3.69.4 \(\int \frac {-10-25 x+5 x^2+5 x^3+e^x (-x^3-x^4)-20 \log (x)}{x^3} \, dx\)

Optimal. Leaf size=30 \[ -e^x x+\frac {5 \left (5+x^2+\frac {\left (2+x^2\right ) (1+\log (x))}{x}\right )}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2176, 2194, 2304} \begin {gather*} \frac {10}{x^2}+\frac {10 \log (x)}{x^2}+5 x+e^x-e^x (x+1)+\frac {25}{x}+5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 25*x + 5*x^2 + 5*x^3 + E^x*(-x^3 - x^4) - 20*Log[x])/x^3,x]

[Out]

E^x + 10/x^2 + 25/x + 5*x - E^x*(1 + x) + 5*Log[x] + (10*Log[x])/x^2

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 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 \left (-e^x (1+x)+\frac {5 \left (-2-5 x+x^2+x^3-4 \log (x)\right )}{x^3}\right ) \, dx\\ &=5 \int \frac {-2-5 x+x^2+x^3-4 \log (x)}{x^3} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)+5 \int \left (\frac {-2-5 x+x^2+x^3}{x^3}-\frac {4 \log (x)}{x^3}\right ) \, dx+\int e^x \, dx\\ &=e^x-e^x (1+x)+5 \int \frac {-2-5 x+x^2+x^3}{x^3} \, dx-20 \int \frac {\log (x)}{x^3} \, dx\\ &=e^x+\frac {5}{x^2}-e^x (1+x)+\frac {10 \log (x)}{x^2}+5 \int \left (1-\frac {2}{x^3}-\frac {5}{x^2}+\frac {1}{x}\right ) \, dx\\ &=e^x+\frac {10}{x^2}+\frac {25}{x}+5 x-e^x (1+x)+5 \log (x)+\frac {10 \log (x)}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.93 \begin {gather*} -\frac {-10-25 x+\left (-5+e^x\right ) x^3-5 \left (2+x^2\right ) \log (x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 25*x + 5*x^2 + 5*x^3 + E^x*(-x^3 - x^4) - 20*Log[x])/x^3,x]

[Out]

-((-10 - 25*x + (-5 + E^x)*x^3 - 5*(2 + x^2)*Log[x])/x^2)

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fricas [A]  time = 0.61, size = 30, normalized size = 1.00 \begin {gather*} -\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, {\left (x^{2} + 2\right )} \log \relax (x) - 25 \, x - 10}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="fricas")

[Out]

-(x^3*e^x - 5*x^3 - 5*(x^2 + 2)*log(x) - 25*x - 10)/x^2

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giac [A]  time = 0.19, size = 32, normalized size = 1.07 \begin {gather*} -\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, x^{2} \log \relax (x) - 25 \, x - 10 \, \log \relax (x) - 10}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="giac")

[Out]

-(x^3*e^x - 5*x^3 - 5*x^2*log(x) - 25*x - 10*log(x) - 10)/x^2

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maple [A]  time = 0.04, size = 31, normalized size = 1.03

method	result	size
default	\(-{\mathrm e}^{x} x +5 x +5 \ln \relax (x )+\frac {25}{x}+\frac {10}{x^{2}}+\frac {10 \ln \relax (x )}{x^{2}}\)	\(31\)
norman	\(\frac {10+25 x +5 x^{3}-{\mathrm e}^{x} x^{3}+10 \ln \relax (x )}{x^{2}}+5 \ln \relax (x )\)	\(31\)
risch	\(\frac {10 \ln \relax (x )}{x^{2}}+\frac {-{\mathrm e}^{x} x^{3}+5 x^{2} \ln \relax (x )+5 x^{3}+25 x +10}{x^{2}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*ln(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x,method=_RETURNVERBOSE)

[Out]

-exp(x)*x+5*x+5*ln(x)+25/x+10/x^2+10*ln(x)/x^2

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maxima [A]  time = 0.38, size = 36, normalized size = 1.20 \begin {gather*} -{\left (x - 1\right )} e^{x} + 5 \, x + \frac {25}{x} + \frac {10 \, \log \relax (x)}{x^{2}} + \frac {10}{x^{2}} - e^{x} + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="maxima")

[Out]

-(x - 1)*e^x + 5*x + 25/x + 10*log(x)/x^2 + 10/x^2 - e^x + 5*log(x)

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mupad [B]  time = 4.24, size = 25, normalized size = 0.83 \begin {gather*} 5\,\ln \relax (x)-x\,\left ({\mathrm {e}}^x-5\right )+\frac {25\,x+10\,\ln \relax (x)+10}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x + 20*log(x) + exp(x)*(x^3 + x^4) - 5*x^2 - 5*x^3 + 10)/x^3,x)

[Out]

5*log(x) - x*(exp(x) - 5) + (25*x + 10*log(x) + 10)/x^2

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sympy [A]  time = 0.30, size = 29, normalized size = 0.97 \begin {gather*} - x e^{x} + 5 x + 5 \log {\relax (x )} + \frac {25 x + 10}{x^{2}} + \frac {10 \log {\relax (x )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*ln(x)+(-x**4-x**3)*exp(x)+5*x**3+5*x**2-25*x-10)/x**3,x)

[Out]

-x*exp(x) + 5*x + 5*log(x) + (25*x + 10)/x**2 + 10*log(x)/x**2

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