\(\int \frac {1}{5} (5 x^3+e^x (1+x)) \, dx\) [3183]

Optimal result

Integrand size = 17, antiderivative size = 17 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=2+\frac {e^x x}{5}+\frac {x^4}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 2207, 2225} \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {x^4}{4}-\frac {e^x}{5}+\frac {1}{5} e^x (x+1) \]

[In]

Int[(5*x^3 + E^x*(1 + x))/5,x]

[Out]

-1/5*E^x + x^4/4 + (E^x*(1 + x))/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {e^x x}{5}+\frac {x^4}{4} \]

[In]

Integrate[(5*x^3 + E^x*(1 + x))/5,x]

[Out]

(E^x*x)/5 + x^4/4

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {{\mathrm e}^{x} x}{5}+\frac {x^{4}}{4}\)	\(12\)
norman	\(\frac {{\mathrm e}^{x} x}{5}+\frac {x^{4}}{4}\)	\(12\)
risch	\(\frac {{\mathrm e}^{x} x}{5}+\frac {x^{4}}{4}\)	\(12\)
parallelrisch	\(\frac {{\mathrm e}^{x} x}{5}+\frac {x^{4}}{4}\)	\(12\)
parts	\(\frac {{\mathrm e}^{x} x}{5}+\frac {x^{4}}{4}\)	\(12\)

[In]

int(1/5*(1+x)*exp(x)+x^3,x,method=_RETURNVERBOSE)

[Out]

1/5*exp(x)*x+1/4*x^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {1}{4} \, x^{4} + \frac {1}{5} \, x e^{x} \]

[In]

integrate(1/5*(1+x)*exp(x)+x^3,x, algorithm="fricas")

[Out]

1/4*x^4 + 1/5*x*e^x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {x^{4}}{4} + \frac {x e^{x}}{5} \]

[In]

integrate(1/5*(1+x)*exp(x)+x**3,x)

[Out]

x**4/4 + x*exp(x)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {1}{4} \, x^{4} + \frac {1}{5} \, {\left (x - 1\right )} e^{x} + \frac {1}{5} \, e^{x} \]

[In]

integrate(1/5*(1+x)*exp(x)+x^3,x, algorithm="maxima")

[Out]

1/4*x^4 + 1/5*(x - 1)*e^x + 1/5*e^x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {1}{4} \, x^{4} + \frac {1}{5} \, x e^{x} \]

[In]

integrate(1/5*(1+x)*exp(x)+x^3,x, algorithm="giac")

[Out]

1/4*x^4 + 1/5*x*e^x

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {1}{5} \left (5 x^3+e^x (1+x)\right ) \, dx=\frac {x\,{\mathrm {e}}^x}{5}+\frac {x^4}{4} \]

[In]

int((exp(x)*(x + 1))/5 + x^3,x)

[Out]

(x*exp(x))/5 + x^4/4