3.4.100 \(\int \frac {1}{3} (-60 x+e^{3+x} (5+5 x)) \, dx\)

Optimal. Leaf size=26 \[ 5 x \left (\frac {e^{3+x}}{3}-x-\frac {-2+x^2}{x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2176, 2194} \begin {gather*} -10 x^2-\frac {5 e^{x+3}}{3}+\frac {5}{3} e^{x+3} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-60*x + E^(3 + x)*(5 + 5*x))/3,x]

[Out]

(-5*E^(3 + x))/3 - 10*x^2 + (5*E^(3 + x)*(1 + x))/3

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}{3} \int \left (-60 x+e^{3+x} (5+5 x)\right ) \, dx\\ &=-10 x^2+\frac {1}{3} \int e^{3+x} (5+5 x) \, dx\\ &=-10 x^2+\frac {5}{3} e^{3+x} (1+x)-\frac {5}{3} \int e^{3+x} \, dx\\ &=-\frac {5 e^{3+x}}{3}-10 x^2+\frac {5}{3} e^{3+x} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.62 \begin {gather*} \frac {5}{3} e^{3+x} x-10 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60*x + E^(3 + x)*(5 + 5*x))/3,x]

[Out]

(5*E^(3 + x)*x)/3 - 10*x^2

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fricas [A]  time = 0.79, size = 13, normalized size = 0.50 \begin {gather*} -10 \, x^{2} + \frac {5}{3} \, x e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(5*x+5)*exp(3+x)-20*x,x, algorithm="fricas")

[Out]

-10*x^2 + 5/3*x*e^(x + 3)

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giac [A]  time = 0.28, size = 13, normalized size = 0.50 \begin {gather*} -10 \, x^{2} + \frac {5}{3} \, x e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(5*x+5)*exp(3+x)-20*x,x, algorithm="giac")

[Out]

-10*x^2 + 5/3*x*e^(x + 3)

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maple [A]  time = 0.03, size = 14, normalized size = 0.54

method	result	size
norman	\(-10 x^{2}+\frac {5 \,{\mathrm e}^{3+x} x}{3}\)	\(14\)
risch	\(-10 x^{2}+\frac {5 \,{\mathrm e}^{3+x} x}{3}\)	\(14\)
default	\(-10 x^{2}+\frac {5 \,{\mathrm e}^{3+x} \left (3+x \right )}{3}-5 \,{\mathrm e}^{3+x}\)	\(22\)
derivativedivides	\(-10 \left (3+x \right )^{2}+180+60 x +\frac {5 \,{\mathrm e}^{3+x} \left (3+x \right )}{3}-5 \,{\mathrm e}^{3+x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(5*x+5)*exp(3+x)-20*x,x,method=_RETURNVERBOSE)

[Out]

-10*x^2+5/3*exp(3+x)*x

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maxima [A]  time = 0.34, size = 25, normalized size = 0.96 \begin {gather*} -10 \, x^{2} + \frac {5}{3} \, {\left (x e^{3} - e^{3}\right )} e^{x} + \frac {5}{3} \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(5*x+5)*exp(3+x)-20*x,x, algorithm="maxima")

[Out]

-10*x^2 + 5/3*(x*e^3 - e^3)*e^x + 5/3*e^(x + 3)

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mupad [B]  time = 0.04, size = 13, normalized size = 0.50 \begin {gather*} -\frac {5\,x\,\left (6\,x-{\mathrm {e}}^{x+3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + 3)*(5*x + 5))/3 - 20*x,x)

[Out]

-(5*x*(6*x - exp(x + 3)))/3

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sympy [A]  time = 0.09, size = 14, normalized size = 0.54 \begin {gather*} - 10 x^{2} + \frac {5 x e^{x + 3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(5*x+5)*exp(3+x)-20*x,x)

[Out]

-10*x**2 + 5*x*exp(x + 3)/3

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