3.16.61 \(\int \frac {1}{4} (4+e^x (3+5 x+x^2)+e^{2 x} (18 x+36 x^2+16 x^3+2 x^4)) \, dx\)

Optimal. Leaf size=22 \[ x+\left (-\frac {1}{4}+e^x \left (x-\frac {1}{2} x (5+x)\right )\right )^2 \]

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Rubi [B]  time = 0.19, antiderivative size = 56, normalized size of antiderivative = 2.55, number of steps used = 26, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 2196, 2194, 2176} \begin {gather*} \frac {1}{4} e^{2 x} x^4+\frac {3}{2} e^{2 x} x^3+\frac {e^x x^2}{4}+\frac {9}{4} e^{2 x} x^2+\frac {3 e^x x}{4}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + E^x*(3 + 5*x + x^2) + E^(2*x)*(18*x + 36*x^2 + 16*x^3 + 2*x^4))/4,x]

[Out]

x + (3*E^x*x)/4 + (E^x*x^2)/4 + (9*E^(2*x)*x^2)/4 + (3*E^(2*x)*x^3)/2 + (E^(2*x)*x^4)/4

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}{4} \int \left (4+e^x \left (3+5 x+x^2\right )+e^{2 x} \left (18 x+36 x^2+16 x^3+2 x^4\right )\right ) \, dx\\ &=x+\frac {1}{4} \int e^x \left (3+5 x+x^2\right ) \, dx+\frac {1}{4} \int e^{2 x} \left (18 x+36 x^2+16 x^3+2 x^4\right ) \, dx\\ &=x+\frac {1}{4} \int \left (3 e^x+5 e^x x+e^x x^2\right ) \, dx+\frac {1}{4} \int \left (18 e^{2 x} x+36 e^{2 x} x^2+16 e^{2 x} x^3+2 e^{2 x} x^4\right ) \, dx\\ &=x+\frac {1}{4} \int e^x x^2 \, dx+\frac {1}{2} \int e^{2 x} x^4 \, dx+\frac {3 \int e^x \, dx}{4}+\frac {5}{4} \int e^x x \, dx+4 \int e^{2 x} x^3 \, dx+\frac {9}{2} \int e^{2 x} x \, dx+9 \int e^{2 x} x^2 \, dx\\ &=\frac {3 e^x}{4}+x+\frac {5 e^x x}{4}+\frac {9}{4} e^{2 x} x+\frac {e^x x^2}{4}+\frac {9}{2} e^{2 x} x^2+2 e^{2 x} x^3+\frac {1}{4} e^{2 x} x^4-\frac {1}{2} \int e^x x \, dx-\frac {5 \int e^x \, dx}{4}-\frac {9}{4} \int e^{2 x} \, dx-6 \int e^{2 x} x^2 \, dx-9 \int e^{2 x} x \, dx-\int e^{2 x} x^3 \, dx\\ &=-\frac {e^x}{2}-\frac {9 e^{2 x}}{8}+x+\frac {3 e^x x}{4}-\frac {9}{4} e^{2 x} x+\frac {e^x x^2}{4}+\frac {3}{2} e^{2 x} x^2+\frac {3}{2} e^{2 x} x^3+\frac {1}{4} e^{2 x} x^4+\frac {\int e^x \, dx}{2}+\frac {3}{2} \int e^{2 x} x^2 \, dx+\frac {9}{2} \int e^{2 x} \, dx+6 \int e^{2 x} x \, dx\\ &=\frac {9 e^{2 x}}{8}+x+\frac {3 e^x x}{4}+\frac {3}{4} e^{2 x} x+\frac {e^x x^2}{4}+\frac {9}{4} e^{2 x} x^2+\frac {3}{2} e^{2 x} x^3+\frac {1}{4} e^{2 x} x^4-\frac {3}{2} \int e^{2 x} x \, dx-3 \int e^{2 x} \, dx\\ &=-\frac {3 e^{2 x}}{8}+x+\frac {3 e^x x}{4}+\frac {e^x x^2}{4}+\frac {9}{4} e^{2 x} x^2+\frac {3}{2} e^{2 x} x^3+\frac {1}{4} e^{2 x} x^4+\frac {3}{4} \int e^{2 x} \, dx\\ &=x+\frac {3 e^x x}{4}+\frac {e^x x^2}{4}+\frac {9}{4} e^{2 x} x^2+\frac {3}{2} e^{2 x} x^3+\frac {1}{4} e^{2 x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 26, normalized size = 1.18 \begin {gather*} \frac {1}{4} x \left (4+e^x (3+x)+e^{2 x} x (3+x)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + E^x*(3 + 5*x + x^2) + E^(2*x)*(18*x + 36*x^2 + 16*x^3 + 2*x^4))/4,x]

[Out]

(x*(4 + E^x*(3 + x) + E^(2*x)*x*(3 + x)^2))/4

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fricas [A]  time = 1.03, size = 33, normalized size = 1.50 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{2} + 3 \, x\right )} e^{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^4+16*x^3+36*x^2+18*x)*exp(x)^2+1/4*(x^2+5*x+3)*exp(x)+1,x, algorithm="fricas")

[Out]

1/4*(x^4 + 6*x^3 + 9*x^2)*e^(2*x) + 1/4*(x^2 + 3*x)*e^x + x

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giac [A]  time = 0.92, size = 33, normalized size = 1.50 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{2} + 3 \, x\right )} e^{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^4+16*x^3+36*x^2+18*x)*exp(x)^2+1/4*(x^2+5*x+3)*exp(x)+1,x, algorithm="giac")

[Out]

1/4*(x^4 + 6*x^3 + 9*x^2)*e^(2*x) + 1/4*(x^2 + 3*x)*e^x + x

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maple [A]  time = 0.03, size = 34, normalized size = 1.55

method	result	size
risch	\(\frac {\left (x^{4}+6 x^{3}+9 x^{2}\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (x^{2}+3 x \right ) {\mathrm e}^{x}}{4}+x\)	\(34\)
default	\(x +\frac {{\mathrm e}^{x} x^{2}}{4}+\frac {3 \,{\mathrm e}^{x} x}{4}+\frac {{\mathrm e}^{2 x} x^{4}}{4}+\frac {3 \,{\mathrm e}^{2 x} x^{3}}{2}+\frac {9 \,{\mathrm e}^{2 x} x^{2}}{4}\)	\(42\)
norman	\(x +\frac {{\mathrm e}^{x} x^{2}}{4}+\frac {3 \,{\mathrm e}^{x} x}{4}+\frac {{\mathrm e}^{2 x} x^{4}}{4}+\frac {3 \,{\mathrm e}^{2 x} x^{3}}{2}+\frac {9 \,{\mathrm e}^{2 x} x^{2}}{4}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(2*x^4+16*x^3+36*x^2+18*x)*exp(x)^2+1/4*(x^2+5*x+3)*exp(x)+1,x,method=_RETURNVERBOSE)

[Out]

1/4*(x^4+6*x^3+9*x^2)*exp(2*x)+1/4*(x^2+3*x)*exp(x)+x

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maxima [B]  time = 0.35, size = 45, normalized size = 2.05 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + \frac {5}{4} \, {\left (x - 1\right )} e^{x} + x + \frac {3}{4} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^4+16*x^3+36*x^2+18*x)*exp(x)^2+1/4*(x^2+5*x+3)*exp(x)+1,x, algorithm="maxima")

[Out]

1/4*(x^4 + 6*x^3 + 9*x^2)*e^(2*x) + 1/4*(x^2 - 2*x + 2)*e^x + 5/4*(x - 1)*e^x + x + 3/4*e^x

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mupad [B]  time = 0.06, size = 37, normalized size = 1.68 \begin {gather*} \frac {x\,\left (3\,{\mathrm {e}}^x+9\,x\,{\mathrm {e}}^{2\,x}+6\,x^2\,{\mathrm {e}}^{2\,x}+x^3\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x+4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(5*x + x^2 + 3))/4 + (exp(2*x)*(18*x + 36*x^2 + 16*x^3 + 2*x^4))/4 + 1,x)

[Out]

(x*(3*exp(x) + 9*x*exp(2*x) + 6*x^2*exp(2*x) + x^3*exp(2*x) + x*exp(x) + 4))/4

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sympy [A]  time = 0.13, size = 36, normalized size = 1.64 \begin {gather*} x + \frac {\left (4 x^{2} + 12 x\right ) e^{x}}{16} + \frac {\left (4 x^{4} + 24 x^{3} + 36 x^{2}\right ) e^{2 x}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x**4+16*x**3+36*x**2+18*x)*exp(x)**2+1/4*(x**2+5*x+3)*exp(x)+1,x)

[Out]

x + (4*x**2 + 12*x)*exp(x)/16 + (4*x**4 + 24*x**3 + 36*x**2)*exp(2*x)/16

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