

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

method	result	size
default	\(x^{4}-\frac {29 x^{3}}{3}+23 x^{2}+\frac {5 x}{4}-\frac {x^{2} \ln \left (x \right )}{3}+{\mathrm e}^{x}\)	\(27\)
norman	\(x^{4}-\frac {29 x^{3}}{3}+23 x^{2}+\frac {5 x}{4}-\frac {x^{2} \ln \left (x \right )}{3}+{\mathrm e}^{x}\)	\(27\)
risch	\(x^{4}-\frac {29 x^{3}}{3}+23 x^{2}+\frac {5 x}{4}-\frac {x^{2} \ln \left (x \right )}{3}+{\mathrm e}^{x}\)	\(27\)
parallelrisch	\(x^{4}-\frac {29 x^{3}}{3}+23 x^{2}+\frac {5 x}{4}-\frac {x^{2} \ln \left (x \right )}{3}+{\mathrm e}^{x}\)	\(27\)
parts	\(x^{4}-\frac {29 x^{3}}{3}+23 x^{2}+\frac {5 x}{4}-\frac {x^{2} \ln \left (x \right )}{3}+{\mathrm e}^{x}\)	\(27\)

[In]

int(-2/3*x*ln(x)+exp(x)+4*x^3-29*x^2+137/3*x+5/4,x,method=_RETURNVERBOSE)

[Out]

x^4-29/3*x^3+23*x^2+5/4*x-1/3*x^2*ln(x)+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=x^{4} - \frac {29}{3} \, x^{3} - \frac {1}{3} \, x^{2} \log \left (x\right ) + 23 \, x^{2} + \frac {5}{4} \, x + e^{x} \]

[In]

integrate(-2/3*x*log(x)+exp(x)+4*x^3-29*x^2+137/3*x+5/4,x, algorithm="fricas")

[Out]

x^4 - 29/3*x^3 - 1/3*x^2*log(x) + 23*x^2 + 5/4*x + e^x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=x^{4} - \frac {29 x^{3}}{3} - \frac {x^{2} \log {\left (x \right )}}{3} + 23 x^{2} + \frac {5 x}{4} + e^{x} \]

[In]

integrate(-2/3*x*ln(x)+exp(x)+4*x**3-29*x**2+137/3*x+5/4,x)

[Out]

x**4 - 29*x**3/3 - x**2*log(x)/3 + 23*x**2 + 5*x/4 + exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=x^{4} - \frac {29}{3} \, x^{3} - \frac {1}{3} \, x^{2} \log \left (x\right ) + 23 \, x^{2} + \frac {5}{4} \, x + e^{x} \]

[In]

integrate(-2/3*x*log(x)+exp(x)+4*x^3-29*x^2+137/3*x+5/4,x, algorithm="maxima")

[Out]

x^4 - 29/3*x^3 - 1/3*x^2*log(x) + 23*x^2 + 5/4*x + e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=x^{4} - \frac {29}{3} \, x^{3} - \frac {1}{3} \, x^{2} \log \left (x\right ) + 23 \, x^{2} + \frac {5}{4} \, x + e^{x} \]

[In]

integrate(-2/3*x*log(x)+exp(x)+4*x^3-29*x^2+137/3*x+5/4,x, algorithm="giac")

[Out]

x^4 - 29/3*x^3 - 1/3*x^2*log(x) + 23*x^2 + 5/4*x + e^x

Mupad [B] (verification not implemented)

Time = 12.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{12} \left (15+12 e^x+548 x-348 x^2+48 x^3-8 x \log (x)\right ) \, dx=\frac {5\,x}{4}+{\mathrm {e}}^x-\frac {x^2\,\ln \left (x\right )}{3}+23\,x^2-\frac {29\,x^3}{3}+x^4 \]

[In]

int((137*x)/3 + exp(x) - (2*x*log(x))/3 - 29*x^2 + 4*x^3 + 5/4,x)

[Out]

(5*x)/4 + exp(x) - (x^2*log(x))/3 + 23*x^2 - (29*x^3)/3 + x^4