3.91.53 \(\int \frac {5+20 x+x^2+8 e^2 x^2+12 x^3+2 x^2 \log (x)}{4 x} \, dx\)

Optimal. Leaf size=17 \[ \left (5+x^2\right ) \left (e^2+x+\frac {\log (x)}{4}\right ) \]

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Rubi [B]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 2.53, number of steps used = 7, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6, 12, 14, 2304} \begin {gather*} x^3+\frac {1}{8} \left (1+8 e^2\right ) x^2-\frac {x^2}{8}+\frac {1}{4} x^2 \log (x)+5 x+\frac {5 \log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 20*x + x^2 + 8*E^2*x^2 + 12*x^3 + 2*x^2*Log[x])/(4*x),x]

[Out]

5*x - x^2/8 + ((1 + 8*E^2)*x^2)/8 + x^3 + (5*Log[x])/4 + (x^2*Log[x])/4

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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} &=\int \frac {5+20 x+\left (1+8 e^2\right ) x^2+12 x^3+2 x^2 \log (x)}{4 x} \, dx\\ &=\frac {1}{4} \int \frac {5+20 x+\left (1+8 e^2\right ) x^2+12 x^3+2 x^2 \log (x)}{x} \, dx\\ &=\frac {1}{4} \int \left (\frac {5+20 x+\left (1+8 e^2\right ) x^2+12 x^3}{x}+2 x \log (x)\right ) \, dx\\ &=\frac {1}{4} \int \frac {5+20 x+\left (1+8 e^2\right ) x^2+12 x^3}{x} \, dx+\frac {1}{2} \int x \log (x) \, dx\\ &=-\frac {x^2}{8}+\frac {1}{4} x^2 \log (x)+\frac {1}{4} \int \left (20+\frac {5}{x}+\left (1+8 e^2\right ) x+12 x^2\right ) \, dx\\ &=5 x-\frac {x^2}{8}+\frac {1}{8} \left (1+8 e^2\right ) x^2+x^3+\frac {5 \log (x)}{4}+\frac {1}{4} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.71 \begin {gather*} 5 x+e^2 x^2+x^3+\frac {5 \log (x)}{4}+\frac {1}{4} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 20*x + x^2 + 8*E^2*x^2 + 12*x^3 + 2*x^2*Log[x])/(4*x),x]

[Out]

5*x + E^2*x^2 + x^3 + (5*Log[x])/4 + (x^2*Log[x])/4

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fricas [A]  time = 0.52, size = 22, normalized size = 1.29 \begin {gather*} x^{3} + x^{2} e^{2} + \frac {1}{4} \, {\left (x^{2} + 5\right )} \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^2*log(x)+8*x^2*exp(2)+12*x^3+x^2+20*x+5)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 24, normalized size = 1.41 \begin {gather*} x^{3} + x^{2} e^{2} + \frac {1}{4} \, x^{2} \log \relax (x) + 5 \, x + \frac {5}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^2*log(x)+8*x^2*exp(2)+12*x^3+x^2+20*x+5)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 1.47

method	result	size
default	\(\frac {x^{2} \ln \relax (x )}{4}+x^{2} {\mathrm e}^{2}+x^{3}+5 x +\frac {5 \ln \relax (x )}{4}\)	\(25\)
norman	\(\frac {x^{2} \ln \relax (x )}{4}+x^{2} {\mathrm e}^{2}+x^{3}+5 x +\frac {5 \ln \relax (x )}{4}\)	\(25\)
risch	\(\frac {x^{2} \ln \relax (x )}{4}+x^{2} {\mathrm e}^{2}+x^{3}+5 x +\frac {5 \ln \relax (x )}{4}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(2*x^2*ln(x)+8*x^2*exp(2)+12*x^3+x^2+20*x+5)/x,x,method=_RETURNVERBOSE)

[Out]

1/4*x^2*ln(x)+x^2*exp(2)+x^3+5*x+5/4*ln(x)

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maxima [A]  time = 0.35, size = 24, normalized size = 1.41 \begin {gather*} x^{3} + x^{2} e^{2} + \frac {1}{4} \, x^{2} \log \relax (x) + 5 \, x + \frac {5}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x^2*log(x)+8*x^2*exp(2)+12*x^3+x^2+20*x+5)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.18, size = 24, normalized size = 1.41 \begin {gather*} 5\,x+\frac {5\,\ln \relax (x)}{4}+\frac {x^2\,\ln \relax (x)}{4}+x^2\,{\mathrm {e}}^2+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + (x^2*log(x))/2 + 2*x^2*exp(2) + x^2/4 + 3*x^3 + 5/4)/x,x)

[Out]

5*x + (5*log(x))/4 + (x^2*log(x))/4 + x^2*exp(2) + x^3

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sympy [A]  time = 0.13, size = 27, normalized size = 1.59 \begin {gather*} x^{3} + \frac {x^{2} \log {\relax (x )}}{4} + x^{2} e^{2} + 5 x + \frac {5 \log {\relax (x )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x**2*ln(x)+8*x**2*exp(2)+12*x**3+x**2+20*x+5)/x,x)

[Out]

x**3 + x**2*log(x)/4 + x**2*exp(2) + 5*x + 5*log(x)/4

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