3.85.83 \(\int \frac {1}{5} (-6 e^x+e^{4 x} (-4-3 x^2-4 x^3)+e^{3 x} (-14+3 x-15 x^2-11 x^3+3 x^4)) \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{5} e^x \left (-6+e^{2 x} \left (-5-e^x+x\right ) \left (1+x^3\right )\right ) \]

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Rubi [B]  time = 0.23, antiderivative size = 68, normalized size of antiderivative = 2.43, number of steps used = 30, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 2194, 2196, 2176} \begin {gather*} \frac {1}{5} e^{3 x} x^4-e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x-\frac {6 e^x}{5}-e^{3 x}-\frac {e^{4 x}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*E^x + E^(4*x)*(-4 - 3*x^2 - 4*x^3) + E^(3*x)*(-14 + 3*x - 15*x^2 - 11*x^3 + 3*x^4))/5,x]

[Out]

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

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]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (-6 e^x+e^{4 x} \left (-4-3 x^2-4 x^3\right )+e^{3 x} \left (-14+3 x-15 x^2-11 x^3+3 x^4\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{4 x} \left (-4-3 x^2-4 x^3\right ) \, dx+\frac {1}{5} \int e^{3 x} \left (-14+3 x-15 x^2-11 x^3+3 x^4\right ) \, dx-\frac {6 \int e^x \, dx}{5}\\ &=-\frac {6 e^x}{5}+\frac {1}{5} \int \left (-4 e^{4 x}-3 e^{4 x} x^2-4 e^{4 x} x^3\right ) \, dx+\frac {1}{5} \int \left (-14 e^{3 x}+3 e^{3 x} x-15 e^{3 x} x^2-11 e^{3 x} x^3+3 e^{3 x} x^4\right ) \, dx\\ &=-\frac {6 e^x}{5}+\frac {3}{5} \int e^{3 x} x \, dx-\frac {3}{5} \int e^{4 x} x^2 \, dx+\frac {3}{5} \int e^{3 x} x^4 \, dx-\frac {4}{5} \int e^{4 x} \, dx-\frac {4}{5} \int e^{4 x} x^3 \, dx-\frac {11}{5} \int e^{3 x} x^3 \, dx-\frac {14}{5} \int e^{3 x} \, dx-3 \int e^{3 x} x^2 \, dx\\ &=-\frac {6 e^x}{5}-\frac {14 e^{3 x}}{15}-\frac {e^{4 x}}{5}+\frac {1}{5} e^{3 x} x-e^{3 x} x^2-\frac {3}{20} e^{4 x} x^2-\frac {11}{15} e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x^4-\frac {1}{5} \int e^{3 x} \, dx+\frac {3}{10} \int e^{4 x} x \, dx+\frac {3}{5} \int e^{4 x} x^2 \, dx-\frac {4}{5} \int e^{3 x} x^3 \, dx+2 \int e^{3 x} x \, dx+\frac {11}{5} \int e^{3 x} x^2 \, dx\\ &=-\frac {6 e^x}{5}-e^{3 x}-\frac {e^{4 x}}{5}+\frac {13}{15} e^{3 x} x+\frac {3}{40} e^{4 x} x-\frac {4}{15} e^{3 x} x^2-e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x^4-\frac {3}{40} \int e^{4 x} \, dx-\frac {3}{10} \int e^{4 x} x \, dx-\frac {2}{3} \int e^{3 x} \, dx+\frac {4}{5} \int e^{3 x} x^2 \, dx-\frac {22}{15} \int e^{3 x} x \, dx\\ &=-\frac {6 e^x}{5}-\frac {11 e^{3 x}}{9}-\frac {7 e^{4 x}}{32}+\frac {17}{45} e^{3 x} x-e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x^4+\frac {3}{40} \int e^{4 x} \, dx+\frac {22}{45} \int e^{3 x} \, dx-\frac {8}{15} \int e^{3 x} x \, dx\\ &=-\frac {6 e^x}{5}-\frac {143 e^{3 x}}{135}-\frac {e^{4 x}}{5}+\frac {1}{5} e^{3 x} x-e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x^4+\frac {8}{45} \int e^{3 x} \, dx\\ &=-\frac {6 e^x}{5}-e^{3 x}-\frac {e^{4 x}}{5}+\frac {1}{5} e^{3 x} x-e^{3 x} x^3-\frac {1}{5} e^{4 x} x^3+\frac {1}{5} e^{3 x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 39, normalized size = 1.39 \begin {gather*} \frac {1}{5} \left (-6 e^x-e^{4 x} \left (1+x^3\right )+e^{3 x} \left (-5+x-5 x^3+x^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E^x + E^(4*x)*(-4 - 3*x^2 - 4*x^3) + E^(3*x)*(-14 + 3*x - 15*x^2 - 11*x^3 + 3*x^4))/5,x]

[Out]

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

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fricas [A]  time = 1.03, size = 33, normalized size = 1.18 \begin {gather*} -\frac {1}{5} \, {\left (x^{3} + 1\right )} e^{\left (4 \, x\right )} + \frac {1}{5} \, {\left (x^{4} - 5 \, x^{3} + x - 5\right )} e^{\left (3 \, x\right )} - \frac {6}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x^3-3*x^2-4)*exp(x)^4+1/5*(3*x^4-11*x^3-15*x^2+3*x-14)*exp(x)^3-6/5*exp(x),x, algorithm="fri
cas")

[Out]

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

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giac [A]  time = 0.12, size = 33, normalized size = 1.18 \begin {gather*} -\frac {1}{5} \, {\left (x^{3} + 1\right )} e^{\left (4 \, x\right )} + \frac {1}{5} \, {\left (x^{4} - 5 \, x^{3} + x - 5\right )} e^{\left (3 \, x\right )} - \frac {6}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x^3-3*x^2-4)*exp(x)^4+1/5*(3*x^4-11*x^3-15*x^2+3*x-14)*exp(x)^3-6/5*exp(x),x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.05, size = 36, normalized size = 1.29

method	result	size
risch	\(\frac {\left (-x^{3}-1\right ) {\mathrm e}^{4 x}}{5}+\frac {\left (x^{4}-5 x^{3}+x -5\right ) {\mathrm e}^{3 x}}{5}-\frac {6 \,{\mathrm e}^{x}}{5}\)	\(36\)
default	\(-\frac {{\mathrm e}^{4 x}}{5}-\frac {x^{3} {\mathrm e}^{4 x}}{5}-{\mathrm e}^{3 x}+\frac {x \,{\mathrm e}^{3 x}}{5}-x^{3} {\mathrm e}^{3 x}+\frac {x^{4} {\mathrm e}^{3 x}}{5}-\frac {6 \,{\mathrm e}^{x}}{5}\)	\(52\)
norman	\(-\frac {{\mathrm e}^{4 x}}{5}-\frac {x^{3} {\mathrm e}^{4 x}}{5}-{\mathrm e}^{3 x}+\frac {x \,{\mathrm e}^{3 x}}{5}-x^{3} {\mathrm e}^{3 x}+\frac {x^{4} {\mathrm e}^{3 x}}{5}-\frac {6 \,{\mathrm e}^{x}}{5}\)	\(52\)
meijerg	\(\frac {12}{5}-\frac {{\mathrm e}^{4 x}}{5}+\frac {\left (-256 x^{3}+192 x^{2}-96 x +24\right ) {\mathrm e}^{4 x}}{1280}-\frac {\left (48 x^{2}-24 x +6\right ) {\mathrm e}^{4 x}}{320}-\frac {14 \,{\mathrm e}^{3 x}}{15}+\frac {\left (405 x^{4}-540 x^{3}+540 x^{2}-360 x +120\right ) {\mathrm e}^{3 x}}{2025}+\frac {11 \left (-108 x^{3}+108 x^{2}-72 x +24\right ) {\mathrm e}^{3 x}}{1620}-\frac {\left (27 x^{2}-18 x +6\right ) {\mathrm e}^{3 x}}{27}-\frac {\left (-6 x +2\right ) {\mathrm e}^{3 x}}{30}-\frac {6 \,{\mathrm e}^{x}}{5}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-4*x^3-3*x^2-4)*exp(x)^4+1/5*(3*x^4-11*x^3-15*x^2+3*x-14)*exp(x)^3-6/5*exp(x),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [B]  time = 0.35, size = 96, normalized size = 3.43 \begin {gather*} -\frac {1}{5} \, {\left (x^{3} + 1\right )} e^{\left (4 \, x\right )} + \frac {1}{135} \, {\left (27 \, x^{4} - 36 \, x^{3} + 36 \, x^{2} - 24 \, x + 8\right )} e^{\left (3 \, x\right )} - \frac {11}{135} \, {\left (9 \, x^{3} - 9 \, x^{2} + 6 \, x - 2\right )} e^{\left (3 \, x\right )} - \frac {1}{9} \, {\left (9 \, x^{2} - 6 \, x + 2\right )} e^{\left (3 \, x\right )} + \frac {1}{15} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} - \frac {14}{15} \, e^{\left (3 \, x\right )} - \frac {6}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x^3-3*x^2-4)*exp(x)^4+1/5*(3*x^4-11*x^3-15*x^2+3*x-14)*exp(x)^3-6/5*exp(x),x, algorithm="max
ima")

[Out]

-1/5*(x^3 + 1)*e^(4*x) + 1/135*(27*x^4 - 36*x^3 + 36*x^2 - 24*x + 8)*e^(3*x) - 11/135*(9*x^3 - 9*x^2 + 6*x - 2
)*e^(3*x) - 1/9*(9*x^2 - 6*x + 2)*e^(3*x) + 1/15*(3*x - 1)*e^(3*x) - 14/15*e^(3*x) - 6/5*e^x

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mupad [B]  time = 5.13, size = 51, normalized size = 1.82 \begin {gather*} \frac {x\,{\mathrm {e}}^{3\,x}}{5}-\frac {{\mathrm {e}}^{4\,x}}{5}-\frac {6\,{\mathrm {e}}^x}{5}-{\mathrm {e}}^{3\,x}-x^3\,{\mathrm {e}}^{3\,x}-\frac {x^3\,{\mathrm {e}}^{4\,x}}{5}+\frac {x^4\,{\mathrm {e}}^{3\,x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- (6*exp(x))/5 - (exp(4*x)*(3*x^2 + 4*x^3 + 4))/5 - (exp(3*x)*(15*x^2 - 3*x + 11*x^3 - 3*x^4 + 14))/5,x)

[Out]

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

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sympy [A]  time = 0.18, size = 42, normalized size = 1.50 \begin {gather*} \frac {\left (- 25 x^{3} - 25\right ) e^{4 x}}{125} + \frac {\left (25 x^{4} - 125 x^{3} + 25 x - 125\right ) e^{3 x}}{125} - \frac {6 e^{x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x**3-3*x**2-4)*exp(x)**4+1/5*(3*x**4-11*x**3-15*x**2+3*x-14)*exp(x)**3-6/5*exp(x),x)

[Out]

(-25*x**3 - 25)*exp(4*x)/125 + (25*x**4 - 125*x**3 + 25*x - 125)*exp(3*x)/125 - 6*exp(x)/5

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