3.64.57 \(\int \frac {e^x x^3+(-15-3 x) \log ^2(x)+(10+x) \log ^3(x)}{x^3} \, dx\)

Optimal. Leaf size=16 \[ e^x-\frac {(5+x) \log ^3(x)}{x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 22, normalized size of antiderivative = 1.38, number of steps used = 19, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {14, 2194, 6742, 2353, 2305, 2304} \begin {gather*} -\frac {5 \log ^3(x)}{x^2}+e^x-\frac {\log ^3(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*x^3 + (-15 - 3*x)*Log[x]^2 + (10 + x)*Log[x]^3)/x^3,x]

[Out]

E^x - (5*Log[x]^3)/x^2 - Log[x]^3/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 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
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x+\frac {\log ^2(x) (-15-3 x+10 \log (x)+x \log (x))}{x^3}\right ) \, dx\\ &=\int e^x \, dx+\int \frac {\log ^2(x) (-15-3 x+10 \log (x)+x \log (x))}{x^3} \, dx\\ &=e^x+\int \left (-\frac {3 (5+x) \log ^2(x)}{x^3}+\frac {(10+x) \log ^3(x)}{x^3}\right ) \, dx\\ &=e^x-3 \int \frac {(5+x) \log ^2(x)}{x^3} \, dx+\int \frac {(10+x) \log ^3(x)}{x^3} \, dx\\ &=e^x-3 \int \left (\frac {5 \log ^2(x)}{x^3}+\frac {\log ^2(x)}{x^2}\right ) \, dx+\int \left (\frac {10 \log ^3(x)}{x^3}+\frac {\log ^3(x)}{x^2}\right ) \, dx\\ &=e^x-3 \int \frac {\log ^2(x)}{x^2} \, dx+10 \int \frac {\log ^3(x)}{x^3} \, dx-15 \int \frac {\log ^2(x)}{x^3} \, dx+\int \frac {\log ^3(x)}{x^2} \, dx\\ &=e^x+\frac {15 \log ^2(x)}{2 x^2}+\frac {3 \log ^2(x)}{x}-\frac {5 \log ^3(x)}{x^2}-\frac {\log ^3(x)}{x}+3 \int \frac {\log ^2(x)}{x^2} \, dx-6 \int \frac {\log (x)}{x^2} \, dx-15 \int \frac {\log (x)}{x^3} \, dx+15 \int \frac {\log ^2(x)}{x^3} \, dx\\ &=e^x+\frac {15}{4 x^2}+\frac {6}{x}+\frac {15 \log (x)}{2 x^2}+\frac {6 \log (x)}{x}-\frac {5 \log ^3(x)}{x^2}-\frac {\log ^3(x)}{x}+6 \int \frac {\log (x)}{x^2} \, dx+15 \int \frac {\log (x)}{x^3} \, dx\\ &=e^x-\frac {5 \log ^3(x)}{x^2}-\frac {\log ^3(x)}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 22, normalized size = 1.38 \begin {gather*} e^x-\frac {5 \log ^3(x)}{x^2}-\frac {\log ^3(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x^3 + (-15 - 3*x)*Log[x]^2 + (10 + x)*Log[x]^3)/x^3,x]

[Out]

E^x - (5*Log[x]^3)/x^2 - Log[x]^3/x

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 21, normalized size = 1.31 \begin {gather*} -\frac {{\left (x + 5\right )} \log \relax (x)^{3} - x^{2} e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(x)^3+(-3*x-15)*log(x)^2+exp(x)*x^3)/x^3,x, algorithm="fricas")

[Out]

-((x + 5)*log(x)^3 - x^2*e^x)/x^2

________________________________________________________________________________________

giac [A]  time = 0.16, size = 25, normalized size = 1.56 \begin {gather*} -\frac {x \log \relax (x)^{3} - x^{2} e^{x} + 5 \, \log \relax (x)^{3}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(x)^3+(-3*x-15)*log(x)^2+exp(x)*x^3)/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 16, normalized size = 1.00

method	result	size
risch	\(-\frac {\ln \relax (x )^{3} \left (5+x \right )}{x^{2}}+{\mathrm e}^{x}\)	\(16\)
default	\(-\frac {\ln \relax (x )^{3}}{x}-\frac {5 \ln \relax (x )^{3}}{x^{2}}+{\mathrm e}^{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+10)*ln(x)^3+(-3*x-15)*ln(x)^2+exp(x)*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.37, size = 79, normalized size = 4.94 \begin {gather*} -\frac {\log \relax (x)^{3} + 3 \, \log \relax (x)^{2} + 6 \, \log \relax (x) + 6}{x} + \frac {3 \, {\left (\log \relax (x)^{2} + 2 \, \log \relax (x) + 2\right )}}{x} - \frac {5 \, {\left (4 \, \log \relax (x)^{3} + 6 \, \log \relax (x)^{2} + 6 \, \log \relax (x) + 3\right )}}{4 \, x^{2}} + \frac {15 \, {\left (2 \, \log \relax (x)^{2} + 2 \, \log \relax (x) + 1\right )}}{4 \, x^{2}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(x)^3+(-3*x-15)*log(x)^2+exp(x)*x^3)/x^3,x, algorithm="maxima")

[Out]

-(log(x)^3 + 3*log(x)^2 + 6*log(x) + 6)/x + 3*(log(x)^2 + 2*log(x) + 2)/x - 5/4*(4*log(x)^3 + 6*log(x)^2 + 6*l
og(x) + 3)/x^2 + 15/4*(2*log(x)^2 + 2*log(x) + 1)/x^2 + e^x

________________________________________________________________________________________

mupad [B]  time = 4.25, size = 21, normalized size = 1.31 \begin {gather*} {\mathrm {e}}^x-\frac {x\,{\ln \relax (x)}^3+5\,{\ln \relax (x)}^3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*exp(x) + log(x)^3*(x + 10) - log(x)^2*(3*x + 15))/x^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 15, normalized size = 0.94 \begin {gather*} e^{x} + \frac {\left (- x - 5\right ) \log {\relax (x )}^{3}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*ln(x)**3+(-3*x-15)*ln(x)**2+exp(x)*x**3)/x**3,x)

[Out]

exp(x) + (-x - 5)*log(x)**3/x**2

________________________________________________________________________________________