\(\int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx\) [9307]

Optimal result

Integrand size = 41, antiderivative size = 28 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=3 \left (4-(2+2 x) \left (5-x^2\right )+e^{10 x^2} \log (x)\right ) \]

[Out]

12-3*(-x^2+5)*(2+2*x)+3*ln(x)*exp(5*x^2)^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {14, 2326} \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6 x^3+6 x^2+3 e^{10 x^2} \log (x)-30 x \]

[In]

Int[(3*E^(10*x^2) - 30*x + 12*x^2 + 18*x^3 + 60*E^(10*x^2)*x^2*Log[x])/x,x]

[Out]

-30*x + 6*x^2 + 6*x^3 + 3*E^(10*x^2)*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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (6 \left (-5+2 x+3 x^2\right )+\frac {3 e^{10 x^2} \left (1+20 x^2 \log (x)\right )}{x}\right ) \, dx \\ & = 3 \int \frac {e^{10 x^2} \left (1+20 x^2 \log (x)\right )}{x} \, dx+6 \int \left (-5+2 x+3 x^2\right ) \, dx \\ & = -30 x+6 x^2+6 x^3+3 e^{10 x^2} \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=3 \left (-10 x+2 x^2+2 x^3+e^{10 x^2} \log (x)\right ) \]

[In]

Integrate[(3*E^(10*x^2) - 30*x + 12*x^2 + 18*x^3 + 60*E^(10*x^2)*x^2*Log[x])/x,x]

[Out]

3*(-10*x + 2*x^2 + 2*x^3 + E^(10*x^2)*Log[x])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
risch	\(-30 x +3 \ln \left (x \right ) {\mathrm e}^{10 x^{2}}+6 x^{2}+6 x^{3}\)	\(25\)
default	\(-30 x +3 \ln \left (x \right ) {\mathrm e}^{10 x^{2}}+6 x^{2}+6 x^{3}\)	\(27\)
parallelrisch	\(-30 x +3 \ln \left (x \right ) {\mathrm e}^{10 x^{2}}+6 x^{2}+6 x^{3}\)	\(27\)
parts	\(-30 x +3 \ln \left (x \right ) {\mathrm e}^{10 x^{2}}+6 x^{2}+6 x^{3}\)	\(27\)

[In]

int((60*x^2*exp(5*x^2)^2*ln(x)+3*exp(5*x^2)^2+18*x^3+12*x^2-30*x)/x,x,method=_RETURNVERBOSE)

[Out]

-30*x+3*ln(x)*exp(10*x^2)+6*x^2+6*x^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6 \, x^{3} + 6 \, x^{2} + 3 \, e^{\left (10 \, x^{2}\right )} \log \left (x\right ) - 30 \, x \]

[In]

integrate((60*x^2*exp(5*x^2)^2*log(x)+3*exp(5*x^2)^2+18*x^3+12*x^2-30*x)/x,x, algorithm="fricas")

[Out]

6*x^3 + 6*x^2 + 3*e^(10*x^2)*log(x) - 30*x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6 x^{3} + 6 x^{2} - 30 x + 3 e^{10 x^{2}} \log {\left (x \right )} \]

[In]

integrate((60*x**2*exp(5*x**2)**2*ln(x)+3*exp(5*x**2)**2+18*x**3+12*x**2-30*x)/x,x)

[Out]

6*x**3 + 6*x**2 - 30*x + 3*exp(10*x**2)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6 \, x^{3} + 6 \, x^{2} + 3 \, e^{\left (10 \, x^{2}\right )} \log \left (x\right ) - 30 \, x \]

[In]

integrate((60*x^2*exp(5*x^2)^2*log(x)+3*exp(5*x^2)^2+18*x^3+12*x^2-30*x)/x,x, algorithm="maxima")

[Out]

6*x^3 + 6*x^2 + 3*e^(10*x^2)*log(x) - 30*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6 \, x^{3} + 6 \, x^{2} + 3 \, e^{\left (10 \, x^{2}\right )} \log \left (x\right ) - 30 \, x \]

[In]

integrate((60*x^2*exp(5*x^2)^2*log(x)+3*exp(5*x^2)^2+18*x^3+12*x^2-30*x)/x,x, algorithm="giac")

[Out]

6*x^3 + 6*x^2 + 3*e^(10*x^2)*log(x) - 30*x

Mupad [B] (verification not implemented)

Time = 13.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{10 x^2}-30 x+12 x^2+18 x^3+60 e^{10 x^2} x^2 \log (x)}{x} \, dx=6\,x^2-30\,x+6\,x^3+3\,{\mathrm {e}}^{10\,x^2}\,\ln \left (x\right ) \]

[In]

int((3*exp(10*x^2) - 30*x + 12*x^2 + 18*x^3 + 60*x^2*exp(10*x^2)*log(x))/x,x)

[Out]

6*x^2 - 30*x + 6*x^3 + 3*exp(10*x^2)*log(x)