3.86.92 \(\int e^3 (6 x^2+72 x^3+(4 x+36 x^2) \log (x)+(4 x+54 x^2) \log ^2(x)) \, dx\)

Optimal. Leaf size=19 \[ 2 e^3 x \left (x+9 x^2\right ) \left (x+\log ^2(x)\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 4.05, number of steps used = 14, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 1593, 43, 2334, 2353, 2305, 2304} \begin {gather*} 18 e^3 x^4+2 e^3 x^3+18 e^3 x^3 \log ^2(x)-12 e^3 x^3 \log (x)+2 e^3 x^2 \log ^2(x)-2 e^3 x^2 \log (x)+2 e^3 \left (6 x^3+x^2\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^3*(6*x^2 + 72*x^3 + (4*x + 36*x^2)*Log[x] + (4*x + 54*x^2)*Log[x]^2),x]

[Out]

2*E^3*x^3 + 18*E^3*x^4 - 2*E^3*x^2*Log[x] - 12*E^3*x^3*Log[x] + 2*E^3*(x^2 + 6*x^3)*Log[x] + 2*E^3*x^2*Log[x]^
2 + 18*E^3*x^3*Log[x]^2

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

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
]))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^3 \int \left (6 x^2+72 x^3+\left (4 x+36 x^2\right ) \log (x)+\left (4 x+54 x^2\right ) \log ^2(x)\right ) \, dx\\ &=2 e^3 x^3+18 e^3 x^4+e^3 \int \left (4 x+36 x^2\right ) \log (x) \, dx+e^3 \int \left (4 x+54 x^2\right ) \log ^2(x) \, dx\\ &=2 e^3 x^3+18 e^3 x^4+e^3 \int x (4+36 x) \log (x) \, dx+e^3 \int x (4+54 x) \log ^2(x) \, dx\\ &=2 e^3 x^3+18 e^3 x^4+2 e^3 \left (x^2+6 x^3\right ) \log (x)-e^3 \int 2 x (1+6 x) \, dx+e^3 \int \left (4 x \log ^2(x)+54 x^2 \log ^2(x)\right ) \, dx\\ &=2 e^3 x^3+18 e^3 x^4+2 e^3 \left (x^2+6 x^3\right ) \log (x)-\left (2 e^3\right ) \int x (1+6 x) \, dx+\left (4 e^3\right ) \int x \log ^2(x) \, dx+\left (54 e^3\right ) \int x^2 \log ^2(x) \, dx\\ &=2 e^3 x^3+18 e^3 x^4+2 e^3 \left (x^2+6 x^3\right ) \log (x)+2 e^3 x^2 \log ^2(x)+18 e^3 x^3 \log ^2(x)-\left (2 e^3\right ) \int \left (x+6 x^2\right ) \, dx-\left (4 e^3\right ) \int x \log (x) \, dx-\left (36 e^3\right ) \int x^2 \log (x) \, dx\\ &=2 e^3 x^3+18 e^3 x^4-2 e^3 x^2 \log (x)-12 e^3 x^3 \log (x)+2 e^3 \left (x^2+6 x^3\right ) \log (x)+2 e^3 x^2 \log ^2(x)+18 e^3 x^3 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 41, normalized size = 2.16 \begin {gather*} 2 e^3 x^3+18 e^3 x^4+2 e^3 x^2 \log ^2(x)+18 e^3 x^3 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^3*(6*x^2 + 72*x^3 + (4*x + 36*x^2)*Log[x] + (4*x + 54*x^2)*Log[x]^2),x]

[Out]

2*E^3*x^3 + 18*E^3*x^4 + 2*E^3*x^2*Log[x]^2 + 18*E^3*x^3*Log[x]^2

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fricas [A]  time = 0.59, size = 31, normalized size = 1.63 \begin {gather*} 2 \, {\left (9 \, x^{3} + x^{2}\right )} e^{3} \log \relax (x)^{2} + 2 \, {\left (9 \, x^{4} + x^{3}\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((54*x^2+4*x)*log(x)^2+(36*x^2+4*x)*log(x)+72*x^3+6*x^2)*exp(log(log(x)^2+x)+3)/(log(x)^2+x),x, algo
rithm="fricas")

[Out]

2*(9*x^3 + x^2)*e^3*log(x)^2 + 2*(9*x^4 + x^3)*e^3

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giac [A]  time = 0.12, size = 37, normalized size = 1.95 \begin {gather*} 18 \, x^{3} e^{3} \log \relax (x)^{2} + 18 \, x^{4} e^{3} + 2 \, x^{2} e^{3} \log \relax (x)^{2} + 2 \, x^{3} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((54*x^2+4*x)*log(x)^2+(36*x^2+4*x)*log(x)+72*x^3+6*x^2)*exp(log(log(x)^2+x)+3)/(log(x)^2+x),x, algo
rithm="giac")

[Out]

18*x^3*e^3*log(x)^2 + 18*x^4*e^3 + 2*x^2*e^3*log(x)^2 + 2*x^3*e^3

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maple [A]  time = 0.04, size = 33, normalized size = 1.74

method	result	size
default	\({\mathrm e}^{3} \left (18 x^{3} \ln \relax (x )^{2}+2 x^{2} \ln \relax (x )^{2}+2 x^{3}+18 x^{4}\right )\)	\(33\)
risch	\({\mathrm e}^{3} \left (18 x^{3}+2 x^{2}\right ) \ln \relax (x )^{2}+18 x^{4} {\mathrm e}^{3}+2 x^{3} {\mathrm e}^{3}\)	\(34\)
norman	\(2 x^{3} {\mathrm e}^{3}+18 x^{4} {\mathrm e}^{3}+2 x^{2} {\mathrm e}^{3} \ln \relax (x )^{2}+18 x^{3} {\mathrm e}^{3} \ln \relax (x )^{2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((54*x^2+4*x)*ln(x)^2+(36*x^2+4*x)*ln(x)+72*x^3+6*x^2)*exp(ln(ln(x)^2+x)+3)/(ln(x)^2+x),x,method=_RETURNVE
RBOSE)

[Out]

exp(3)*(18*x^3*ln(x)^2+2*x^2*ln(x)^2+2*x^3+18*x^4)

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maxima [B]  time = 0.35, size = 65, normalized size = 3.42 \begin {gather*} {\left (2 \, {\left (9 \, \log \relax (x)^{2} - 6 \, \log \relax (x) + 2\right )} x^{3} + 18 \, x^{4} + {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} - 2 \, x^{3} - x^{2} + 2 \, {\left (6 \, x^{3} + x^{2}\right )} \log \relax (x)\right )} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((54*x^2+4*x)*log(x)^2+(36*x^2+4*x)*log(x)+72*x^3+6*x^2)*exp(log(log(x)^2+x)+3)/(log(x)^2+x),x, algo
rithm="maxima")

[Out]

(2*(9*log(x)^2 - 6*log(x) + 2)*x^3 + 18*x^4 + (2*log(x)^2 - 2*log(x) + 1)*x^2 - 2*x^3 - x^2 + 2*(6*x^3 + x^2)*
log(x))*e^3

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mupad [B]  time = 5.33, size = 18, normalized size = 0.95 \begin {gather*} 2\,x^2\,{\mathrm {e}}^3\,\left ({\ln \relax (x)}^2+x\right )\,\left (9\,x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x + log(x)^2) + 3)*(log(x)^2*(4*x + 54*x^2) + log(x)*(4*x + 36*x^2) + 6*x^2 + 72*x^3))/(x + log(x
)^2),x)

[Out]

2*x^2*exp(3)*(x + log(x)^2)*(9*x + 1)

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sympy [A]  time = 0.13, size = 37, normalized size = 1.95 \begin {gather*} 18 x^{4} e^{3} + 2 x^{3} e^{3} + \left (18 x^{3} e^{3} + 2 x^{2} e^{3}\right ) \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((54*x**2+4*x)*ln(x)**2+(36*x**2+4*x)*ln(x)+72*x**3+6*x**2)*exp(ln(ln(x)**2+x)+3)/(ln(x)**2+x),x)

[Out]

18*x**4*exp(3) + 2*x**3*exp(3) + (18*x**3*exp(3) + 2*x**2*exp(3))*log(x)**2

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