3.49.53 \(\int (1+3 x+4 x^3+e^3 (12 x+16 x^3)+(2 x+8 e^3 x) \log (x)) \, dx\)

Optimal. Leaf size=20 \[ 3+x+x \left (x+4 e^3 x\right ) \left (1+x^2+\log (x)\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 2.75, number of steps used = 5, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 2304} \begin {gather*} 4 e^3 x^4+x^4-\frac {1}{2} \left (1+4 e^3\right ) x^2+6 e^3 x^2+\frac {3 x^2}{2}+\left (1+4 e^3\right ) x^2 \log (x)+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + 3*x + 4*x^3 + E^3*(12*x + 16*x^3) + (2*x + 8*E^3*x)*Log[x],x]

[Out]

x + (3*x^2)/2 + 6*E^3*x^2 - ((1 + 4*E^3)*x^2)/2 + x^4 + 4*E^3*x^4 + (1 + 4*E^3)*x^2*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x+\frac {3 x^2}{2}+x^4+e^3 \int \left (12 x+16 x^3\right ) \, dx+\int \left (2 x+8 e^3 x\right ) \log (x) \, dx\\ &=x+\frac {3 x^2}{2}+6 e^3 x^2+x^4+4 e^3 x^4+\int \left (2+8 e^3\right ) x \log (x) \, dx\\ &=x+\frac {3 x^2}{2}+6 e^3 x^2+x^4+4 e^3 x^4+\left (2 \left (1+4 e^3\right )\right ) \int x \log (x) \, dx\\ &=x+\frac {3 x^2}{2}+6 e^3 x^2-\frac {1}{2} \left (1+4 e^3\right ) x^2+x^4+4 e^3 x^4+\left (1+4 e^3\right ) x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 2.00 \begin {gather*} x+x^2+4 e^3 x^2+x^4+4 e^3 x^4+x^2 \log (x)+4 e^3 x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + 3*x + 4*x^3 + E^3*(12*x + 16*x^3) + (2*x + 8*E^3*x)*Log[x],x]

[Out]

x + x^2 + 4*E^3*x^2 + x^4 + 4*E^3*x^4 + x^2*Log[x] + 4*E^3*x^2*Log[x]

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fricas [A]  time = 0.68, size = 33, normalized size = 1.65 \begin {gather*} x^{4} + x^{2} + 4 \, {\left (x^{4} + x^{2}\right )} e^{3} + {\left (4 \, x^{2} e^{3} + x^{2}\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*exp(3)+2*x)*log(x)+(16*x^3+12*x)*exp(3)+4*x^3+3*x+1,x, algorithm="fricas")

[Out]

x^4 + x^2 + 4*(x^4 + x^2)*e^3 + (4*x^2*e^3 + x^2)*log(x) + x

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giac [B]  time = 0.21, size = 45, normalized size = 2.25 \begin {gather*} x^{4} + 4 \, x^{2} e^{3} \log \relax (x) - 2 \, x^{2} e^{3} + x^{2} \log \relax (x) + x^{2} + 2 \, {\left (2 \, x^{4} + 3 \, x^{2}\right )} e^{3} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*exp(3)+2*x)*log(x)+(16*x^3+12*x)*exp(3)+4*x^3+3*x+1,x, algorithm="giac")

[Out]

x^4 + 4*x^2*e^3*log(x) - 2*x^2*e^3 + x^2*log(x) + x^2 + 2*(2*x^4 + 3*x^2)*e^3 + x

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maple [A]  time = 0.04, size = 35, normalized size = 1.75

method	result	size
norman	\(x +\left (4 \,{\mathrm e}^{3}+1\right ) x^{2}+\left (4 \,{\mathrm e}^{3}+1\right ) x^{4}+\left (4 \,{\mathrm e}^{3}+1\right ) x^{2} \ln \relax (x )\)	\(35\)
risch	\(4 x^{4} {\mathrm e}^{3}+4 \,{\mathrm e}^{3} x^{2} \ln \relax (x )+x^{4}+x^{2} \ln \relax (x )+4 x^{2} {\mathrm e}^{3}+x^{2}+x\)	\(38\)
default	\(x +4 \,{\mathrm e}^{3} x^{2} \ln \relax (x )-2 x^{2} {\mathrm e}^{3}+x^{2} \ln \relax (x )+x^{2}+{\mathrm e}^{3} \left (4 x^{4}+6 x^{2}\right )+x^{4}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*exp(3)+2*x)*ln(x)+(16*x^3+12*x)*exp(3)+4*x^3+3*x+1,x,method=_RETURNVERBOSE)

[Out]

x+(4*exp(3)+1)*x^2+(4*exp(3)+1)*x^4+(4*exp(3)+1)*x^2*ln(x)

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maxima [B]  time = 0.35, size = 50, normalized size = 2.50 \begin {gather*} x^{4} - \frac {1}{2} \, x^{2} {\left (4 \, e^{3} + 1\right )} + \frac {3}{2} \, x^{2} + 2 \, {\left (2 \, x^{4} + 3 \, x^{2}\right )} e^{3} + {\left (4 \, x^{2} e^{3} + x^{2}\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*exp(3)+2*x)*log(x)+(16*x^3+12*x)*exp(3)+4*x^3+3*x+1,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.35, size = 31, normalized size = 1.55 \begin {gather*} x+x^4\,\left (4\,{\mathrm {e}}^3+1\right )+x^2\,\left (4\,{\mathrm {e}}^3+\ln \relax (x)\,\left (4\,{\mathrm {e}}^3+1\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*x + exp(3)*(12*x + 16*x^3) + log(x)*(2*x + 8*x*exp(3)) + 4*x^3 + 1,x)

[Out]

x + x^4*(4*exp(3) + 1) + x^2*(4*exp(3) + log(x)*(4*exp(3) + 1) + 1)

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sympy [A]  time = 0.12, size = 36, normalized size = 1.80 \begin {gather*} x^{4} \left (1 + 4 e^{3}\right ) + x^{2} \left (1 + 4 e^{3}\right ) + x + \left (x^{2} + 4 x^{2} e^{3}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*exp(3)+2*x)*ln(x)+(16*x**3+12*x)*exp(3)+4*x**3+3*x+1,x)

[Out]

x**4*(1 + 4*exp(3)) + x**2*(1 + 4*exp(3)) + x + (x**2 + 4*x**2*exp(3))*log(x)

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