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

Optimal. Leaf size=17 \[ 2+\frac {e^x x}{5}+\frac {x^4}{4} \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.47, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 2176, 2194} \begin {gather*} \frac {x^4}{4}-\frac {e^x}{5}+\frac {1}{5} e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2194

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} \frac {e^x x}{5}+\frac {x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^3 + E^x*(1 + x))/5,x]

[Out]

(E^x*x)/5 + x^4/4

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fricas [A]  time = 0.52, size = 11, normalized size = 0.65 \begin {gather*} \frac {1}{4} \, x^{4} + \frac {1}{5} \, x e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/5*x*e^x

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giac [A]  time = 0.21, size = 11, normalized size = 0.65 \begin {gather*} \frac {1}{4} \, x^{4} + \frac {1}{5} \, x e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/5*x*e^x

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maple [A]  time = 0.02, size = 12, normalized size = 0.71

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{4} \, x^{4} + \frac {1}{5} \, {\left (x - 1\right )} e^{x} + \frac {1}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4 + 1/5*(x - 1)*e^x + 1/5*e^x

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mupad [B]  time = 1.85, size = 11, normalized size = 0.65 \begin {gather*} \frac {x\,{\mathrm {e}}^x}{5}+\frac {x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x + 1))/5 + x^3,x)

[Out]

(x*exp(x))/5 + x^4/4

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sympy [A]  time = 0.08, size = 10, normalized size = 0.59 \begin {gather*} \frac {x^{4}}{4} + \frac {x e^{x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4/4 + x*exp(x)/5

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