3.66.78 \(\int \frac {120 x-9 x^2 \log (5)+(-30 x+2 x^2 \log (5)) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx\)

Optimal. Leaf size=20 \[ \frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \]

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Rubi [B]  time = 0.76, antiderivative size = 89, normalized size of antiderivative = 4.45, number of steps used = 23, number of rules used = 15, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.168, Rules used = {6688, 12, 6742, 2411, 2353, 2297, 2299, 2178, 2302, 30, 2306, 2309, 2399, 2389, 2390} \begin {gather*} -\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120*x - 9*x^2*Log[5] + (-30*x + 2*x^2*Log[5])*Log[15 - x*Log[5]])/(-240*Log[3] + 16*x*Log[3]*Log[5] + (12
0*Log[3] - 8*x*Log[3]*Log[5])*Log[15 - x*Log[5]] + (-15*Log[3] + x*Log[3]*Log[5])*Log[15 - x*Log[5]]^2),x]

[Out]

-225/(Log[3]*Log[5]^2*(4 - Log[15 - x*Log[5]])) + (30*(15 - x*Log[5]))/(Log[3]*Log[5]^2*(4 - Log[15 - x*Log[5]
])) - (15 - x*Log[5])^2/(Log[3]*Log[5]^2*(4 - Log[15 - x*Log[5]]))

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2297

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

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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} &=\int \frac {x (-120+9 x \log (5)-2 (-15+x \log (5)) \log (15-x \log (5)))}{\log (3) (15-x \log (5)) (4-\log (15-x \log (5)))^2} \, dx\\ &=\frac {\int \frac {x (-120+9 x \log (5)-2 (-15+x \log (5)) \log (15-x \log (5)))}{(15-x \log (5)) (4-\log (15-x \log (5)))^2} \, dx}{\log (3)}\\ &=\frac {\int \left (-\frac {x^2 \log (5)}{(-15+x \log (5)) (-4+\log (15-x \log (5)))^2}+\frac {2 x}{-4+\log (15-x \log (5))}\right ) \, dx}{\log (3)}\\ &=\frac {2 \int \frac {x}{-4+\log (15-x \log (5))} \, dx}{\log (3)}-\frac {\log (5) \int \frac {x^2}{(-15+x \log (5)) (-4+\log (15-x \log (5)))^2} \, dx}{\log (3)}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {15}{\log (5)}-\frac {x}{\log (5)}\right )^2}{x (-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3)}+\frac {2 \int \left (\frac {15}{\log (5) (-4+\log (15-x \log (5)))}-\frac {15-x \log (5)}{\log (5) (-4+\log (15-x \log (5)))}\right ) \, dx}{\log (3)}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {30}{\log ^2(5) (-4+\log (x))^2}+\frac {225}{x \log ^2(5) (-4+\log (x))^2}+\frac {x}{\log ^2(5) (-4+\log (x))^2}\right ) \, dx,x,15-x \log (5)\right )}{\log (3)}-\frac {2 \int \frac {15-x \log (5)}{-4+\log (15-x \log (5))} \, dx}{\log (3) \log (5)}+\frac {30 \int \frac {1}{-4+\log (15-x \log (5))} \, dx}{\log (3) \log (5)}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x}{(-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}+\frac {2 \operatorname {Subst}\left (\int \frac {x}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}+\frac {30 \operatorname {Subst}\left (\int \frac {1}{(-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {30 \operatorname {Subst}\left (\int \frac {1}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {225 \operatorname {Subst}\left (\int \frac {1}{x (-4+\log (x))^2} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}\\ &=\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}-\frac {2 \operatorname {Subst}\left (\int \frac {x}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {30 \operatorname {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}+\frac {30 \operatorname {Subst}\left (\int \frac {1}{-4+\log (x)} \, dx,x,15-x \log (5)\right )}{\log (3) \log ^2(5)}-\frac {225 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-4+\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}\\ &=\frac {2 e^8 \text {Ei}(-2 (4-\log (15-x \log (5))))}{\log (3) \log ^2(5)}-\frac {30 e^4 \text {Ei}(-4+\log (15-x \log (5)))}{\log (3) \log ^2(5)}-\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}+\frac {30 \operatorname {Subst}\left (\int \frac {e^x}{-4+x} \, dx,x,\log (15-x \log (5))\right )}{\log (3) \log ^2(5)}\\ &=-\frac {225}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}+\frac {30 (15-x \log (5))}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}-\frac {(15-x \log (5))^2}{\log (3) \log ^2(5) (4-\log (15-x \log (5)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 20, normalized size = 1.00 \begin {gather*} \frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120*x - 9*x^2*Log[5] + (-30*x + 2*x^2*Log[5])*Log[15 - x*Log[5]])/(-240*Log[3] + 16*x*Log[3]*Log[5]
+ (120*Log[3] - 8*x*Log[3]*Log[5])*Log[15 - x*Log[5]] + (-15*Log[3] + x*Log[3]*Log[5])*Log[15 - x*Log[5]]^2),
x]

[Out]

x^2/(Log[3]*(-4 + Log[15 - x*Log[5]]))

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fricas [A]  time = 0.53, size = 22, normalized size = 1.10 \begin {gather*} \frac {x^{2}}{\log \relax (3) \log \left (-x \log \relax (5) + 15\right ) - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*log(3)*log(5)-15*log(3))*log(-x*log(5
)+15)^2+(-8*x*log(3)*log(5)+120*log(3))*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="fricas"
)

[Out]

x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))

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giac [A]  time = 0.19, size = 22, normalized size = 1.10 \begin {gather*} \frac {x^{2}}{\log \relax (3) \log \left (-x \log \relax (5) + 15\right ) - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*log(3)*log(5)-15*log(3))*log(-x*log(5
)+15)^2+(-8*x*log(3)*log(5)+120*log(3))*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="giac")

[Out]

x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))

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maple [A]  time = 0.18, size = 21, normalized size = 1.05

method	result	size
norman	\(\frac {x^{2}}{\left (\ln \left (-x \ln \relax (5)+15\right )-4\right ) \ln \relax (3)}\)	\(21\)
risch	\(\frac {x^{2}}{\left (\ln \left (-x \ln \relax (5)+15\right )-4\right ) \ln \relax (3)}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2*ln(5)-30*x)*ln(-x*ln(5)+15)-9*x^2*ln(5)+120*x)/((x*ln(3)*ln(5)-15*ln(3))*ln(-x*ln(5)+15)^2+(-8*x*l
n(3)*ln(5)+120*ln(3))*ln(-x*ln(5)+15)+16*x*ln(3)*ln(5)-240*ln(3)),x,method=_RETURNVERBOSE)

[Out]

x^2/(ln(-x*ln(5)+15)-4)/ln(3)

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maxima [A]  time = 0.50, size = 22, normalized size = 1.10 \begin {gather*} \frac {x^{2}}{\log \relax (3) \log \left (-x \log \relax (5) + 15\right ) - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*log(3)*log(5)-15*log(3))*log(-x*log(5
)+15)^2+(-8*x*log(3)*log(5)+120*log(3))*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="maxima"
)

[Out]

x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))

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mupad [B]  time = 0.38, size = 20, normalized size = 1.00 \begin {gather*} \frac {x^2}{\ln \relax (3)\,\left (\ln \left (15-x\,\ln \relax (5)\right )-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(15 - x*log(5))*(30*x - 2*x^2*log(5)) - 120*x + 9*x^2*log(5))/(240*log(3) - log(15 - x*log(5))*(120*lo
g(3) - 8*x*log(3)*log(5)) + log(15 - x*log(5))^2*(15*log(3) - x*log(3)*log(5)) - 16*x*log(3)*log(5)),x)

[Out]

x^2/(log(3)*(log(15 - x*log(5)) - 4))

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sympy [A]  time = 0.14, size = 19, normalized size = 0.95 \begin {gather*} \frac {x^{2}}{\log {\relax (3 )} \log {\left (- x \log {\relax (5 )} + 15 \right )} - 4 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2*ln(5)-30*x)*ln(-x*ln(5)+15)-9*x**2*ln(5)+120*x)/((x*ln(3)*ln(5)-15*ln(3))*ln(-x*ln(5)+15)**
2+(-8*x*ln(3)*ln(5)+120*ln(3))*ln(-x*ln(5)+15)+16*x*ln(3)*ln(5)-240*ln(3)),x)

[Out]

x**2/(log(3)*log(-x*log(5) + 15) - 4*log(3))

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