3.10.27 \(\int \frac {e^3+45 x+252 x^3+(-60 x-192 x^3) \log (x)+(15 x+36 x^3) \log ^2(x)}{e^3 x} \, dx\)

Optimal. Leaf size=31 \[ 4+\frac {3 x^2 \left (-x+5 \left (\frac {1}{x}+x\right )\right ) (3-\log (x))^2}{e^3}+\log (x) \]

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Rubi [B]  time = 0.13, antiderivative size = 73, normalized size of antiderivative = 2.35, number of steps used = 13, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {12, 14, 2313, 2330, 2296, 2295, 2305, 2304} \begin {gather*} \frac {108 x^3}{e^3}+\frac {12 x^3 \log ^2(x)}{e^3}-\frac {8 x^3 \log (x)}{e^3}-\frac {4 \left (16 x^3+15 x\right ) \log (x)}{e^3}+\frac {135 x}{e^3}+\frac {15 x \log ^2(x)}{e^3}-\frac {30 x \log (x)}{e^3}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3 + 45*x + 252*x^3 + (-60*x - 192*x^3)*Log[x] + (15*x + 36*x^3)*Log[x]^2)/(E^3*x),x]

[Out]

(135*x)/E^3 + (108*x^3)/E^3 + Log[x] - (30*x*Log[x])/E^3 - (8*x^3*Log[x])/E^3 - (4*(15*x + 16*x^3)*Log[x])/E^3
+ (15*x*Log[x]^2)/E^3 + (12*x^3*Log[x]^2)/E^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rule 2330

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^3+45 x+252 x^3+\left (-60 x-192 x^3\right ) \log (x)+\left (15 x+36 x^3\right ) \log ^2(x)}{x} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^3+45 x+252 x^3}{x}-12 \left (5+16 x^2\right ) \log (x)+3 \left (5+12 x^2\right ) \log ^2(x)\right ) \, dx}{e^3}\\ &=\frac {\int \frac {e^3+45 x+252 x^3}{x} \, dx}{e^3}+\frac {3 \int \left (5+12 x^2\right ) \log ^2(x) \, dx}{e^3}-\frac {12 \int \left (5+16 x^2\right ) \log (x) \, dx}{e^3}\\ &=-\frac {4 \left (15 x+16 x^3\right ) \log (x)}{e^3}+\frac {\int \left (45+\frac {e^3}{x}+252 x^2\right ) \, dx}{e^3}+\frac {3 \int \left (5 \log ^2(x)+12 x^2 \log ^2(x)\right ) \, dx}{e^3}+\frac {12 \int \left (5+\frac {16 x^2}{3}\right ) \, dx}{e^3}\\ &=\frac {105 x}{e^3}+\frac {316 x^3}{3 e^3}+\log (x)-\frac {4 \left (15 x+16 x^3\right ) \log (x)}{e^3}+\frac {15 \int \log ^2(x) \, dx}{e^3}+\frac {36 \int x^2 \log ^2(x) \, dx}{e^3}\\ &=\frac {105 x}{e^3}+\frac {316 x^3}{3 e^3}+\log (x)-\frac {4 \left (15 x+16 x^3\right ) \log (x)}{e^3}+\frac {15 x \log ^2(x)}{e^3}+\frac {12 x^3 \log ^2(x)}{e^3}-\frac {24 \int x^2 \log (x) \, dx}{e^3}-\frac {30 \int \log (x) \, dx}{e^3}\\ &=\frac {135 x}{e^3}+\frac {108 x^3}{e^3}+\log (x)-\frac {30 x \log (x)}{e^3}-\frac {8 x^3 \log (x)}{e^3}-\frac {4 \left (15 x+16 x^3\right ) \log (x)}{e^3}+\frac {15 x \log ^2(x)}{e^3}+\frac {12 x^3 \log ^2(x)}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 47, normalized size = 1.52 \begin {gather*} \frac {135 x+108 x^3+e^3 \log (x)-90 x \log (x)-72 x^3 \log (x)+15 x \log ^2(x)+12 x^3 \log ^2(x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3 + 45*x + 252*x^3 + (-60*x - 192*x^3)*Log[x] + (15*x + 36*x^3)*Log[x]^2)/(E^3*x),x]

[Out]

(135*x + 108*x^3 + E^3*Log[x] - 90*x*Log[x] - 72*x^3*Log[x] + 15*x*Log[x]^2 + 12*x^3*Log[x]^2)/E^3

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fricas [A]  time = 0.51, size = 44, normalized size = 1.42 \begin {gather*} {\left (108 \, x^{3} + 3 \, {\left (4 \, x^{3} + 5 \, x\right )} \log \relax (x)^{2} - {\left (72 \, x^{3} + 90 \, x - e^{3}\right )} \log \relax (x) + 135 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3+15*x)*log(x)^2+(-192*x^3-60*x)*log(x)+exp(3)+252*x^3+45*x)/x/exp(3),x, algorithm="fricas")

[Out]

(108*x^3 + 3*(4*x^3 + 5*x)*log(x)^2 - (72*x^3 + 90*x - e^3)*log(x) + 135*x)*e^(-3)

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giac [A]  time = 0.32, size = 45, normalized size = 1.45 \begin {gather*} {\left (12 \, x^{3} \log \relax (x)^{2} - 72 \, x^{3} \log \relax (x) + 108 \, x^{3} + 15 \, x \log \relax (x)^{2} - 90 \, x \log \relax (x) + e^{3} \log \relax (x) + 135 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3+15*x)*log(x)^2+(-192*x^3-60*x)*log(x)+exp(3)+252*x^3+45*x)/x/exp(3),x, algorithm="giac")

[Out]

(12*x^3*log(x)^2 - 72*x^3*log(x) + 108*x^3 + 15*x*log(x)^2 - 90*x*log(x) + e^3*log(x) + 135*x)*e^(-3)

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maple [A]  time = 0.04, size = 46, normalized size = 1.48

method	result	size
risch	\({\mathrm e}^{-3} \left (12 x^{3}+15 x \right ) \ln \relax (x )^{2}+{\mathrm e}^{-3} \left (-72 x^{3}-90 x \right ) \ln \relax (x )+108 \,{\mathrm e}^{-3} x^{3}+135 x \,{\mathrm e}^{-3}+\ln \relax (x )\)	\(46\)
default	\({\mathrm e}^{-3} \left (12 x^{3} \ln \relax (x )^{2}-72 x^{3} \ln \relax (x )+108 x^{3}+15 x \ln \relax (x )^{2}-90 x \ln \relax (x )+135 x +\ln \relax (x ) {\mathrm e}^{3}\right )\)	\(48\)
norman	\(\ln \relax (x )+135 x \,{\mathrm e}^{-3}+108 \,{\mathrm e}^{-3} x^{3}-90 x \,{\mathrm e}^{-3} \ln \relax (x )+15 x \,{\mathrm e}^{-3} \ln \relax (x )^{2}-72 \,{\mathrm e}^{-3} x^{3} \ln \relax (x )+12 \,{\mathrm e}^{-3} x^{3} \ln \relax (x )^{2}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x^3+15*x)*ln(x)^2+(-192*x^3-60*x)*ln(x)+exp(3)+252*x^3+45*x)/x/exp(3),x,method=_RETURNVERBOSE)

[Out]

exp(-3)*(12*x^3+15*x)*ln(x)^2+exp(-3)*(-72*x^3-90*x)*ln(x)+108*exp(-3)*x^3+135*x*exp(-3)+ln(x)

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maxima [B]  time = 0.57, size = 61, normalized size = 1.97 \begin {gather*} \frac {1}{3} \, {\left (4 \, {\left (9 \, \log \relax (x)^{2} - 6 \, \log \relax (x) + 2\right )} x^{3} - 192 \, x^{3} \log \relax (x) + 316 \, x^{3} + 45 \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - 180 \, x \log \relax (x) + 3 \, e^{3} \log \relax (x) + 315 \, x\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3+15*x)*log(x)^2+(-192*x^3-60*x)*log(x)+exp(3)+252*x^3+45*x)/x/exp(3),x, algorithm="maxima")

[Out]

1/3*(4*(9*log(x)^2 - 6*log(x) + 2)*x^3 - 192*x^3*log(x) + 316*x^3 + 45*(log(x)^2 - 2*log(x) + 2)*x - 180*x*log
(x) + 3*e^3*log(x) + 315*x)*e^(-3)

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mupad [B]  time = 0.78, size = 51, normalized size = 1.65 \begin {gather*} 12\,{\mathrm {e}}^{-3}\,x^3\,{\ln \relax (x)}^2-72\,{\mathrm {e}}^{-3}\,x^3\,\ln \relax (x)+108\,{\mathrm {e}}^{-3}\,x^3+15\,{\mathrm {e}}^{-3}\,x\,{\ln \relax (x)}^2-90\,{\mathrm {e}}^{-3}\,x\,\ln \relax (x)+135\,{\mathrm {e}}^{-3}\,x+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*(45*x + exp(3) + log(x)^2*(15*x + 36*x^3) - log(x)*(60*x + 192*x^3) + 252*x^3))/x,x)

[Out]

log(x) + 135*x*exp(-3) + 108*x^3*exp(-3) - 90*x*exp(-3)*log(x) + 15*x*exp(-3)*log(x)^2 - 72*x^3*exp(-3)*log(x)
+ 12*x^3*exp(-3)*log(x)^2

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sympy [A]  time = 0.20, size = 51, normalized size = 1.65 \begin {gather*} \frac {\left (- 72 x^{3} - 90 x\right ) \log {\relax (x )}}{e^{3}} + \frac {\left (12 x^{3} + 15 x\right ) \log {\relax (x )}^{2}}{e^{3}} + \frac {108 x^{3} + 135 x + e^{3} \log {\relax (x )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x**3+15*x)*ln(x)**2+(-192*x**3-60*x)*ln(x)+exp(3)+252*x**3+45*x)/x/exp(3),x)

[Out]

(-72*x**3 - 90*x)*exp(-3)*log(x) + (12*x**3 + 15*x)*exp(-3)*log(x)**2 + (108*x**3 + 135*x + exp(3)*log(x))*exp
(-3)

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