3.26.52 \(\int \frac {1}{5} (-5-20 e^{4 x}+e^3 (-6+2 x)) \, dx\)

Optimal. Leaf size=32 \[ -1-e^{-4+4 (1+x)}-x+e^3 \left (-x+\frac {1}{5} (-1+x) x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2194} \begin {gather*} \frac {1}{5} e^3 (3-x)^2-e^{4 x}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 20*E^(4*x) + E^3*(-6 + 2*x))/5,x]

[Out]

-E^(4*x) + (E^3*(3 - x)^2)/5 - 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 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 (-5-20 e^{4 x}+e^3 (-6+2 x)\right ) \, dx\\ &=\frac {1}{5} e^3 (3-x)^2-x-4 \int e^{4 x} \, dx\\ &=-e^{4 x}+\frac {1}{5} e^3 (3-x)^2-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.91 \begin {gather*} -e^{4 x}-x-\frac {6 e^3 x}{5}+\frac {e^3 x^2}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 20*E^(4*x) + E^3*(-6 + 2*x))/5,x]

[Out]

-E^(4*x) - x - (6*E^3*x)/5 + (E^3*x^2)/5

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fricas [A]  time = 0.95, size = 21, normalized size = 0.66 \begin {gather*} \frac {1}{5} \, {\left (x^{2} - 6 \, x\right )} e^{3} - x - e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(4*x)+1/5*(2*x-6)*exp(3)-1,x, algorithm="fricas")

[Out]

1/5*(x^2 - 6*x)*e^3 - x - e^(4*x)

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giac [A]  time = 0.19, size = 21, normalized size = 0.66 \begin {gather*} \frac {1}{5} \, {\left (x^{2} - 6 \, x\right )} e^{3} - x - e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(4*x)+1/5*(2*x-6)*exp(3)-1,x, algorithm="giac")

[Out]

1/5*(x^2 - 6*x)*e^3 - x - e^(4*x)

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maple [A]  time = 0.02, size = 22, normalized size = 0.69

method	result	size
default	\(-x +\frac {{\mathrm e}^{3} \left (x^{2}-6 x \right )}{5}-{\mathrm e}^{4 x}\)	\(22\)
norman	\(\left (-\frac {6 \,{\mathrm e}^{3}}{5}-1\right ) x +\frac {x^{2} {\mathrm e}^{3}}{5}-{\mathrm e}^{4 x}\)	\(23\)
risch	\(-{\mathrm e}^{4 x}+\frac {x^{2} {\mathrm e}^{3}}{5}-\frac {6 x \,{\mathrm e}^{3}}{5}-x\)	\(23\)
derivativedivides	\(-x +\frac {{\mathrm e}^{3} \left (4 x^{2}-24 x \right )}{20}-{\mathrm e}^{4 x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4*exp(4*x)+1/5*(2*x-6)*exp(3)-1,x,method=_RETURNVERBOSE)

[Out]

-x+1/5*exp(3)*(x^2-6*x)-exp(4*x)

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maxima [A]  time = 0.44, size = 21, normalized size = 0.66 \begin {gather*} \frac {1}{5} \, {\left (x^{2} - 6 \, x\right )} e^{3} - x - e^{\left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(4*x)+1/5*(2*x-6)*exp(3)-1,x, algorithm="maxima")

[Out]

1/5*(x^2 - 6*x)*e^3 - x - e^(4*x)

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mupad [B]  time = 1.41, size = 23, normalized size = 0.72 \begin {gather*} \frac {x^2\,{\mathrm {e}}^3}{5}-{\mathrm {e}}^{4\,x}-x\,\left (\frac {6\,{\mathrm {e}}^3}{5}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(2*x - 6))/5 - 4*exp(4*x) - 1,x)

[Out]

(x^2*exp(3))/5 - exp(4*x) - x*((6*exp(3))/5 + 1)

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sympy [A]  time = 0.09, size = 24, normalized size = 0.75 \begin {gather*} \frac {x^{2} e^{3}}{5} + x \left (- \frac {6 e^{3}}{5} - 1\right ) - e^{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(4*x)+1/5*(2*x-6)*exp(3)-1,x)

[Out]

x**2*exp(3)/5 + x*(-6*exp(3)/5 - 1) - exp(4*x)

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