3.88.54 \(\int (30 x-20 x^3+e^{\frac {2}{5} (28-4 x)} (-10 x+8 x^2)+e^{\frac {1}{5} (28-4 x)} (-30 x^2+8 x^3)) \, dx\)

Optimal. Leaf size=27 \[ 5 x^2 \left (3-\left (e^{3-x+\frac {13+x}{5}}+x\right )^2\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 19, number of rules used = 4, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} -5 x^4-10 e^{\frac {4 (7-x)}{5}} x^3-5 e^{\frac {8 (7-x)}{5}} x^2+15 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[30*x - 20*x^3 + E^((2*(28 - 4*x))/5)*(-10*x + 8*x^2) + E^((28 - 4*x)/5)*(-30*x^2 + 8*x^3),x]

[Out]

15*x^2 - 5*E^((8*(7 - x))/5)*x^2 - 10*E^((4*(7 - x))/5)*x^3 - 5*x^4

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=15 x^2-5 x^4+\int e^{\frac {2}{5} (28-4 x)} \left (-10 x+8 x^2\right ) \, dx+\int e^{\frac {1}{5} (28-4 x)} \left (-30 x^2+8 x^3\right ) \, dx\\ &=15 x^2-5 x^4+\int e^{\frac {1}{5} (28-4 x)} x^2 (-30+8 x) \, dx+\int e^{\frac {2}{5} (28-4 x)} x (-10+8 x) \, dx\\ &=15 x^2-5 x^4+\int \left (-10 e^{\frac {2}{5} (28-4 x)} x+8 e^{\frac {2}{5} (28-4 x)} x^2\right ) \, dx+\int \left (-30 e^{\frac {1}{5} (28-4 x)} x^2+8 e^{\frac {1}{5} (28-4 x)} x^3\right ) \, dx\\ &=15 x^2-5 x^4+8 \int e^{\frac {2}{5} (28-4 x)} x^2 \, dx+8 \int e^{\frac {1}{5} (28-4 x)} x^3 \, dx-10 \int e^{\frac {2}{5} (28-4 x)} x \, dx-30 \int e^{\frac {1}{5} (28-4 x)} x^2 \, dx\\ &=\frac {25}{4} e^{\frac {8 (7-x)}{5}} x+15 x^2+\frac {75}{2} e^{\frac {4 (7-x)}{5}} x^2-5 e^{\frac {8 (7-x)}{5}} x^2-10 e^{\frac {4 (7-x)}{5}} x^3-5 x^4-\frac {25}{4} \int e^{\frac {2}{5} (28-4 x)} \, dx+10 \int e^{\frac {2}{5} (28-4 x)} x \, dx+30 \int e^{\frac {1}{5} (28-4 x)} x^2 \, dx-75 \int e^{\frac {1}{5} (28-4 x)} x \, dx\\ &=\frac {125}{32} e^{\frac {8 (7-x)}{5}}+\frac {375}{4} e^{\frac {4 (7-x)}{5}} x+15 x^2-5 e^{\frac {8 (7-x)}{5}} x^2-10 e^{\frac {4 (7-x)}{5}} x^3-5 x^4+\frac {25}{4} \int e^{\frac {2}{5} (28-4 x)} \, dx+75 \int e^{\frac {1}{5} (28-4 x)} x \, dx-\frac {375}{4} \int e^{\frac {1}{5} (28-4 x)} \, dx\\ &=\frac {1875}{16} e^{\frac {4 (7-x)}{5}}+15 x^2-5 e^{\frac {8 (7-x)}{5}} x^2-10 e^{\frac {4 (7-x)}{5}} x^3-5 x^4+\frac {375}{4} \int e^{\frac {1}{5} (28-4 x)} \, dx\\ &=15 x^2-5 e^{\frac {8 (7-x)}{5}} x^2-10 e^{\frac {4 (7-x)}{5}} x^3-5 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 43, normalized size = 1.59 \begin {gather*} -5 e^{-8 x/5} x^2 \left (e^{56/5}+2 e^{\frac {4 (7+x)}{5}} x+e^{8 x/5} \left (-3+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[30*x - 20*x^3 + E^((2*(28 - 4*x))/5)*(-10*x + 8*x^2) + E^((28 - 4*x)/5)*(-30*x^2 + 8*x^3),x]

[Out]

(-5*x^2*(E^(56/5) + 2*E^((4*(7 + x))/5)*x + E^((8*x)/5)*(-3 + x^2)))/E^((8*x)/5)

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fricas [A]  time = 0.59, size = 33, normalized size = 1.22 \begin {gather*} -5 \, x^{4} - 10 \, x^{3} e^{\left (-\frac {4}{5} \, x + \frac {28}{5}\right )} - 5 \, x^{2} e^{\left (-\frac {8}{5} \, x + \frac {56}{5}\right )} + 15 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-10*x)*exp(-4/5*x+28/5)^2+(8*x^3-30*x^2)*exp(-4/5*x+28/5)-20*x^3+30*x,x, algorithm="fricas")

[Out]

-5*x^4 - 10*x^3*e^(-4/5*x + 28/5) - 5*x^2*e^(-8/5*x + 56/5) + 15*x^2

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giac [A]  time = 0.13, size = 33, normalized size = 1.22 \begin {gather*} -5 \, x^{4} - 10 \, x^{3} e^{\left (-\frac {4}{5} \, x + \frac {28}{5}\right )} - 5 \, x^{2} e^{\left (-\frac {8}{5} \, x + \frac {56}{5}\right )} + 15 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-10*x)*exp(-4/5*x+28/5)^2+(8*x^3-30*x^2)*exp(-4/5*x+28/5)-20*x^3+30*x,x, algorithm="giac")

[Out]

-5*x^4 - 10*x^3*e^(-4/5*x + 28/5) - 5*x^2*e^(-8/5*x + 56/5) + 15*x^2

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maple [A]  time = 0.09, size = 34, normalized size = 1.26

method	result	size
risch	\(15 x^{2}-5 x^{4}-10 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} x^{3}-5 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} x^{2}\)	\(34\)
norman	\(15 x^{2}-5 x^{4}-10 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} x^{3}-5 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} x^{2}\)	\(36\)
default	\(-245 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}}+\frac {175 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )}{2}-\frac {125 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{2}}{16}+\frac {3675 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )}{2}-3430 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}}-\frac {2625 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{2}}{8}+\frac {625 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{3}}{32}+15 x^{2}-5 x^{4}\)	\(105\)
derivativedivides	\(210 x -1470+\frac {375 \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{2}}{16}-5 x^{4}+\frac {3675 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )}{2}-3430 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}}-\frac {2625 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{2}}{8}+\frac {625 \,{\mathrm e}^{-\frac {4 x}{5}+\frac {28}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{3}}{32}-245 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}}+\frac {175 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )}{2}-\frac {125 \,{\mathrm e}^{-\frac {8 x}{5}+\frac {56}{5}} \left (-\frac {4 x}{5}+\frac {28}{5}\right )^{2}}{16}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2-10*x)*exp(-4/5*x+28/5)^2+(8*x^3-30*x^2)*exp(-4/5*x+28/5)-20*x^3+30*x,x,method=_RETURNVERBOSE)

[Out]

15*x^2-5*x^4-10*exp(-4/5*x+28/5)*x^3-5*exp(-8/5*x+56/5)*x^2

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maxima [A]  time = 0.38, size = 33, normalized size = 1.22 \begin {gather*} -5 \, x^{4} - 10 \, x^{3} e^{\left (-\frac {4}{5} \, x + \frac {28}{5}\right )} - 5 \, x^{2} e^{\left (-\frac {8}{5} \, x + \frac {56}{5}\right )} + 15 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2-10*x)*exp(-4/5*x+28/5)^2+(8*x^3-30*x^2)*exp(-4/5*x+28/5)-20*x^3+30*x,x, algorithm="maxima")

[Out]

-5*x^4 - 10*x^3*e^(-4/5*x + 28/5) - 5*x^2*e^(-8/5*x + 56/5) + 15*x^2

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mupad [B]  time = 0.07, size = 25, normalized size = 0.93 \begin {gather*} -5\,x^2\,\left ({\mathrm {e}}^{\frac {56}{5}-\frac {8\,x}{5}}+2\,x\,{\mathrm {e}}^{\frac {28}{5}-\frac {4\,x}{5}}+x^2-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(30*x - exp(56/5 - (8*x)/5)*(10*x - 8*x^2) - exp(28/5 - (4*x)/5)*(30*x^2 - 8*x^3) - 20*x^3,x)

[Out]

-5*x^2*(exp(56/5 - (8*x)/5) + 2*x*exp(28/5 - (4*x)/5) + x^2 - 3)

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sympy [B]  time = 0.12, size = 39, normalized size = 1.44 \begin {gather*} - 5 x^{4} - 10 x^{3} e^{\frac {28}{5} - \frac {4 x}{5}} - 5 x^{2} e^{\frac {56}{5} - \frac {8 x}{5}} + 15 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**2-10*x)*exp(-4/5*x+28/5)**2+(8*x**3-30*x**2)*exp(-4/5*x+28/5)-20*x**3+30*x,x)

[Out]

-5*x**4 - 10*x**3*exp(28/5 - 4*x/5) - 5*x**2*exp(56/5 - 8*x/5) + 15*x**2

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