3.8.100 \(\int \frac {100+250 x-37 x^2-3 x^3+e^x (-75 x+27 x^2+x^3)+(-25 x-2 x^2) \log (x)}{25 x} \, dx\)

Optimal. Leaf size=23 \[ \left (4-x-\frac {x^2}{25}\right ) \left (-7-e^x+x+\log (x)\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 55, normalized size of antiderivative = 2.39, number of steps used = 17, number of rules used = 6, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 14, 2196, 2194, 2176, 2313} \begin {gather*} -\frac {x^3}{25}+\frac {e^x x^2}{25}-\frac {18 x^2}{25}-\frac {1}{25} \left (x^2+25 x\right ) \log (x)+e^x x+11 x-4 e^x+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 + 250*x - 37*x^2 - 3*x^3 + E^x*(-75*x + 27*x^2 + x^3) + (-25*x - 2*x^2)*Log[x])/(25*x),x]

[Out]

-4*E^x + 11*x + E^x*x - (18*x^2)/25 + (E^x*x^2)/25 - x^3/25 + 4*Log[x] - ((25*x + x^2)*Log[x])/25

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 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

Rule 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {100+250 x-37 x^2-3 x^3+e^x \left (-75 x+27 x^2+x^3\right )+\left (-25 x-2 x^2\right ) \log (x)}{x} \, dx\\ &=\frac {1}{25} \int \left (e^x \left (-75+27 x+x^2\right )+\frac {100+250 x-37 x^2-3 x^3-25 x \log (x)-2 x^2 \log (x)}{x}\right ) \, dx\\ &=\frac {1}{25} \int e^x \left (-75+27 x+x^2\right ) \, dx+\frac {1}{25} \int \frac {100+250 x-37 x^2-3 x^3-25 x \log (x)-2 x^2 \log (x)}{x} \, dx\\ &=\frac {1}{25} \int \left (-75 e^x+27 e^x x+e^x x^2\right ) \, dx+\frac {1}{25} \int \left (\frac {100+250 x-37 x^2-3 x^3}{x}-(25+2 x) \log (x)\right ) \, dx\\ &=\frac {1}{25} \int e^x x^2 \, dx+\frac {1}{25} \int \frac {100+250 x-37 x^2-3 x^3}{x} \, dx-\frac {1}{25} \int (25+2 x) \log (x) \, dx+\frac {27}{25} \int e^x x \, dx-3 \int e^x \, dx\\ &=-3 e^x+\frac {27 e^x x}{25}+\frac {e^x x^2}{25}-\frac {1}{25} \left (25 x+x^2\right ) \log (x)+\frac {1}{25} \int (25+x) \, dx+\frac {1}{25} \int \left (250+\frac {100}{x}-37 x-3 x^2\right ) \, dx-\frac {2}{25} \int e^x x \, dx-\frac {27 \int e^x \, dx}{25}\\ &=-\frac {102 e^x}{25}+11 x+e^x x-\frac {18 x^2}{25}+\frac {e^x x^2}{25}-\frac {x^3}{25}+4 \log (x)-\frac {1}{25} \left (25 x+x^2\right ) \log (x)+\frac {2 \int e^x \, dx}{25}\\ &=-4 e^x+11 x+e^x x-\frac {18 x^2}{25}+\frac {e^x x^2}{25}-\frac {x^3}{25}+4 \log (x)-\frac {1}{25} \left (25 x+x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 42, normalized size = 1.83 \begin {gather*} \frac {1}{25} \left (275 x-18 x^2-x^3+e^x \left (-100+25 x+x^2\right )+100 \log (x)-x (25+x) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 + 250*x - 37*x^2 - 3*x^3 + E^x*(-75*x + 27*x^2 + x^3) + (-25*x - 2*x^2)*Log[x])/(25*x),x]

[Out]

(275*x - 18*x^2 - x^3 + E^x*(-100 + 25*x + x^2) + 100*Log[x] - x*(25 + x)*Log[x])/25

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fricas [A]  time = 0.61, size = 38, normalized size = 1.65 \begin {gather*} -\frac {1}{25} \, x^{3} - \frac {18}{25} \, x^{2} + \frac {1}{25} \, {\left (x^{2} + 25 \, x - 100\right )} e^{x} - \frac {1}{25} \, {\left (x^{2} + 25 \, x - 100\right )} \log \relax (x) + 11 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((-2*x^2-25*x)*log(x)+(x^3+27*x^2-75*x)*exp(x)-3*x^3-37*x^2+250*x+100)/x,x, algorithm="fricas")

[Out]

-1/25*x^3 - 18/25*x^2 + 1/25*(x^2 + 25*x - 100)*e^x - 1/25*(x^2 + 25*x - 100)*log(x) + 11*x

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giac [B]  time = 0.27, size = 45, normalized size = 1.96 \begin {gather*} -\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} e^{x} - \frac {1}{25} \, x^{2} \log \relax (x) - \frac {18}{25} \, x^{2} + x e^{x} - x \log \relax (x) + 11 \, x - 4 \, e^{x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((-2*x^2-25*x)*log(x)+(x^3+27*x^2-75*x)*exp(x)-3*x^3-37*x^2+250*x+100)/x,x, algorithm="giac")

[Out]

-1/25*x^3 + 1/25*x^2*e^x - 1/25*x^2*log(x) - 18/25*x^2 + x*e^x - x*log(x) + 11*x - 4*e^x + 4*log(x)

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maple [B]  time = 0.03, size = 46, normalized size = 2.00

method	result	size
default	\(-\frac {x^{2} \ln \relax (x )}{25}-\frac {18 x^{2}}{25}-x \ln \relax (x )+11 x +\frac {{\mathrm e}^{x} x^{2}}{25}+{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-\frac {x^{3}}{25}+4 \ln \relax (x )\)	\(46\)
norman	\(-\frac {x^{2} \ln \relax (x )}{25}-\frac {18 x^{2}}{25}-x \ln \relax (x )+11 x +\frac {{\mathrm e}^{x} x^{2}}{25}+{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-\frac {x^{3}}{25}+4 \ln \relax (x )\)	\(46\)
risch	\(\frac {\left (-x^{2}-25 x \right ) \ln \relax (x )}{25}-\frac {x^{3}}{25}+\frac {{\mathrm e}^{x} x^{2}}{25}-\frac {18 x^{2}}{25}+{\mathrm e}^{x} x +4 \ln \relax (x )+11 x -4 \,{\mathrm e}^{x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((-2*x^2-25*x)*ln(x)+(x^3+27*x^2-75*x)*exp(x)-3*x^3-37*x^2+250*x+100)/x,x,method=_RETURNVERBOSE)

[Out]

-1/25*x^2*ln(x)-18/25*x^2-x*ln(x)+11*x+1/25*exp(x)*x^2+exp(x)*x-4*exp(x)-1/25*x^3+4*ln(x)

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maxima [B]  time = 0.44, size = 53, normalized size = 2.30 \begin {gather*} -\frac {1}{25} \, x^{3} - \frac {1}{25} \, x^{2} \log \relax (x) - \frac {18}{25} \, x^{2} + \frac {1}{25} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + \frac {27}{25} \, {\left (x - 1\right )} e^{x} - x \log \relax (x) + 11 \, x - 3 \, e^{x} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((-2*x^2-25*x)*log(x)+(x^3+27*x^2-75*x)*exp(x)-3*x^3-37*x^2+250*x+100)/x,x, algorithm="maxima")

[Out]

-1/25*x^3 - 1/25*x^2*log(x) - 18/25*x^2 + 1/25*(x^2 - 2*x + 2)*e^x + 27/25*(x - 1)*e^x - x*log(x) + 11*x - 3*e
^x + 4*log(x)

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mupad [B]  time = 0.67, size = 45, normalized size = 1.96 \begin {gather*} 11\,x-4\,{\mathrm {e}}^x+4\,\ln \relax (x)+\frac {x^2\,{\mathrm {e}}^x}{25}-\frac {x^2\,\ln \relax (x)}{25}+x\,{\mathrm {e}}^x-x\,\ln \relax (x)-\frac {18\,x^2}{25}-\frac {x^3}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + (exp(x)*(27*x^2 - 75*x + x^3))/25 - (log(x)*(25*x + 2*x^2))/25 - (37*x^2)/25 - (3*x^3)/25 + 4)/x,x
)

[Out]

11*x - 4*exp(x) + 4*log(x) + (x^2*exp(x))/25 - (x^2*log(x))/25 + x*exp(x) - x*log(x) - (18*x^2)/25 - x^3/25

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sympy [B]  time = 0.32, size = 44, normalized size = 1.91 \begin {gather*} - \frac {x^{3}}{25} - \frac {18 x^{2}}{25} + 11 x + \left (- \frac {x^{2}}{25} - x\right ) \log {\relax (x )} + \frac {\left (x^{2} + 25 x - 100\right ) e^{x}}{25} + 4 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((-2*x**2-25*x)*ln(x)+(x**3+27*x**2-75*x)*exp(x)-3*x**3-37*x**2+250*x+100)/x,x)

[Out]

-x**3/25 - 18*x**2/25 + 11*x + (-x**2/25 - x)*log(x) + (x**2 + 25*x - 100)*exp(x)/25 + 4*log(x)

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