3.43.40 \(\int \frac {120 x+(-28+8 x) \log ^2(-7+2 x)}{(-7 x+2 x^2) \log ^2(-7+2 x)} \, dx\)

Optimal. Leaf size=19 \[ 4 \left (\log \left (\frac {x}{2}\right )-\frac {15}{\log (-7+2 x)}\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1593, 6688, 12, 2390, 2302, 30} \begin {gather*} 4 \log (x)-\frac {60}{\log (2 x-7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120*x + (-28 + 8*x)*Log[-7 + 2*x]^2)/((-7*x + 2*x^2)*Log[-7 + 2*x]^2),x]

[Out]

4*Log[x] - 60/Log[-7 + 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 30

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

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 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 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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {120 x+(-28+8 x) \log ^2(-7+2 x)}{x (-7+2 x) \log ^2(-7+2 x)} \, dx\\ &=\int 4 \left (\frac {1}{x}+\frac {30}{(-7+2 x) \log ^2(-7+2 x)}\right ) \, dx\\ &=4 \int \left (\frac {1}{x}+\frac {30}{(-7+2 x) \log ^2(-7+2 x)}\right ) \, dx\\ &=4 \log (x)+120 \int \frac {1}{(-7+2 x) \log ^2(-7+2 x)} \, dx\\ &=4 \log (x)+60 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-7+2 x\right )\\ &=4 \log (x)+60 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-7+2 x)\right )\\ &=4 \log (x)-\frac {60}{\log (-7+2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.79 \begin {gather*} 4 \left (\log (x)-\frac {15}{\log (-7+2 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120*x + (-28 + 8*x)*Log[-7 + 2*x]^2)/((-7*x + 2*x^2)*Log[-7 + 2*x]^2),x]

[Out]

4*(Log[x] - 15/Log[-7 + 2*x])

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fricas [A]  time = 0.50, size = 21, normalized size = 1.11 \begin {gather*} \frac {4 \, {\left (\log \left (2 \, x - 7\right ) \log \relax (x) - 15\right )}}{\log \left (2 \, x - 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-28)*log(2*x-7)^2+120*x)/(2*x^2-7*x)/log(2*x-7)^2,x, algorithm="fricas")

[Out]

4*(log(2*x - 7)*log(x) - 15)/log(2*x - 7)

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giac [A]  time = 0.15, size = 15, normalized size = 0.79 \begin {gather*} -\frac {60}{\log \left (2 \, x - 7\right )} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-28)*log(2*x-7)^2+120*x)/(2*x^2-7*x)/log(2*x-7)^2,x, algorithm="giac")

[Out]

-60/log(2*x - 7) + 4*log(x)

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maple [A]  time = 0.08, size = 16, normalized size = 0.84

method	result	size
norman	\(-\frac {60}{\ln \left (2 x -7\right )}+4 \ln \relax (x )\)	\(16\)
risch	\(-\frac {60}{\ln \left (2 x -7\right )}+4 \ln \relax (x )\)	\(16\)
derivativedivides	\(4 \ln \left (2 x \right )-\frac {60}{\ln \left (2 x -7\right )}\)	\(18\)
default	\(4 \ln \left (2 x \right )-\frac {60}{\ln \left (2 x -7\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x-28)*ln(2*x-7)^2+120*x)/(2*x^2-7*x)/ln(2*x-7)^2,x,method=_RETURNVERBOSE)

[Out]

-60/ln(2*x-7)+4*ln(x)

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maxima [A]  time = 0.38, size = 15, normalized size = 0.79 \begin {gather*} -\frac {60}{\log \left (2 \, x - 7\right )} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-28)*log(2*x-7)^2+120*x)/(2*x^2-7*x)/log(2*x-7)^2,x, algorithm="maxima")

[Out]

-60/log(2*x - 7) + 4*log(x)

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mupad [B]  time = 0.13, size = 15, normalized size = 0.79 \begin {gather*} 4\,\ln \relax (x)-\frac {60}{\ln \left (2\,x-7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(120*x + log(2*x - 7)^2*(8*x - 28))/(log(2*x - 7)^2*(7*x - 2*x^2)),x)

[Out]

4*log(x) - 60/log(2*x - 7)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.63 \begin {gather*} 4 \log {\relax (x )} - \frac {60}{\log {\left (2 x - 7 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-28)*ln(2*x-7)**2+120*x)/(2*x**2-7*x)/ln(2*x-7)**2,x)

[Out]

4*log(x) - 60/log(2*x - 7)

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