3.80.32 \(\int \frac {1}{2} (e^{3 x} (5+15 x)+e^{2 x} (-25-40 x+10 x^2)) \, dx\)

Optimal. Leaf size=16 \[ \frac {5}{2} e^{2 x} x \left (-5+e^x+x\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 2.88, number of steps used = 12, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2176, 2194, 2196} \begin {gather*} \frac {5}{2} e^{2 x} x^2-\frac {25}{2} e^{2 x} x-\frac {5 e^{3 x}}{6}+\frac {5}{6} e^{3 x} (3 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3*x)*(5 + 15*x) + E^(2*x)*(-25 - 40*x + 10*x^2))/2,x]

[Out]

(-5*E^(3*x))/6 - (25*E^(2*x)*x)/2 + (5*E^(2*x)*x^2)/2 + (5*E^(3*x)*(1 + 3*x))/6

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}{2} \int \left (e^{3 x} (5+15 x)+e^{2 x} \left (-25-40 x+10 x^2\right )\right ) \, dx\\ &=\frac {1}{2} \int e^{3 x} (5+15 x) \, dx+\frac {1}{2} \int e^{2 x} \left (-25-40 x+10 x^2\right ) \, dx\\ &=\frac {5}{6} e^{3 x} (1+3 x)+\frac {1}{2} \int \left (-25 e^{2 x}-40 e^{2 x} x+10 e^{2 x} x^2\right ) \, dx-\frac {5}{2} \int e^{3 x} \, dx\\ &=-\frac {5 e^{3 x}}{6}+\frac {5}{6} e^{3 x} (1+3 x)+5 \int e^{2 x} x^2 \, dx-\frac {25}{2} \int e^{2 x} \, dx-20 \int e^{2 x} x \, dx\\ &=-\frac {25 e^{2 x}}{4}-\frac {5 e^{3 x}}{6}-10 e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x)-5 \int e^{2 x} x \, dx+10 \int e^{2 x} \, dx\\ &=-\frac {5 e^{2 x}}{4}-\frac {5 e^{3 x}}{6}-\frac {25}{2} e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x)+\frac {5}{2} \int e^{2 x} \, dx\\ &=-\frac {5 e^{3 x}}{6}-\frac {25}{2} e^{2 x} x+\frac {5}{2} e^{2 x} x^2+\frac {5}{6} e^{3 x} (1+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 16, normalized size = 1.00 \begin {gather*} \frac {5}{2} e^{2 x} x \left (-5+e^x+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*x)*(5 + 15*x) + E^(2*x)*(-25 - 40*x + 10*x^2))/2,x]

[Out]

(5*E^(2*x)*x*(-5 + E^x + x))/2

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fricas [A]  time = 0.68, size = 21, normalized size = 1.31 \begin {gather*} \frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{2} \, {\left (x^{2} - 5 \, x\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x+5)*exp(x)^3+1/2*(10*x^2-40*x-25)*exp(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 21, normalized size = 1.31 \begin {gather*} \frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{2} \, {\left (x^{2} - 5 \, x\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x+5)*exp(x)^3+1/2*(10*x^2-40*x-25)*exp(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 24, normalized size = 1.50

method	result	size
risch	\(\frac {5 x \,{\mathrm e}^{3 x}}{2}+\frac {\left (5 x^{2}-25 x \right ) {\mathrm e}^{2 x}}{2}\)	\(24\)
default	\(\frac {5 x \,{\mathrm e}^{3 x}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}\)	\(25\)
norman	\(\frac {5 x \,{\mathrm e}^{3 x}}{2}-\frac {25 x \,{\mathrm e}^{2 x}}{2}+\frac {5 \,{\mathrm e}^{2 x} x^{2}}{2}\)	\(25\)
meijerg	\(\frac {5 \,{\mathrm e}^{3 x}}{6}-\frac {5 \left (-6 x +2\right ) {\mathrm e}^{3 x}}{12}-\frac {25 \,{\mathrm e}^{2 x}}{4}+\frac {5 \left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{24}+\frac {5 \left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(15*x+5)*exp(x)^3+1/2*(10*x^2-40*x-25)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

5/2*x*exp(3*x)+1/2*(5*x^2-25*x)*exp(2*x)

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maxima [B]  time = 0.46, size = 41, normalized size = 2.56 \begin {gather*} \frac {5}{2} \, x e^{\left (3 \, x\right )} + \frac {5}{4} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 5 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {25}{4} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x+5)*exp(x)^3+1/2*(10*x^2-40*x-25)*exp(x)^2,x, algorithm="maxima")

[Out]

5/2*x*e^(3*x) + 5/4*(2*x^2 - 2*x + 1)*e^(2*x) - 5*(2*x - 1)*e^(2*x) - 25/4*e^(2*x)

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mupad [B]  time = 0.09, size = 12, normalized size = 0.75 \begin {gather*} \frac {5\,x\,{\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^x-5\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x)*(15*x + 5))/2 - (exp(2*x)*(40*x - 10*x^2 + 25))/2,x)

[Out]

(5*x*exp(2*x)*(x + exp(x) - 5))/2

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sympy [A]  time = 0.13, size = 24, normalized size = 1.50 \begin {gather*} \frac {5 x e^{3 x}}{2} + \frac {\left (10 x^{2} - 50 x\right ) e^{2 x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x+5)*exp(x)**3+1/2*(10*x**2-40*x-25)*exp(x)**2,x)

[Out]

5*x*exp(3*x)/2 + (10*x**2 - 50*x)*exp(2*x)/4

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