\(\int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx\) [31]

Optimal result

Integrand size = 36, antiderivative size = 19 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=\frac {9}{x^2}+3 x+e^x \log ^2(5 x) \]

[Out]

3*x+ln(5*x)^2*exp(x)+9/x^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14, 2233} \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=\frac {9}{x^2}+3 x+e^x \log ^2(5 x) \]

[In]

Int[(-18 + 3*x^3 + 2*E^x*x^2*Log[5*x] + E^x*x^3*Log[5*x]^2)/x^3,x]

[Out]

9/x^2 + 3*x + E^x*Log[5*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 2233

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(x_)^(m_.)*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) +
(g_.)*(x_))), x_Symbol] :> Simp[e*x^(m + 1)*F^(c*(a + b*x))*(Log[d*x]^(n + 1)/(n + 1)), x] /; FreeQ[{F, a, b,
c, d, e, f, g, h, m, n}, x] && EqQ[e*(m + 1) - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n,
-1]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 \left (-6+x^3\right )}{x^3}+\frac {e^x \log (5 x) (2+x \log (5 x))}{x}\right ) \, dx \\ & = 3 \int \frac {-6+x^3}{x^3} \, dx+\int \frac {e^x \log (5 x) (2+x \log (5 x))}{x} \, dx \\ & = e^x \log ^2(5 x)+3 \int \left (1-\frac {6}{x^3}\right ) \, dx \\ & = \frac {9}{x^2}+3 x+e^x \log ^2(5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=\frac {9}{x^2}+3 x+e^x \log ^2(5 x) \]

[In]

Integrate[(-18 + 3*x^3 + 2*E^x*x^2*Log[5*x] + E^x*x^3*Log[5*x]^2)/x^3,x]

[Out]

9/x^2 + 3*x + E^x*Log[5*x]^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
default	\(3 x +\ln \left (5 x \right )^{2} {\mathrm e}^{x}+\frac {9}{x^{2}}\)	\(19\)
parts	\(3 x +\ln \left (5 x \right )^{2} {\mathrm e}^{x}+\frac {9}{x^{2}}\)	\(19\)
risch	\(\ln \left (5 x \right )^{2} {\mathrm e}^{x}+\frac {3 x^{3}+9}{x^{2}}\)	\(21\)
parallelrisch	\(-\frac {-x^{2} {\mathrm e}^{x} \ln \left (5 x \right )^{2}-9-3 x^{3}}{x^{2}}\)	\(26\)

[In]

int((x^3*exp(x)*ln(5*x)^2+2*x^2*exp(x)*ln(5*x)+3*x^3-18)/x^3,x,method=_RETURNVERBOSE)

[Out]

3*x+ln(5*x)^2*exp(x)+9/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=\frac {x^{2} e^{x} \log \left (5 \, x\right )^{2} + 3 \, x^{3} + 9}{x^{2}} \]

[In]

integrate((x^3*exp(x)*log(5*x)^2+2*x^2*exp(x)*log(5*x)+3*x^3-18)/x^3,x, algorithm="fricas")

[Out]

(x^2*e^x*log(5*x)^2 + 3*x^3 + 9)/x^2

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=3 x + e^{x} \log {\left (5 x \right )}^{2} + \frac {9}{x^{2}} \]

[In]

integrate((x**3*exp(x)*ln(5*x)**2+2*x**2*exp(x)*ln(5*x)+3*x**3-18)/x**3,x)

[Out]

3*x + exp(x)*log(5*x)**2 + 9/x**2

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx={\left (\log \left (5\right )^{2} + 2 \, \log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2}\right )} e^{x} + 3 \, x + \frac {9}{x^{2}} \]

[In]

integrate((x^3*exp(x)*log(5*x)^2+2*x^2*exp(x)*log(5*x)+3*x^3-18)/x^3,x, algorithm="maxima")

[Out]

(log(5)^2 + 2*log(5)*log(x) + log(x)^2)*e^x + 3*x + 9/x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=\frac {x^{2} e^{x} \log \left (5 \, x\right )^{2} + 3 \, x^{3} + 9}{x^{2}} \]

[In]

integrate((x^3*exp(x)*log(5*x)^2+2*x^2*exp(x)*log(5*x)+3*x^3-18)/x^3,x, algorithm="giac")

[Out]

(x^2*e^x*log(5*x)^2 + 3*x^3 + 9)/x^2

Mupad [B] (verification not implemented)

Time = 7.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-18+3 x^3+2 e^x x^2 \log (5 x)+e^x x^3 \log ^2(5 x)}{x^3} \, dx=3\,x+\frac {9}{x^2}+{\ln \left (5\,x\right )}^2\,{\mathrm {e}}^x \]

[In]

int((3*x^3 + 2*x^2*log(5*x)*exp(x) + x^3*log(5*x)^2*exp(x) - 18)/x^3,x)

[Out]

3*x + 9/x^2 + log(5*x)^2*exp(x)