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

Optimal. Leaf size=23 \[ \frac {1}{x}+x \left (\frac {1}{9} e^{2 x^2 (1+x)}+\log (x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.96, number of steps used = 9, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 14, 2288, 2295} \begin {gather*} \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) \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 2295

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{x}+\frac {1}{9} e^{2 x^2 (1+x)} x+x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

________________________________________________________________________________________

fricas [A]  time = 1.45, size = 30, normalized size = 1.30 \begin {gather*} \frac {x^{2} e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + 9 \, x^{2} \log \relax (x) + 9}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [A]  time = 0.20, size = 30, normalized size = 1.30 \begin {gather*} \frac {x^{2} e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + 9 \, x^{2} \log \relax (x) + 9}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 24, normalized size = 1.04

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{9} \, x e^{\left (2 \, x^{3} + 2 \, x^{2}\right )} + x \log \relax (x) + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 5.33, size = 26, normalized size = 1.13 \begin {gather*} x+\frac {x\,{\mathrm {e}}^{2\,x^3+2\,x^2}}{9}+x\,\left (\ln \relax (x)-1\right )+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.31, size = 22, normalized size = 0.96 \begin {gather*} \frac {x e^{2 x^{3} + 2 x^{2}}}{9} + x \log {\relax (x )} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________