\(\int \frac {-10-25 x+5 x^2+5 x^3+e^x (-x^3-x^4)-20 \log (x)}{x^3} \, dx\) [6804]

Optimal result

Integrand size = 38, antiderivative size = 30 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=-e^x x+\frac {5 \left (5+x^2+\frac {\left (2+x^2\right ) (1+\log (x))}{x}\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2207, 2225, 2341} \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=\frac {10}{x^2}+\frac {10 \log (x)}{x^2}+5 x+e^x-e^x (x+1)+\frac {25}{x}+5 \log (x) \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2341

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=-\frac {-10-25 x+\left (-5+e^x\right ) x^3-5 \left (2+x^2\right ) \log (x)}{x^2} \]

[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)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
default	\(-{\mathrm e}^{x} x +5 x +5 \ln \left (x \right )+\frac {25}{x}+\frac {10}{x^{2}}+\frac {10 \ln \left (x \right )}{x^{2}}\)	\(31\)
parts	\(-{\mathrm e}^{x} x +5 x +5 \ln \left (x \right )+\frac {25}{x}+\frac {10}{x^{2}}+\frac {10 \ln \left (x \right )}{x^{2}}\)	\(31\)
norman	\(\frac {10+5 x^{2} \ln \left (x \right )+25 x +5 x^{3}-{\mathrm e}^{x} x^{3}+10 \ln \left (x \right )}{x^{2}}\)	\(33\)
parallelrisch	\(-\frac {{\mathrm e}^{x} x^{3}-5 x^{2} \ln \left (x \right )-5 x^{3}-10-25 x -10 \ln \left (x \right )}{x^{2}}\)	\(33\)
risch	\(\frac {10 \ln \left (x \right )}{x^{2}}+\frac {-{\mathrm e}^{x} x^{3}+5 x^{2} \ln \left (x \right )+5 x^{3}+25 x +10}{x^{2}}\)	\(37\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=-\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, {\left (x^{2} + 2\right )} \log \left (x\right ) - 25 \, x - 10}{x^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=- x e^{x} + 5 x + 5 \log {\left (x \right )} + \frac {25 x + 10}{x^{2}} + \frac {10 \log {\left (x \right )}}{x^{2}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=-{\left (x - 1\right )} e^{x} + 5 \, x + \frac {25}{x} + \frac {10 \, \log \left (x\right )}{x^{2}} + \frac {10}{x^{2}} - e^{x} + 5 \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=-\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, x^{2} \log \left (x\right ) - 25 \, x - 10 \, \log \left (x\right ) - 10}{x^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 12.94 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-10-25 x+5 x^2+5 x^3+e^x \left (-x^3-x^4\right )-20 \log (x)}{x^3} \, dx=5\,\ln \left (x\right )-x\,\left ({\mathrm {e}}^x-5\right )+\frac {25\,x+10\,\ln \left (x\right )+10}{x^2} \]

[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