\(\int \frac {-9+9 x^2+e^{2 x^2+2 x^3} (x^2+4 x^4+6 x^5)+9 x^2 \log (x)}{9 x^2} \, dx\) [7795]

Optimal result

Integrand size = 49, antiderivative size = 23 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {1}{x}+x \left (\frac {1}{9} e^{2 x^2 (1+x)}+\log (x)\right ) \]

[Out]

x*(ln(x)+1/9*exp(2*x^2*(1+x)))+1/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 14, 2326, 2332} \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {e^{2 x^2 (x+1)} \left (3 x^3+2 x^2\right )}{9 \left (x^2+2 (x+1) x\right )}+\frac {1}{x}+x \log (x) \]

[In]

Int[(-9 + 9*x^2 + E^(2*x^2 + 2*x^3)*(x^2 + 4*x^4 + 6*x^5) + 9*x^2*Log[x])/(9*x^2),x]

[Out]

x^(-1) + (E^(2*x^2*(1 + x))*(2*x^2 + 3*x^3))/(9*(x^2 + 2*x*(1 + x))) + x*Log[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 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]

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}& = \frac {1}{9} \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{x^2} \, dx \\ & = \frac {1}{9} \int \left (e^{2 x^2 (1+x)} \left (1+4 x^2+6 x^3\right )+\frac {9 \left (-1+x^2+x^2 \log (x)\right )}{x^2}\right ) \, dx \\ & = \frac {1}{9} \int e^{2 x^2 (1+x)} \left (1+4 x^2+6 x^3\right ) \, dx+\int \frac {-1+x^2+x^2 \log (x)}{x^2} \, dx \\ & = \frac {e^{2 x^2 (1+x)} \left (2 x^2+3 x^3\right )}{9 \left (x^2+2 x (1+x)\right )}+\int \left (\frac {-1+x^2}{x^2}+\log (x)\right ) \, dx \\ & = \frac {e^{2 x^2 (1+x)} \left (2 x^2+3 x^3\right )}{9 \left (x^2+2 x (1+x)\right )}+\int \frac {-1+x^2}{x^2} \, dx+\int \log (x) \, dx \\ & = -x+\frac {e^{2 x^2 (1+x)} \left (2 x^2+3 x^3\right )}{9 \left (x^2+2 x (1+x)\right )}+x \log (x)+\int \left (1-\frac {1}{x^2}\right ) \, dx \\ & = \frac {1}{x}+\frac {e^{2 x^2 (1+x)} \left (2 x^2+3 x^3\right )}{9 \left (x^2+2 x (1+x)\right )}+x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {1}{x}+\frac {1}{9} e^{2 x^2 (1+x)} x+x \log (x) \]

[In]

Integrate[(-9 + 9*x^2 + E^(2*x^2 + 2*x^3)*(x^2 + 4*x^4 + 6*x^5) + 9*x^2*Log[x])/(9*x^2),x]

[Out]

x^(-1) + (E^(2*x^2*(1 + x))*x)/9 + x*Log[x]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {x \,{\mathrm e}^{2 x^{3}+2 x^{2}}}{9}+\frac {1}{x}+x \ln \left (x \right )\)	\(24\)
parts	\(\frac {x \,{\mathrm e}^{2 x^{3}+2 x^{2}}}{9}+\frac {1}{x}+x \ln \left (x \right )\)	\(24\)
risch	\(x \ln \left (x \right )+\frac {{\mathrm e}^{2 x^{2} \left (1+x \right )} x^{2}+9}{9 x}\)	\(26\)
parallelrisch	\(-\frac {-9 x^{2} \ln \left (x \right )-{\mathrm e}^{2 x^{3}+2 x^{2}} x^{2}-9}{9 x}\)	\(32\)

[In]

int(1/9*(9*x^2*ln(x)+(6*x^5+4*x^4+x^2)*exp(2*x^3+2*x^2)+9*x^2-9)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/9*x*exp(2*x^3+2*x^2)+1/x+x*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {x^{2} e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + 9 \, x^{2} \log \left (x\right ) + 9}{9 \, x} \]

[In]

integrate(1/9*(9*x^2*log(x)+(6*x^5+4*x^4+x^2)*exp(2*x^3+2*x^2)+9*x^2-9)/x^2,x, algorithm="fricas")

[Out]

1/9*(x^2*e^(2*x^3 + 2*x^2) + 9*x^2*log(x) + 9)/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {x e^{2 x^{3} + 2 x^{2}}}{9} + x \log {\left (x \right )} + \frac {1}{x} \]

[In]

integrate(1/9*(9*x**2*ln(x)+(6*x**5+4*x**4+x**2)*exp(2*x**3+2*x**2)+9*x**2-9)/x**2,x)

[Out]

x*exp(2*x**3 + 2*x**2)/9 + x*log(x) + 1/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {1}{9} \, x e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + x \log \left (x\right ) + \frac {1}{x} \]

[In]

integrate(1/9*(9*x^2*log(x)+(6*x^5+4*x^4+x^2)*exp(2*x^3+2*x^2)+9*x^2-9)/x^2,x, algorithm="maxima")

[Out]

1/9*x*e^(2*x^3 + 2*x^2) + x*log(x) + 1/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=\frac {x^{2} e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + 9 \, x^{2} \log \left (x\right ) + 9}{9 \, x} \]

[In]

integrate(1/9*(9*x^2*log(x)+(6*x^5+4*x^4+x^2)*exp(2*x^3+2*x^2)+9*x^2-9)/x^2,x, algorithm="giac")

[Out]

1/9*(x^2*e^(2*x^3 + 2*x^2) + 9*x^2*log(x) + 9)/x

Mupad [B] (verification not implemented)

Time = 13.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-9+9 x^2+e^{2 x^2+2 x^3} \left (x^2+4 x^4+6 x^5\right )+9 x^2 \log (x)}{9 x^2} \, dx=x+\frac {x\,{\mathrm {e}}^{2\,x^3+2\,x^2}}{9}+x\,\left (\ln \left (x\right )-1\right )+\frac {1}{x} \]

[In]

int((x^2*log(x) + (exp(2*x^2 + 2*x^3)*(x^2 + 4*x^4 + 6*x^5))/9 + x^2 - 1)/x^2,x)

[Out]

x + (x*exp(2*x^2 + 2*x^3))/9 + x*(log(x) - 1) + 1/x