\(\int \frac {3-18 x+e^{25-11 x+x^2} (11 x-2 x^2)-18 x \log (x)}{x} \, dx\) [4022]

Optimal result

Integrand size = 34, antiderivative size = 22 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-e^{(-5+x)^2-x}+(3-18 x) \log (x) \]

[Out]

ln(x)*(-18*x+3)-exp((-5+x)^2-x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2268, 45, 2332} \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-e^{x^2-11 x+25}-18 x \log (x)+3 \log (x) \]

[In]

Int[(3 - 18*x + E^(25 - 11*x + x^2)*(11*x - 2*x^2) - 18*x*Log[x])/x,x]

[Out]

-E^(25 - 11*x + x^2) + 3*Log[x] - 18*x*Log[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 2268

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 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 \left (-e^{25-11 x+x^2} (-11+2 x)-\frac {3 (-1+6 x+6 x \log (x))}{x}\right ) \, dx \\ & = -\left (3 \int \frac {-1+6 x+6 x \log (x)}{x} \, dx\right )-\int e^{25-11 x+x^2} (-11+2 x) \, dx \\ & = -e^{25-11 x+x^2}-3 \int \left (\frac {-1+6 x}{x}+6 \log (x)\right ) \, dx \\ & = -e^{25-11 x+x^2}-3 \int \frac {-1+6 x}{x} \, dx-18 \int \log (x) \, dx \\ & = -e^{25-11 x+x^2}+18 x-18 x \log (x)-3 \int \left (6-\frac {1}{x}\right ) \, dx \\ & = -e^{25-11 x+x^2}+3 \log (x)-18 x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-e^{25-11 x+x^2}+3 \log (x)-18 x \log (x) \]

[In]

Integrate[(3 - 18*x + E^(25 - 11*x + x^2)*(11*x - 2*x^2) - 18*x*Log[x])/x,x]

[Out]

-E^(25 - 11*x + x^2) + 3*Log[x] - 18*x*Log[x]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
norman	\(3 \ln \left (x \right )-18 x \ln \left (x \right )-{\mathrm e}^{x^{2}-11 x +25}\)	\(22\)
risch	\(3 \ln \left (x \right )-18 x \ln \left (x \right )-{\mathrm e}^{x^{2}-11 x +25}\)	\(22\)
parallelrisch	\(3 \ln \left (x \right )-18 x \ln \left (x \right )-{\mathrm e}^{x^{2}-11 x +25}\)	\(22\)
default	\(-\frac {11 i {\mathrm e}^{25} \sqrt {\pi }\, {\mathrm e}^{-\frac {121}{4}} \operatorname {erf}\left (i x -\frac {11}{2} i\right )}{2}-2 \,{\mathrm e}^{25} \left (\frac {{\mathrm e}^{x^{2}-11 x}}{2}-\frac {11 i \sqrt {\pi }\, {\mathrm e}^{-\frac {121}{4}} \operatorname {erf}\left (i x -\frac {11}{2} i\right )}{4}\right )+3 \ln \left (x \right )-18 x \ln \left (x \right )\)	\(60\)
parts	\(-\frac {11 i {\mathrm e}^{25} \sqrt {\pi }\, {\mathrm e}^{-\frac {121}{4}} \operatorname {erf}\left (i x -\frac {11}{2} i\right )}{2}-2 \,{\mathrm e}^{25} \left (\frac {{\mathrm e}^{x^{2}-11 x}}{2}-\frac {11 i \sqrt {\pi }\, {\mathrm e}^{-\frac {121}{4}} \operatorname {erf}\left (i x -\frac {11}{2} i\right )}{4}\right )+3 \ln \left (x \right )-18 x \ln \left (x \right )\)	\(60\)

[In]

int((-18*x*ln(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x,method=_RETURNVERBOSE)

[Out]

3*ln(x)-18*x*ln(x)-exp(x^2-11*x+25)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-3 \, {\left (6 \, x - 1\right )} \log \left (x\right ) - e^{\left (x^{2} - 11 \, x + 25\right )} \]

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="fricas")

[Out]

-3*(6*x - 1)*log(x) - e^(x^2 - 11*x + 25)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=- 18 x \log {\left (x \right )} - e^{x^{2} - 11 x + 25} + 3 \log {\left (x \right )} \]

[In]

integrate((-18*x*ln(x)+(-2*x**2+11*x)*exp(x**2-11*x+25)-18*x+3)/x,x)

[Out]

-18*x*log(x) - exp(x**2 - 11*x + 25) + 3*log(x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 3.50 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-\frac {11}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - \frac {11}{2} i\right ) e^{\left (-\frac {21}{4}\right )} - \frac {1}{2} \, {\left (\frac {11 \, \sqrt {\pi } {\left (2 \, x - 11\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x - 11\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x - 11\right )}^{2}}} + 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x - 11\right )}^{2}\right )}\right )} e^{\left (-\frac {21}{4}\right )} - 18 \, x \log \left (x\right ) + 3 \, \log \left (x\right ) \]

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="maxima")

[Out]

-11/2*I*sqrt(pi)*erf(I*x - 11/2*I)*e^(-21/4) - 1/2*(11*sqrt(pi)*(2*x - 11)*(erf(1/2*sqrt(-(2*x - 11)^2)) - 1)/
sqrt(-(2*x - 11)^2) + 2*e^(1/4*(2*x - 11)^2))*e^(-21/4) - 18*x*log(x) + 3*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=-18 \, x \log \left (x\right ) - e^{\left (x^{2} - 11 \, x + 25\right )} + 3 \, \log \left (x\right ) \]

[In]

integrate((-18*x*log(x)+(-2*x^2+11*x)*exp(x^2-11*x+25)-18*x+3)/x,x, algorithm="giac")

[Out]

-18*x*log(x) - e^(x^2 - 11*x + 25) + 3*log(x)

Mupad [B] (verification not implemented)

Time = 9.66 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {3-18 x+e^{25-11 x+x^2} \left (11 x-2 x^2\right )-18 x \log (x)}{x} \, dx=3\,\ln \left (x\right )-18\,x\,\ln \left (x\right )-{\mathrm {e}}^{-11\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{25} \]

[In]

int(-(18*x - exp(x^2 - 11*x + 25)*(11*x - 2*x^2) + 18*x*log(x) - 3)/x,x)

[Out]

3*log(x) - 18*x*log(x) - exp(-11*x)*exp(x^2)*exp(25)