3.98.63 \(\int \frac {1}{5} (15+e^5 (-4 x-9 x^2+4 x^3)) \, dx\)

Optimal. Leaf size=24 \[ 3 x-\frac {1}{5} e^5 \left (3+\frac {2}{x}-x\right ) x^3 \]

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12} \begin {gather*} \frac {e^5 x^4}{5}-\frac {3 e^5 x^3}{5}-\frac {2 e^5 x^2}{5}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + E^5*(-4*x - 9*x^2 + 4*x^3))/5,x]

[Out]

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

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (15+e^5 \left (-4 x-9 x^2+4 x^3\right )\right ) \, dx\\ &=3 x+\frac {1}{5} e^5 \int \left (-4 x-9 x^2+4 x^3\right ) \, dx\\ &=3 x-\frac {2 e^5 x^2}{5}-\frac {3 e^5 x^3}{5}+\frac {e^5 x^4}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.42 \begin {gather*} 3 x-\frac {2 e^5 x^2}{5}-\frac {3 e^5 x^3}{5}+\frac {e^5 x^4}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + E^5*(-4*x - 9*x^2 + 4*x^3))/5,x]

[Out]

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

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fricas [A]  time = 0.67, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{5} \, {\left (x^{4} - 3 \, x^{3} - 2 \, x^{2}\right )} e^{5} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3-9*x^2-4*x)*exp(5)+3,x, algorithm="fricas")

[Out]

1/5*(x^4 - 3*x^3 - 2*x^2)*e^5 + 3*x

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giac [A]  time = 0.15, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{5} \, {\left (x^{4} - 3 \, x^{3} - 2 \, x^{2}\right )} e^{5} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3-9*x^2-4*x)*exp(5)+3,x, algorithm="giac")

[Out]

1/5*(x^4 - 3*x^3 - 2*x^2)*e^5 + 3*x

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maple [A]  time = 0.02, size = 23, normalized size = 0.96

method	result	size
default	\(\frac {{\mathrm e}^{5} \left (x^{4}-3 x^{3}-2 x^{2}\right )}{5}+3 x\)	\(23\)
gosper	\(\frac {x \left (x^{3} {\mathrm e}^{5}-3 x^{2} {\mathrm e}^{5}-2 x \,{\mathrm e}^{5}+15\right )}{5}\)	\(24\)
norman	\(3 x -\frac {2 x^{2} {\mathrm e}^{5}}{5}-\frac {3 x^{3} {\mathrm e}^{5}}{5}+\frac {x^{4} {\mathrm e}^{5}}{5}\)	\(26\)
risch	\(3 x -\frac {2 x^{2} {\mathrm e}^{5}}{5}-\frac {3 x^{3} {\mathrm e}^{5}}{5}+\frac {x^{4} {\mathrm e}^{5}}{5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*x^3-9*x^2-4*x)*exp(5)+3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{5} \, {\left (x^{4} - 3 \, x^{3} - 2 \, x^{2}\right )} e^{5} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x^3-9*x^2-4*x)*exp(5)+3,x, algorithm="maxima")

[Out]

1/5*(x^4 - 3*x^3 - 2*x^2)*e^5 + 3*x

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mupad [B]  time = 0.05, size = 25, normalized size = 1.04 \begin {gather*} \frac {{\mathrm {e}}^5\,x^4}{5}-\frac {3\,{\mathrm {e}}^5\,x^3}{5}-\frac {2\,{\mathrm {e}}^5\,x^2}{5}+3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3 - (exp(5)*(4*x + 9*x^2 - 4*x^3))/5,x)

[Out]

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

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sympy [A]  time = 0.06, size = 31, normalized size = 1.29 \begin {gather*} \frac {x^{4} e^{5}}{5} - \frac {3 x^{3} e^{5}}{5} - \frac {2 x^{2} e^{5}}{5} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*x**3-9*x**2-4*x)*exp(5)+3,x)

[Out]

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

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