3.27.78 \(\int (-15-5 e^x+5 x^2+(-3-e^x+x^2) \log (x)+(-18+e^x (-6-5 x)+16 x^2+(-3+e^x (-1-x)+3 x^2) \log (x)) \log (2 x)) \, dx\)

Optimal. Leaf size=20 \[ x \left (-3-e^x+x^2\right ) (5+\log (x)) \log (2 x) \]

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Rubi [B]  time = 1.17, antiderivative size = 87, normalized size of antiderivative = 4.35, number of steps used = 46, number of rules used = 16, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.246, Rules used = {2194, 2554, 2178, 2176, 6742, 2288, 2199, 2557, 6483, 6475, 14, 2295, 2304, 2361, 2366, 12} \begin {gather*} -\frac {x^3}{9}+\frac {1}{9} x^3 (1-3 \log (x))+\frac {1}{3} x^3 \log (x)+x^3 \log (x) \log (2 x)+5 x^3 \log (2 x)-5 e^x x \log (2 x)-e^x x \log (x) \log (2 x)-3 x \log (x) \log (2 x)-15 x \log (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-15 - 5*E^x + 5*x^2 + (-3 - E^x + x^2)*Log[x] + (-18 + E^x*(-6 - 5*x) + 16*x^2 + (-3 + E^x*(-1 - x) + 3*x^
2)*Log[x])*Log[2*x],x]

[Out]

-1/9*x^3 + (x^3*(1 - 3*Log[x]))/9 + (x^3*Log[x])/3 - 15*x*Log[2*x] - 5*E^x*x*Log[2*x] + 5*x^3*Log[2*x] - 3*x*L
og[x]*Log[2*x] - E^x*x*Log[x]*Log[2*x] + x^3*Log[x]*Log[2*x]

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 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 2361

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

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 6475

Int[ExpIntegralE[1, (b_.)*(x_)]/(x_), x_Symbol] :> Simp[b*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, -(b*x)], x
] + (-Simp[EulerGamma*Log[x], x] - Simp[(1*Log[b*x]^2)/2, x]) /; FreeQ[b, x]

Rule 6483

Int[ExpIntegralEi[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[Log[x]*(ExpIntegralEi[b*x] + ExpIntegralE[1, -(b*x)]), x
] - Int[ExpIntegralE[1, -(b*x)]/x, x] /; FreeQ[b, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-15 x+\frac {5 x^3}{3}-5 \int e^x \, dx+\int \left (-3-e^x+x^2\right ) \log (x) \, dx+\int \left (-18+e^x (-6-5 x)+16 x^2+\left (-3+e^x (-1-x)+3 x^2\right ) \log (x)\right ) \log (2 x) \, dx\\ &=-5 e^x-15 x+\frac {5 x^3}{3}-e^x \log (x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)-\int \left (-3-\frac {e^x}{x}+\frac {x^2}{3}\right ) \, dx+\int \left (-e^x (6+5 x+\log (x)+x \log (x)) \log (2 x)+\left (-18+16 x^2-3 \log (x)+3 x^2 \log (x)\right ) \log (2 x)\right ) \, dx\\ &=-5 e^x-12 x+\frac {14 x^3}{9}-e^x \log (x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)+\int \frac {e^x}{x} \, dx-\int e^x (6+5 x+\log (x)+x \log (x)) \log (2 x) \, dx+\int \left (-18+16 x^2-3 \log (x)+3 x^2 \log (x)\right ) \log (2 x) \, dx\\ &=-5 e^x-12 x+\frac {14 x^3}{9}+\text {Ei}(x)-e^x \log (x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)-\int \left (6 e^x \log (2 x)+5 e^x x \log (2 x)+e^x \log (x) \log (2 x)+e^x x \log (x) \log (2 x)\right ) \, dx+\int \left (-18 \log (2 x)+16 x^2 \log (2 x)-3 \log (x) \log (2 x)+3 x^2 \log (x) \log (2 x)\right ) \, dx\\ &=-5 e^x-12 x+\frac {14 x^3}{9}+\text {Ei}(x)-e^x \log (x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)-3 \int \log (x) \log (2 x) \, dx+3 \int x^2 \log (x) \log (2 x) \, dx-5 \int e^x x \log (2 x) \, dx-6 \int e^x \log (2 x) \, dx+16 \int x^2 \log (2 x) \, dx-18 \int \log (2 x) \, dx-\int e^x \log (x) \log (2 x) \, dx-\int e^x x \log (x) \log (2 x) \, dx\\ &=-5 e^x+6 x-\frac {2 x^3}{9}+\text {Ei}(x)-e^x \log (x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)-e^x \log (2 x)-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)+3 \int (-1+\log (x)) \, dx-3 \int \frac {1}{9} x^2 (-1+3 \log (x)) \, dx+5 \int \frac {e^x (-1+x)}{x} \, dx+6 \int \frac {e^x}{x} \, dx+\int \frac {e^x \log (x)}{x} \, dx+\int \frac {e^x (-1+x) \log (x)}{x} \, dx+\int \frac {e^x \log (2 x)}{x} \, dx+\int \frac {e^x (-1+x) \log (2 x)}{x} \, dx\\ &=-5 e^x+3 x-\frac {2 x^3}{9}+7 \text {Ei}(x)-3 x \log (x)+\frac {1}{3} x^3 \log (x)-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)-\frac {1}{3} \int x^2 (-1+3 \log (x)) \, dx+3 \int \log (x) \, dx+5 \int \left (e^x-\frac {e^x}{x}\right ) \, dx-2 \int \frac {e^x-\text {Ei}(x)}{x} \, dx-2 \int \frac {\text {Ei}(x)}{x} \, dx\\ &=-5 e^x-\frac {x^3}{9}+7 \text {Ei}(x)+\frac {1}{9} x^3 (1-3 \log (x))+\frac {1}{3} x^3 \log (x)-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)+5 \int e^x \, dx-5 \int \frac {e^x}{x} \, dx-2 \left ((E_1(-x)+\text {Ei}(x)) \log (x)-\int \frac {E_1(-x)}{x} \, dx\right )-2 \int \left (\frac {e^x}{x}-\frac {\text {Ei}(x)}{x}\right ) \, dx\\ &=-\frac {x^3}{9}+2 \text {Ei}(x)+\frac {1}{9} x^3 (1-3 \log (x))+\frac {1}{3} x^3 \log (x)-2 \left (x \, _3F_3(1,1,1;2,2,2;x)+\frac {1}{2} \log ^2(-x)+\gamma \log (x)+(E_1(-x)+\text {Ei}(x)) \log (x)\right )-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)-2 \left (\int \frac {e^x}{x} \, dx-\int \frac {\text {Ei}(x)}{x} \, dx\right )\\ &=-\frac {x^3}{9}+2 \text {Ei}(x)+\frac {1}{9} x^3 (1-3 \log (x))+\frac {1}{3} x^3 \log (x)-2 \left (x \, _3F_3(1,1,1;2,2,2;x)+\frac {1}{2} \log ^2(-x)+\gamma \log (x)+(E_1(-x)+\text {Ei}(x)) \log (x)\right )-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)-2 \left (\text {Ei}(x)-(E_1(-x)+\text {Ei}(x)) \log (x)+\int \frac {E_1(-x)}{x} \, dx\right )\\ &=-\frac {x^3}{9}+2 \text {Ei}(x)+\frac {1}{9} x^3 (1-3 \log (x))+\frac {1}{3} x^3 \log (x)-2 \left (\text {Ei}(x)-x \, _3F_3(1,1,1;2,2,2;x)-\frac {1}{2} \log ^2(-x)-\gamma \log (x)-(E_1(-x)+\text {Ei}(x)) \log (x)\right )-2 \left (x \, _3F_3(1,1,1;2,2,2;x)+\frac {1}{2} \log ^2(-x)+\gamma \log (x)+(E_1(-x)+\text {Ei}(x)) \log (x)\right )-15 x \log (2 x)-5 e^x x \log (2 x)+5 x^3 \log (2 x)-3 x \log (x) \log (2 x)-e^x x \log (x) \log (2 x)+x^3 \log (x) \log (2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} x \left (-3-e^x+x^2\right ) (5+\log (x)) \log (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-15 - 5*E^x + 5*x^2 + (-3 - E^x + x^2)*Log[x] + (-18 + E^x*(-6 - 5*x) + 16*x^2 + (-3 + E^x*(-1 - x)
+ 3*x^2)*Log[x])*Log[2*x],x]

[Out]

x*(-3 - E^x + x^2)*(5 + Log[x])*Log[2*x]

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fricas [B]  time = 1.15, size = 70, normalized size = 3.50 \begin {gather*} -5 \, x e^{x} \log \relax (2) + {\left (x^{3} - x e^{x} - 3 \, x\right )} \log \relax (x)^{2} + 5 \, {\left (x^{3} - 3 \, x\right )} \log \relax (2) + {\left (5 \, x^{3} - {\left (x \log \relax (2) + 5 \, x\right )} e^{x} + {\left (x^{3} - 3 \, x\right )} \log \relax (2) - 15 \, x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(x)+3*x^2-3)*log(x)+(-5*x-6)*exp(x)+16*x^2-18)*log(2*x)+(-exp(x)+x^2-3)*log(x)-5*exp(x)+
5*x^2-15,x, algorithm="fricas")

[Out]

-5*x*e^x*log(2) + (x^3 - x*e^x - 3*x)*log(x)^2 + 5*(x^3 - 3*x)*log(2) + (5*x^3 - (x*log(2) + 5*x)*e^x + (x^3 -
3*x)*log(2) - 15*x)*log(x)

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giac [B]  time = 0.25, size = 107, normalized size = 5.35 \begin {gather*} x^{3} \log \relax (2) \log \relax (x) + x^{3} \log \relax (x)^{2} + 5 \, x^{3} \log \relax (2) + \frac {14}{3} \, x^{3} \log \relax (x) - x e^{x} \log \relax (2) \log \relax (x) - x e^{x} \log \relax (x)^{2} - 5 \, x e^{x} \log \relax (2) - 5 \, x e^{x} \log \relax (x) - 3 \, x \log \relax (2) \log \relax (x) - 3 \, x \log \relax (x)^{2} - 15 \, x \log \relax (2) + \frac {1}{3} \, {\left (x^{3} - 9 \, x - 3 \, e^{x}\right )} \log \relax (x) - 12 \, x \log \relax (x) + e^{x} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(x)+3*x^2-3)*log(x)+(-5*x-6)*exp(x)+16*x^2-18)*log(2*x)+(-exp(x)+x^2-3)*log(x)-5*exp(x)+
5*x^2-15,x, algorithm="giac")

[Out]

x^3*log(2)*log(x) + x^3*log(x)^2 + 5*x^3*log(2) + 14/3*x^3*log(x) - x*e^x*log(2)*log(x) - x*e^x*log(x)^2 - 5*x
*e^x*log(2) - 5*x*e^x*log(x) - 3*x*log(2)*log(x) - 3*x*log(x)^2 - 15*x*log(2) + 1/3*(x^3 - 9*x - 3*e^x)*log(x)
- 12*x*log(x) + e^x*log(x)

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maple [B]  time = 0.09, size = 88, normalized size = 4.40

method	result	size
risch	\(x^{3} \ln \relax (x )^{2}+\ln \relax (x ) \ln \relax (2) x^{3}-x \,{\mathrm e}^{x} \ln \relax (x )^{2}-x \ln \relax (2) {\mathrm e}^{x} \ln \relax (x )+5 x^{3} \ln \relax (x )+5 x^{3} \ln \relax (2)-3 x \ln \relax (x )^{2}-5 x \,{\mathrm e}^{x} \ln \relax (x )-3 x \ln \relax (2) \ln \relax (x )-5 x \ln \relax (2) {\mathrm e}^{x}-15 x \ln \relax (x )-15 x \ln \relax (2)\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x-1)*exp(x)+3*x^2-3)*ln(x)+(-5*x-6)*exp(x)+16*x^2-18)*ln(2*x)+(-exp(x)+x^2-3)*ln(x)-5*exp(x)+5*x^2-15,
x,method=_RETURNVERBOSE)

[Out]

x^3*ln(x)^2+ln(x)*ln(2)*x^3-x*exp(x)*ln(x)^2-x*ln(2)*exp(x)*ln(x)+5*x^3*ln(x)+5*x^3*ln(2)-3*x*ln(x)^2-5*x*exp(
x)*ln(x)-3*x*ln(2)*ln(x)-5*x*ln(2)*exp(x)-15*x*ln(x)-15*x*ln(2)

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maxima [B]  time = 0.53, size = 87, normalized size = 4.35 \begin {gather*} \frac {5}{3} \, x^{3} {\left (3 \, \log \relax (2) - 1\right )} + \frac {5}{3} \, x^{3} + {\left (x^{3} - 3 \, x\right )} \log \relax (x)^{2} - 15 \, x {\left (\log \relax (2) - 1\right )} - {\left (x {\left (\log \relax (2) + 5\right )} \log \relax (x) + x \log \relax (x)^{2} + 5 \, x \log \relax (2) - 5\right )} e^{x} + {\left (x^{3} {\left (\log \relax (2) + 5\right )} - 3 \, x {\left (\log \relax (2) + 5\right )}\right )} \log \relax (x) - 15 \, x - 5 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(x)+3*x^2-3)*log(x)+(-5*x-6)*exp(x)+16*x^2-18)*log(2*x)+(-exp(x)+x^2-3)*log(x)-5*exp(x)+
5*x^2-15,x, algorithm="maxima")

[Out]

5/3*x^3*(3*log(2) - 1) + 5/3*x^3 + (x^3 - 3*x)*log(x)^2 - 15*x*(log(2) - 1) - (x*(log(2) + 5)*log(x) + x*log(x
)^2 + 5*x*log(2) - 5)*e^x + (x^3*(log(2) + 5) - 3*x*(log(2) + 5))*log(x) - 15*x - 5*e^x

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mupad [B]  time = 1.55, size = 21, normalized size = 1.05 \begin {gather*} -x\,\left (\ln \relax (x)+5\right )\,\left (\ln \relax (2)+\ln \relax (x)\right )\,\left ({\mathrm {e}}^x-x^2+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*x^2 - log(2*x)*(exp(x)*(5*x + 6) - 16*x^2 + log(x)*(exp(x)*(x + 1) - 3*x^2 + 3) + 18) - log(x)*(exp(x) -
x^2 + 3) - 5*exp(x) - 15,x)

[Out]

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

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sympy [B]  time = 0.47, size = 85, normalized size = 4.25 \begin {gather*} 5 x^{3} \log {\relax (2 )} - 15 x \log {\relax (2 )} + \left (x^{3} - 3 x\right ) \log {\relax (x )}^{2} + \left (- x \log {\relax (x )}^{2} - 5 x \log {\relax (x )} - x \log {\relax (2 )} \log {\relax (x )} - 5 x \log {\relax (2 )}\right ) e^{x} + \left (x^{3} \log {\relax (2 )} + 5 x^{3} - 15 x - 3 x \log {\relax (2 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(x)+3*x**2-3)*ln(x)+(-5*x-6)*exp(x)+16*x**2-18)*ln(2*x)+(-exp(x)+x**2-3)*ln(x)-5*exp(x)+
5*x**2-15,x)

[Out]

5*x**3*log(2) - 15*x*log(2) + (x**3 - 3*x)*log(x)**2 + (-x*log(x)**2 - 5*x*log(x) - x*log(2)*log(x) - 5*x*log(
2))*exp(x) + (x**3*log(2) + 5*x**3 - 15*x - 3*x*log(2))*log(x)

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