3.100.60 \(\int \frac {35 x+8 e^2 x+7 x^2+26 x^3+(-5 x-e^2 x-2 x^2-3 x^3) \log (x)+e^3 x (5-17 x+2 x \log (x))}{x} \, dx\)

Optimal. Leaf size=29 \[ \left (-5-e^2-x+e^3 x-x^2\right ) (5+x (-9+\log (x))) \]

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Rubi [B]  time = 0.10, antiderivative size = 85, normalized size of antiderivative = 2.93, number of steps used = 8, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 14, 2356, 2295, 2304} \begin {gather*} 9 x^3+x^3 (-\log (x))+\frac {1}{2} \left (1-e^3\right ) x^2+\frac {1}{2} \left (7-17 e^3\right ) x^2-\left (1-e^3\right ) x^2 \log (x)+\left (35+e^2 (8+5 e)\right ) x+\left (5+e^2\right ) x-\left (5+e^2\right ) x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(35*x + 8*E^2*x + 7*x^2 + 26*x^3 + (-5*x - E^2*x - 2*x^2 - 3*x^3)*Log[x] + E^3*x*(5 - 17*x + 2*x*Log[x]))/
x,x]

[Out]

(5 + E^2)*x + (35 + E^2*(8 + 5*E))*x + ((7 - 17*E^3)*x^2)/2 + ((1 - E^3)*x^2)/2 + 9*x^3 - (5 + E^2)*x*Log[x] -
(1 - E^3)*x^2*Log[x] - x^3*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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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
]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (35+8 e^2\right ) x+7 x^2+26 x^3+\left (-5 x-e^2 x-2 x^2-3 x^3\right ) \log (x)+e^3 x (5-17 x+2 x \log (x))}{x} \, dx\\ &=\int \left (35 \left (1+\frac {1}{35} e^2 (8+5 e)\right )+7 \left (1-\frac {17 e^3}{7}\right ) x+26 x^2+\left (-5-e^2-2 \left (1-e^3\right ) x-3 x^2\right ) \log (x)\right ) \, dx\\ &=\left (35+e^2 (8+5 e)\right ) x+\frac {1}{2} \left (7-17 e^3\right ) x^2+\frac {26 x^3}{3}+\int \left (-5-e^2-2 \left (1-e^3\right ) x-3 x^2\right ) \log (x) \, dx\\ &=\left (35+e^2 (8+5 e)\right ) x+\frac {1}{2} \left (7-17 e^3\right ) x^2+\frac {26 x^3}{3}+\int \left (-5 \left (1+\frac {e^2}{5}\right ) \log (x)+2 (-1+e) \left (1+e+e^2\right ) x \log (x)-3 x^2 \log (x)\right ) \, dx\\ &=\left (35+e^2 (8+5 e)\right ) x+\frac {1}{2} \left (7-17 e^3\right ) x^2+\frac {26 x^3}{3}-3 \int x^2 \log (x) \, dx-\left (5+e^2\right ) \int \log (x) \, dx+\left (2 (-1+e) \left (1+e+e^2\right )\right ) \int x \log (x) \, dx\\ &=\left (5+e^2\right ) x+\left (35+e^2 (8+5 e)\right ) x+\frac {1}{2} (1-e) \left (1+e+e^2\right ) x^2+\frac {1}{2} \left (7-17 e^3\right ) x^2+9 x^3-\left (5+e^2\right ) x \log (x)-(1-e) \left (1+e+e^2\right ) x^2 \log (x)-x^3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 70, normalized size = 2.41 \begin {gather*} 40 x+9 e^2 x+5 e^3 x+4 x^2-9 e^3 x^2+9 x^3-5 x \log (x)-e^2 x \log (x)-x^2 \log (x)+e^3 x^2 \log (x)-x^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(35*x + 8*E^2*x + 7*x^2 + 26*x^3 + (-5*x - E^2*x - 2*x^2 - 3*x^3)*Log[x] + E^3*x*(5 - 17*x + 2*x*Log
[x]))/x,x]

[Out]

40*x + 9*E^2*x + 5*E^3*x + 4*x^2 - 9*E^3*x^2 + 9*x^3 - 5*x*Log[x] - E^2*x*Log[x] - x^2*Log[x] + E^3*x^2*Log[x]
- x^3*Log[x]

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fricas [B]  time = 0.85, size = 57, normalized size = 1.97 \begin {gather*} 9 \, x^{3} + 4 \, x^{2} - {\left (9 \, x^{2} - 5 \, x\right )} e^{3} + 9 \, x e^{2} - {\left (x^{3} - x^{2} e^{3} + x^{2} + x e^{2} + 5 \, x\right )} \log \relax (x) + 40 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)-17*x+5)*exp(3+log(x))+(-exp(2)*x-3*x^3-2*x^2-5*x)*log(x)+8*exp(2)*x+26*x^3+7*x^2+35*x)/
x,x, algorithm="fricas")

[Out]

9*x^3 + 4*x^2 - (9*x^2 - 5*x)*e^3 + 9*x*e^2 - (x^3 - x^2*e^3 + x^2 + x*e^2 + 5*x)*log(x) + 40*x

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giac [B]  time = 0.13, size = 65, normalized size = 2.24 \begin {gather*} -x^{3} \log \relax (x) + x^{2} e^{3} \log \relax (x) + 9 \, x^{3} - 9 \, x^{2} e^{3} - x^{2} \log \relax (x) - x e^{2} \log \relax (x) + 4 \, x^{2} + 5 \, x e^{3} + 9 \, x e^{2} - 5 \, x \log \relax (x) + 40 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)-17*x+5)*exp(3+log(x))+(-exp(2)*x-3*x^3-2*x^2-5*x)*log(x)+8*exp(2)*x+26*x^3+7*x^2+35*x)/
x,x, algorithm="giac")

[Out]

-x^3*log(x) + x^2*e^3*log(x) + 9*x^3 - 9*x^2*e^3 - x^2*log(x) - x*e^2*log(x) + 4*x^2 + 5*x*e^3 + 9*x*e^2 - 5*x
*log(x) + 40*x

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maple [A]  time = 0.04, size = 56, normalized size = 1.93

method	result	size
norman	\(\left (4-9 \,{\mathrm e}^{3}\right ) x^{2}+\left (40+9 \,{\mathrm e}^{2}+5 \,{\mathrm e}^{3}\right ) x +\left (-{\mathrm e}^{2}-5\right ) x \ln \relax (x )+\left ({\mathrm e}^{3}-1\right ) x^{2} \ln \relax (x )+9 x^{3}-x^{3} \ln \relax (x )\)	\(56\)
risch	\(\left (x^{2} {\mathrm e}^{3}-{\mathrm e}^{2} x -x^{3}-x^{2}-5 x \right ) \ln \relax (x )-9 x^{2} {\mathrm e}^{3}+5 x \,{\mathrm e}^{3}+9 \,{\mathrm e}^{2} x +9 x^{3}+4 x^{2}+40 x\)	\(60\)
default	\(40 x +{\mathrm e}^{3} \ln \relax (x ) x^{2}-9 x^{2} {\mathrm e}^{3}+5 x \,{\mathrm e}^{3}-x^{3} \ln \relax (x )+9 x^{3}-x \,{\mathrm e}^{2} \ln \relax (x )+9 \,{\mathrm e}^{2} x -x^{2} \ln \relax (x )+4 x^{2}-5 x \ln \relax (x )\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*ln(x)-17*x+5)*exp(3+ln(x))+(-exp(2)*x-3*x^3-2*x^2-5*x)*ln(x)+8*exp(2)*x+26*x^3+7*x^2+35*x)/x,x,metho
d=_RETURNVERBOSE)

[Out]

(4-9*exp(3))*x^2+(40+9*exp(2)+5*exp(3))*x+(-exp(2)-5)*x*ln(x)+(exp(3)-1)*x^2*ln(x)+9*x^3-x^3*ln(x)

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maxima [B]  time = 0.36, size = 79, normalized size = 2.72 \begin {gather*} -x^{3} \log \relax (x) + 9 \, x^{3} - \frac {17}{2} \, x^{2} e^{3} - x^{2} \log \relax (x) + 4 \, x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{3} + 5 \, x e^{3} - {\left (x \log \relax (x) - x\right )} e^{2} + 8 \, x e^{2} - 5 \, x \log \relax (x) + 40 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*log(x)-17*x+5)*exp(3+log(x))+(-exp(2)*x-3*x^3-2*x^2-5*x)*log(x)+8*exp(2)*x+26*x^3+7*x^2+35*x)/
x,x, algorithm="maxima")

[Out]

-x^3*log(x) + 9*x^3 - 17/2*x^2*e^3 - x^2*log(x) + 4*x^2 + 1/2*(2*x^2*log(x) - x^2)*e^3 + 5*x*e^3 - (x*log(x) -
x)*e^2 + 8*x*e^2 - 5*x*log(x) + 40*x

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mupad [B]  time = 8.54, size = 47, normalized size = 1.62 \begin {gather*} x^2\,\left (\ln \relax (x)\,\left ({\mathrm {e}}^3-1\right )-9\,{\mathrm {e}}^3+4\right )-x^3\,\left (\ln \relax (x)-9\right )+x\,\left (9\,{\mathrm {e}}^2+5\,{\mathrm {e}}^3-\ln \relax (x)\,\left ({\mathrm {e}}^2+5\right )+40\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((35*x - log(x)*(5*x + x*exp(2) + 2*x^2 + 3*x^3) + 8*x*exp(2) + 7*x^2 + 26*x^3 + exp(log(x) + 3)*(2*x*log(x
) - 17*x + 5))/x,x)

[Out]

x^2*(log(x)*(exp(3) - 1) - 9*exp(3) + 4) - x^3*(log(x) - 9) + x*(9*exp(2) + 5*exp(3) - log(x)*(exp(2) + 5) + 4
0)

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sympy [B]  time = 0.15, size = 53, normalized size = 1.83 \begin {gather*} 9 x^{3} + x^{2} \left (4 - 9 e^{3}\right ) + x \left (40 + 9 e^{2} + 5 e^{3}\right ) + \left (- x^{3} - x^{2} + x^{2} e^{3} - x e^{2} - 5 x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*ln(x)-17*x+5)*exp(3+ln(x))+(-exp(2)*x-3*x**3-2*x**2-5*x)*ln(x)+8*exp(2)*x+26*x**3+7*x**2+35*x)
/x,x)

[Out]

9*x**3 + x**2*(4 - 9*exp(3)) + x*(40 + 9*exp(2) + 5*exp(3)) + (-x**3 - x**2 + x**2*exp(3) - x*exp(2) - 5*x)*lo
g(x)

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