3.9.9 \(\int \frac {-156-65 x+e^{2 x/3} (-36+12 x)+e^{x/3} (-150-6 x+10 x^2)+(25+e^{x/3} (12-2 x)+5 x) \log (x)-\log ^2(x)}{2 x^3} \, dx\)

Optimal. Leaf size=30 \[ \frac {\left (6 \left (2+e^{x/3}+\frac {5 x}{6}\right )-\log (x)\right )^2}{4 x^2} \]

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Rubi [B]  time = 0.65, antiderivative size = 117, normalized size of antiderivative = 3.90, number of steps used = 28, number of rules used = 13, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {12, 14, 2197, 6742, 37, 2334, 43, 2305, 2304, 2199, 2177, 2178, 2554} \begin {gather*} \frac {13 (5 x+12)^2}{48 x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {3}{x^2}+\frac {\log ^2(x)}{4 x^2}-\frac {(x+5)^2 \log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}+\frac {\log (x)}{4 x^2}+\frac {15 e^{x/3}}{x}-\frac {5}{2 x}+\frac {\log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-156 - 65*x + E^((2*x)/3)*(-36 + 12*x) + E^(x/3)*(-150 - 6*x + 10*x^2) + (25 + E^(x/3)*(12 - 2*x) + 5*x)*
Log[x] - Log[x]^2)/(2*x^3),x]

[Out]

-3/x^2 + (36*E^(x/3))/x^2 + (9*E^((2*x)/3))/x^2 - 5/(2*x) + (15*E^(x/3))/x + (13*(12 + 5*x)^2)/(48*x^2) + Log[
x]/4 + Log[x]/(4*x^2) - (3*E^(x/3)*Log[x])/x^2 - ((5 + x)^2*Log[x])/(4*x^2) + Log[x]^2/(4*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 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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-156-65 x+e^{2 x/3} (-36+12 x)+e^{x/3} \left (-150-6 x+10 x^2\right )+\left (25+e^{x/3} (12-2 x)+5 x\right ) \log (x)-\log ^2(x)}{x^3} \, dx\\ &=\frac {1}{2} \int \left (\frac {12 e^{2 x/3} (-3+x)}{x^3}+\frac {(12+5 x-\log (x)) (-13+\log (x))}{x^3}+\frac {2 e^{x/3} \left (-75-3 x+5 x^2+6 \log (x)-x \log (x)\right )}{x^3}\right ) \, dx\\ &=\frac {1}{2} \int \frac {(12+5 x-\log (x)) (-13+\log (x))}{x^3} \, dx+6 \int \frac {e^{2 x/3} (-3+x)}{x^3} \, dx+\int \frac {e^{x/3} \left (-75-3 x+5 x^2+6 \log (x)-x \log (x)\right )}{x^3} \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}+\frac {1}{2} \int \left (-\frac {13 (12+5 x)}{x^3}+\frac {5 (5+x) \log (x)}{x^3}-\frac {\log ^2(x)}{x^3}\right ) \, dx+\int \left (\frac {e^{x/3} \left (-75-3 x+5 x^2\right )}{x^3}-\frac {e^{x/3} (-6+x) \log (x)}{x^3}\right ) \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}-\frac {1}{2} \int \frac {\log ^2(x)}{x^3} \, dx+\frac {5}{2} \int \frac {(5+x) \log (x)}{x^3} \, dx-\frac {13}{2} \int \frac {12+5 x}{x^3} \, dx+\int \frac {e^{x/3} \left (-75-3 x+5 x^2\right )}{x^3} \, dx-\int \frac {e^{x/3} (-6+x) \log (x)}{x^3} \, dx\\ &=\frac {9 e^{2 x/3}}{x^2}+\frac {13 (12+5 x)^2}{48 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}-\frac {1}{2} \int \frac {\log (x)}{x^3} \, dx-\frac {5}{2} \int -\frac {(5+x)^2}{10 x^3} \, dx+\int \left (-\frac {75 e^{x/3}}{x^3}-\frac {3 e^{x/3}}{x^2}+\frac {5 e^{x/3}}{x}\right ) \, dx+\int \frac {3 e^{x/3}}{x^3} \, dx\\ &=\frac {1}{8 x^2}+\frac {9 e^{2 x/3}}{x^2}+\frac {13 (12+5 x)^2}{48 x^2}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{4} \int \frac {(5+x)^2}{x^3} \, dx+3 \int \frac {e^{x/3}}{x^3} \, dx-3 \int \frac {e^{x/3}}{x^2} \, dx+5 \int \frac {e^{x/3}}{x} \, dx-75 \int \frac {e^{x/3}}{x^3} \, dx\\ &=\frac {1}{8 x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}+\frac {3 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+5 \text {Ei}\left (\frac {x}{3}\right )+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{4} \int \left (\frac {25}{x^3}+\frac {10}{x^2}+\frac {1}{x}\right ) \, dx+\frac {1}{2} \int \frac {e^{x/3}}{x^2} \, dx-\frac {25}{2} \int \frac {e^{x/3}}{x^2} \, dx-\int \frac {e^{x/3}}{x} \, dx\\ &=-\frac {3}{x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {5}{2 x}+\frac {15 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+4 \text {Ei}\left (\frac {x}{3}\right )+\frac {\log (x)}{4}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}+\frac {1}{6} \int \frac {e^{x/3}}{x} \, dx-\frac {25}{6} \int \frac {e^{x/3}}{x} \, dx\\ &=-\frac {3}{x^2}+\frac {36 e^{x/3}}{x^2}+\frac {9 e^{2 x/3}}{x^2}-\frac {5}{2 x}+\frac {15 e^{x/3}}{x}+\frac {13 (12+5 x)^2}{48 x^2}+\frac {\log (x)}{4}+\frac {\log (x)}{4 x^2}-\frac {3 e^{x/3} \log (x)}{x^2}-\frac {(5+x)^2 \log (x)}{4 x^2}+\frac {\log ^2(x)}{4 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 41, normalized size = 1.37 \begin {gather*} \frac {\left (6 \left (2+e^{x/3}\right )-\log (x)\right ) \left (12+6 e^{x/3}+10 x-\log (x)\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-156 - 65*x + E^((2*x)/3)*(-36 + 12*x) + E^(x/3)*(-150 - 6*x + 10*x^2) + (25 + E^(x/3)*(12 - 2*x) +
5*x)*Log[x] - Log[x]^2)/(2*x^3),x]

[Out]

((6*(2 + E^(x/3)) - Log[x])*(12 + 6*E^(x/3) + 10*x - Log[x]))/(4*x^2)

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fricas [B]  time = 0.57, size = 46, normalized size = 1.53 \begin {gather*} \frac {12 \, {\left (5 \, x + 12\right )} e^{\left (\frac {1}{3} \, x\right )} - 2 \, {\left (5 \, x + 6 \, e^{\left (\frac {1}{3} \, x\right )} + 12\right )} \log \relax (x) + \log \relax (x)^{2} + 120 \, x + 36 \, e^{\left (\frac {2}{3} \, x\right )} + 144}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(x)^2+((-2*x+12)*exp(1/3*x)+25+5*x)*log(x)+(12*x-36)*exp(1/3*x)^2+(10*x^2-6*x-150)*exp(1/3*
x)-65*x-156)/x^3,x, algorithm="fricas")

[Out]

1/4*(12*(5*x + 12)*e^(1/3*x) - 2*(5*x + 6*e^(1/3*x) + 12)*log(x) + log(x)^2 + 120*x + 36*e^(2/3*x) + 144)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {12 \, {\left (x - 3\right )} e^{\left (\frac {2}{3} \, x\right )} + 2 \, {\left (5 \, x^{2} - 3 \, x - 75\right )} e^{\left (\frac {1}{3} \, x\right )} - {\left (2 \, {\left (x - 6\right )} e^{\left (\frac {1}{3} \, x\right )} - 5 \, x - 25\right )} \log \relax (x) - \log \relax (x)^{2} - 65 \, x - 156}{2 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(x)^2+((-2*x+12)*exp(1/3*x)+25+5*x)*log(x)+(12*x-36)*exp(1/3*x)^2+(10*x^2-6*x-150)*exp(1/3*
x)-65*x-156)/x^3,x, algorithm="giac")

[Out]

integrate(1/2*(12*(x - 3)*e^(2/3*x) + 2*(5*x^2 - 3*x - 75)*e^(1/3*x) - (2*(x - 6)*e^(1/3*x) - 5*x - 25)*log(x)
- log(x)^2 - 65*x - 156)/x^3, x)

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maple [B]  time = 0.08, size = 58, normalized size = 1.93

method	result	size
risch	\(\frac {\ln \relax (x )^{2}}{4 x^{2}}-\frac {\left (5 x +12+6 \,{\mathrm e}^{\frac {x}{3}}\right ) \ln \relax (x )}{2 x^{2}}+\frac {15 x \,{\mathrm e}^{\frac {x}{3}}+9 \,{\mathrm e}^{\frac {2 x}{3}}+30 x +36 \,{\mathrm e}^{\frac {x}{3}}+36}{x^{2}}\)	\(58\)
default	\(\frac {30 x \,{\mathrm e}^{\frac {x}{3}}-6 \ln \relax (x ) {\mathrm e}^{\frac {x}{3}}+72 \,{\mathrm e}^{\frac {x}{3}}}{2 x^{2}}+\frac {\ln \relax (x )^{2}}{4 x^{2}}-\frac {6 \ln \relax (x )}{x^{2}}+\frac {36}{x^{2}}+\frac {30}{x}-\frac {5 \ln \relax (x )}{2 x}+\frac {9 \,{\mathrm e}^{\frac {2 x}{3}}}{x^{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-ln(x)^2+((-2*x+12)*exp(1/3*x)+25+5*x)*ln(x)+(12*x-36)*exp(1/3*x)^2+(10*x^2-6*x-150)*exp(1/3*x)-65*x-
156)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/4*ln(x)^2/x^2-1/2*(5*x+12+6*exp(1/3*x))/x^2*ln(x)+3*(5*x*exp(1/3*x)+3*exp(2/3*x)+10*x+12*exp(1/3*x)+12)/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {5 \, \log \relax (x)}{2 \, x} - \frac {24 \, e^{\left (\frac {1}{3} \, x\right )} \log \relax (x) - 2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) - 1}{8 \, x^{2}} + \frac {30}{x} - \frac {25 \, \log \relax (x)}{4 \, x^{2}} + \frac {287}{8 \, x^{2}} + 5 \, {\rm Ei}\left (\frac {1}{3} \, x\right ) - \Gamma \left (-1, -\frac {1}{3} \, x\right ) + 4 \, \Gamma \left (-1, -\frac {2}{3} \, x\right ) + \frac {25}{3} \, \Gamma \left (-2, -\frac {1}{3} \, x\right ) + 8 \, \Gamma \left (-2, -\frac {2}{3} \, x\right ) + 3 \, \int \frac {e^{\left (\frac {1}{3} \, x\right )}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(x)^2+((-2*x+12)*exp(1/3*x)+25+5*x)*log(x)+(12*x-36)*exp(1/3*x)^2+(10*x^2-6*x-150)*exp(1/3*
x)-65*x-156)/x^3,x, algorithm="maxima")

[Out]

-5/2*log(x)/x - 1/8*(24*e^(1/3*x)*log(x) - 2*log(x)^2 - 2*log(x) - 1)/x^2 + 30/x - 25/4*log(x)/x^2 + 287/8/x^2
+ 5*Ei(1/3*x) - gamma(-1, -1/3*x) + 4*gamma(-1, -2/3*x) + 25/3*gamma(-2, -1/3*x) + 8*gamma(-2, -2/3*x) + 3*in
tegrate(e^(1/3*x)/x^3, x)

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mupad [B]  time = 0.69, size = 32, normalized size = 1.07 \begin {gather*} \frac {\left (6\,{\mathrm {e}}^{x/3}-\ln \relax (x)+12\right )\,\left (10\,x+6\,{\mathrm {e}}^{x/3}-\ln \relax (x)+12\right )}{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((65*x)/2 + (exp(x/3)*(6*x - 10*x^2 + 150))/2 + log(x)^2/2 - (log(x)*(5*x - exp(x/3)*(2*x - 12) + 25))/2
- (exp((2*x)/3)*(12*x - 36))/2 + 78)/x^3,x)

[Out]

((6*exp(x/3) - log(x) + 12)*(10*x + 6*exp(x/3) - log(x) + 12))/(4*x^2)

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sympy [B]  time = 0.42, size = 73, normalized size = 2.43 \begin {gather*} - \frac {- 30 x - 36}{x^{2}} + \frac {\left (- 5 x - 12\right ) \log {\relax (x )}}{2 x^{2}} + \frac {\log {\relax (x )}^{2}}{4 x^{2}} + \frac {9 x^{2} e^{\frac {2 x}{3}} + \left (15 x^{3} - 3 x^{2} \log {\relax (x )} + 36 x^{2}\right ) e^{\frac {x}{3}}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-ln(x)**2+((-2*x+12)*exp(1/3*x)+25+5*x)*ln(x)+(12*x-36)*exp(1/3*x)**2+(10*x**2-6*x-150)*exp(1/3
*x)-65*x-156)/x**3,x)

[Out]

-(-30*x - 36)/x**2 + (-5*x - 12)*log(x)/(2*x**2) + log(x)**2/(4*x**2) + (9*x**2*exp(2*x/3) + (15*x**3 - 3*x**2
*log(x) + 36*x**2)*exp(x/3))/x**4

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