\(\int \frac {-e^2+14 x^3+e^x (2 x^3+2 x^4)}{2 x^3} \, dx\) [7000]

Optimal result

Integrand size = 33, antiderivative size = 27 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=5-e^3+\frac {e^2}{4 x^2}+2 x+\left (5+e^x\right ) x \]

[Out]

2*x+(exp(x)+5)*x+1/4*exp(2)/x^2+5-exp(3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 14, 2207, 2225} \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=\frac {e^2}{4 x^2}+7 x-e^x+e^x (x+1) \]

[In]

Int[(-E^2 + 14*x^3 + E^x*(2*x^3 + 2*x^4))/(2*x^3),x]

[Out]

-E^x + E^2/(4*x^2) + 7*x + E^x*(1 + 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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{x^3} \, dx \\ & = \frac {1}{2} \int \left (2 e^x (1+x)+\frac {-e^2+14 x^3}{x^3}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-e^2+14 x^3}{x^3} \, dx+\int e^x (1+x) \, dx \\ & = e^x (1+x)+\frac {1}{2} \int \left (14-\frac {e^2}{x^3}\right ) \, dx-\int e^x \, dx \\ & = -e^x+\frac {e^2}{4 x^2}+7 x+e^x (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=\frac {e^2}{4 x^2}+7 x+e^x x \]

[In]

Integrate[(-E^2 + 14*x^3 + E^x*(2*x^3 + 2*x^4))/(2*x^3),x]

[Out]

E^2/(4*x^2) + 7*x + E^x*x

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.59

method	result	size
default	\(7 x +\frac {{\mathrm e}^{2}}{4 x^{2}}+{\mathrm e}^{x} x\)	\(16\)
risch	\(7 x +\frac {{\mathrm e}^{2}}{4 x^{2}}+{\mathrm e}^{x} x\)	\(16\)
parts	\(7 x +\frac {{\mathrm e}^{2}}{4 x^{2}}+{\mathrm e}^{x} x\)	\(16\)
norman	\(\frac {{\mathrm e}^{x} x^{3}+7 x^{3}+\frac {{\mathrm e}^{2}}{4}}{x^{2}}\)	\(21\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{x} x^{3}+28 x^{3}+{\mathrm e}^{2}}{4 x^{2}}\)	\(21\)

[In]

int(1/2*((2*x^4+2*x^3)*exp(x)-exp(2)+14*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

7*x+1/4*exp(2)/x^2+exp(x)*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=\frac {4 \, x^{3} e^{x} + 28 \, x^{3} + e^{2}}{4 \, x^{2}} \]

[In]

integrate(1/2*((2*x^4+2*x^3)*exp(x)-exp(2)+14*x^3)/x^3,x, algorithm="fricas")

[Out]

1/4*(4*x^3*e^x + 28*x^3 + e^2)/x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=x e^{x} + 7 x + \frac {e^{2}}{4 x^{2}} \]

[In]

integrate(1/2*((2*x**4+2*x**3)*exp(x)-exp(2)+14*x**3)/x**3,x)

[Out]

x*exp(x) + 7*x + exp(2)/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx={\left (x - 1\right )} e^{x} + 7 \, x + \frac {e^{2}}{4 \, x^{2}} + e^{x} \]

[In]

integrate(1/2*((2*x^4+2*x^3)*exp(x)-exp(2)+14*x^3)/x^3,x, algorithm="maxima")

[Out]

(x - 1)*e^x + 7*x + 1/4*e^2/x^2 + e^x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=\frac {4 \, x^{3} e^{x} + 28 \, x^{3} + e^{2}}{4 \, x^{2}} \]

[In]

integrate(1/2*((2*x^4+2*x^3)*exp(x)-exp(2)+14*x^3)/x^3,x, algorithm="giac")

[Out]

1/4*(4*x^3*e^x + 28*x^3 + e^2)/x^2

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {-e^2+14 x^3+e^x \left (2 x^3+2 x^4\right )}{2 x^3} \, dx=x\,\left ({\mathrm {e}}^x+7\right )+\frac {{\mathrm {e}}^2}{4\,x^2} \]

[In]

int(((exp(x)*(2*x^3 + 2*x^4))/2 - exp(2)/2 + 7*x^3)/x^3,x)

[Out]

x*(exp(x) + 7) + exp(2)/(4*x^2)