\(\int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} (-9+18 x^3)}{x^2} \, dx\) [6783]

Optimal result

Integrand size = 37, antiderivative size = 19 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=9 \left (-3+e^x+e^{x \left (\frac {1}{x^2}+x\right )}-x\right ) \]

[Out]

9*exp(x)-27+9*exp(x*(1/x^2+x))-9*x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {14, 2225, 6838} \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=9 e^{x^2+\frac {1}{x}}-9 x+9 e^x \]

[In]

Int[(-9*x^2 + 9*E^x*x^2 + E^((1 + x^3)/x)*(-9 + 18*x^3))/x^2,x]

[Out]

9*E^x + 9*E^(x^(-1) + x^2) - 9*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 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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (9 \left (-1+e^x\right )+\frac {9 e^{\frac {1}{x}+x^2} \left (-1+2 x^3\right )}{x^2}\right ) \, dx \\ & = 9 \int \left (-1+e^x\right ) \, dx+9 \int \frac {e^{\frac {1}{x}+x^2} \left (-1+2 x^3\right )}{x^2} \, dx \\ & = 9 e^{\frac {1}{x}+x^2}-9 x+9 \int e^x \, dx \\ & = 9 e^x+9 e^{\frac {1}{x}+x^2}-9 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=9 \left (e^x+e^{\frac {1}{x}+x^2}-x\right ) \]

[In]

Integrate[(-9*x^2 + 9*E^x*x^2 + E^((1 + x^3)/x)*(-9 + 18*x^3))/x^2,x]

[Out]

9*(E^x + E^(x^(-1) + x^2) - x)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(-9 x +9 \,{\mathrm e}^{x}+9 \,{\mathrm e}^{\frac {x^{3}+1}{x}}\)	\(21\)
parts	\(-9 x +9 \,{\mathrm e}^{x}+9 \,{\mathrm e}^{\frac {x^{3}+1}{x}}\)	\(21\)
risch	\(-9 x +9 \,{\mathrm e}^{x}+9 \,{\mathrm e}^{\frac {\left (1+x \right ) \left (x^{2}-x +1\right )}{x}}\)	\(27\)
norman	\(\frac {-9 x^{2}+9 \,{\mathrm e}^{x} x +9 \,{\mathrm e}^{\frac {x^{3}+1}{x}} x}{x}\)	\(29\)

[In]

int(((18*x^3-9)*exp((x^3+1)/x)+9*exp(x)*x^2-9*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=-9 \, x + 9 \, e^{x} + 9 \, e^{\left (\frac {x^{3} + 1}{x}\right )} \]

[In]

integrate(((18*x^3-9)*exp((x^3+1)/x)+9*exp(x)*x^2-9*x^2)/x^2,x, algorithm="fricas")

[Out]

-9*x + 9*e^x + 9*e^((x^3 + 1)/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=- 9 x + 9 e^{x} + 9 e^{\frac {x^{3} + 1}{x}} \]

[In]

integrate(((18*x**3-9)*exp((x**3+1)/x)+9*exp(x)*x**2-9*x**2)/x**2,x)

[Out]

-9*x + 9*exp(x) + 9*exp((x**3 + 1)/x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=-9 \, x + 9 \, e^{\left (x^{2} + \frac {1}{x}\right )} + 9 \, e^{x} \]

[In]

integrate(((18*x^3-9)*exp((x^3+1)/x)+9*exp(x)*x^2-9*x^2)/x^2,x, algorithm="maxima")

[Out]

-9*x + 9*e^(x^2 + 1/x) + 9*e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=-9 \, x + 9 \, e^{x} + 9 \, e^{\left (\frac {x^{3} + 1}{x}\right )} \]

[In]

integrate(((18*x^3-9)*exp((x^3+1)/x)+9*exp(x)*x^2-9*x^2)/x^2,x, algorithm="giac")

[Out]

-9*x + 9*e^x + 9*e^((x^3 + 1)/x)

Mupad [B] (verification not implemented)

Time = 11.85 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-9 x^2+9 e^x x^2+e^{\frac {1+x^3}{x}} \left (-9+18 x^3\right )}{x^2} \, dx=9\,{\mathrm {e}}^{\frac {1}{x}+x^2}-9\,x+9\,{\mathrm {e}}^x \]

[In]

int((9*x^2*exp(x) + exp((x^3 + 1)/x)*(18*x^3 - 9) - 9*x^2)/x^2,x)

[Out]

9*exp(1/x + x^2) - 9*x + 9*exp(x)