\(\int \frac {e^{10} (246-250 e^2)-x^3-250 e^{10} \log (x^2)}{x^3} \, dx\) [6537]

Optimal result

Integrand size = 30, antiderivative size = 23 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=-x+\frac {e^{10} \left (2+125 \left (e^2+\log \left (x^2\right )\right )\right )}{x^2} \]

[Out]

exp(5)^2*(2+125*exp(1)^2+125*ln(x^2))/x^2-x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2341} \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=-\frac {e^{10} \left (123-125 e^2\right )}{x^2}+\frac {125 e^{10}}{x^2}+\frac {125 e^{10} \log \left (x^2\right )}{x^2}-x \]

[In]

Int[(E^10*(246 - 250*E^2) - x^3 - 250*E^10*Log[x^2])/x^3,x]

[Out]

(125*E^10)/x^2 - (E^10*(123 - 125*E^2))/x^2 - x + (125*E^10*Log[x^2])/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 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 (\frac {246 e^{10}-250 e^{12}-x^3}{x^3}-\frac {250 e^{10} \log \left (x^2\right )}{x^3}\right ) \, dx \\ & = -\left (\left (250 e^{10}\right ) \int \frac {\log \left (x^2\right )}{x^3} \, dx\right )+\int \frac {246 e^{10}-250 e^{12}-x^3}{x^3} \, dx \\ & = \frac {125 e^{10}}{x^2}+\frac {125 e^{10} \log \left (x^2\right )}{x^2}+\int \left (-1+\frac {2 e^{10} \left (123-125 e^2\right )}{x^3}\right ) \, dx \\ & = \frac {125 e^{10}}{x^2}-\frac {e^{10} \left (123-125 e^2\right )}{x^2}-x+\frac {125 e^{10} \log \left (x^2\right )}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=\frac {2 e^{10}}{x^2}+\frac {125 e^{12}}{x^2}-x+\frac {125 e^{10} \log \left (x^2\right )}{x^2} \]

[In]

Integrate[(E^10*(246 - 250*E^2) - x^3 - 250*E^10*Log[x^2])/x^3,x]

[Out]

(2*E^10)/x^2 + (125*E^12)/x^2 - x + (125*E^10*Log[x^2])/x^2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {125 \,{\mathrm e}^{10} \ln \left (x^{2}\right )}{x^{2}}+\frac {125 \,{\mathrm e}^{12}+2 \,{\mathrm e}^{10}-x^{3}}{x^{2}}\)	\(31\)
parallelrisch	\(\frac {250 \,{\mathrm e}^{10} {\mathrm e}^{2}+250 \,{\mathrm e}^{10} \ln \left (x^{2}\right )-2 x^{3}+4 \,{\mathrm e}^{10}}{2 x^{2}}\)	\(38\)
default	\(-x +\frac {250 \,{\mathrm e}^{12}-246 \,{\mathrm e}^{10}}{2 x^{2}}+\frac {125 \,{\mathrm e}^{10} \ln \left (x^{2}\right )}{x^{2}}+\frac {125 \,{\mathrm e}^{10}}{x^{2}}\)	\(41\)
parts	\(-x +\frac {250 \,{\mathrm e}^{12}-246 \,{\mathrm e}^{10}}{2 x^{2}}+\frac {125 \,{\mathrm e}^{10} \ln \left (x^{2}\right )}{x^{2}}+\frac {125 \,{\mathrm e}^{10}}{x^{2}}\)	\(41\)

[In]

int((-250*exp(5)^2*ln(x^2)+(-250*exp(1)^2+246)*exp(5)^2-x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

125*exp(10)/x^2*ln(x^2)+(125*exp(12)+2*exp(10)-x^3)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=-\frac {x^{3} - 125 \, e^{10} \log \left (x^{2}\right ) - 125 \, e^{12} - 2 \, e^{10}}{x^{2}} \]

[In]

integrate((-250*exp(5)^2*log(x^2)+(-250*exp(1)^2+246)*exp(5)^2-x^3)/x^3,x, algorithm="fricas")

[Out]

-(x^3 - 125*e^10*log(x^2) - 125*e^12 - 2*e^10)/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=- x + \frac {125 e^{10} \log {\left (x^{2} \right )}}{x^{2}} - \frac {- 125 e^{12} - 2 e^{10}}{x^{2}} \]

[In]

integrate((-250*exp(5)**2*ln(x**2)+(-250*exp(1)**2+246)*exp(5)**2-x**3)/x**3,x)

[Out]

-x + 125*exp(10)*log(x**2)/x**2 - (-125*exp(12) - 2*exp(10))/x**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=125 \, {\left (\frac {\log \left (x^{2}\right )}{x^{2}} + \frac {1}{x^{2}}\right )} e^{10} - x + \frac {125 \, e^{12}}{x^{2}} - \frac {123 \, e^{10}}{x^{2}} \]

[In]

integrate((-250*exp(5)^2*log(x^2)+(-250*exp(1)^2+246)*exp(5)^2-x^3)/x^3,x, algorithm="maxima")

[Out]

125*(log(x^2)/x^2 + 1/x^2)*e^10 - x + 125*e^12/x^2 - 123*e^10/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=-\frac {x^{3} - 125 \, e^{10} \log \left (x^{2}\right ) - 125 \, e^{12} - 2 \, e^{10}}{x^{2}} \]

[In]

integrate((-250*exp(5)^2*log(x^2)+(-250*exp(1)^2+246)*exp(5)^2-x^3)/x^3,x, algorithm="giac")

[Out]

-(x^3 - 125*e^10*log(x^2) - 125*e^12 - 2*e^10)/x^2

Mupad [B] (verification not implemented)

Time = 12.83 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{10} \left (246-250 e^2\right )-x^3-250 e^{10} \log \left (x^2\right )}{x^3} \, dx=\frac {125\,\ln \left (x^2\right )\,{\mathrm {e}}^{10}+{\mathrm {e}}^{10}\,\left (125\,{\mathrm {e}}^2+2\right )}{x^2}-x \]

[In]

int(-(250*log(x^2)*exp(10) + x^3 + exp(10)*(250*exp(2) - 246))/x^3,x)

[Out]

(125*log(x^2)*exp(10) + exp(10)*(125*exp(2) + 2))/x^2 - x