3.94 \(\int \log ^2(\frac {c (b+a x)}{x}) \, dx\)

Optimal. Leaf size=67 \[ \frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )\right )}{a}-\frac {2 b \text {Li}_2\left (\frac {b}{a x}+1\right )}{a} \]

[Out]

(a*x+b)*ln(a*c+b*c/x)^2/a-2*b*ln(c*(a+b/x))*ln(-b/a/x)/a-2*b*polylog(2,1+b/a/x)/a

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Rubi [A]  time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2453, 2449, 2454, 2394, 2315} \[ -\frac {2 b \text {PolyLog}\left (2,\frac {b}{a x}+1\right )}{a}+\frac {(a x+b) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[(c*(b + a*x))/x]^2,x]

[Out]

((b + a*x)*Log[a*c + (b*c)/x]^2)/a - (2*b*Log[c*(a + b/x)]*Log[-(b/(a*x))])/a - (2*b*PolyLog[2, 1 + b/(a*x)])/
a

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2394

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

Rule 2449

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

Rule 2453

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.), x_Symbol] :> Int[(a + b*Log[c*ExpandToSum[v, x]^p])^q, x] /;
FreeQ[{a, b, c, p, q}, x] && BinomialQ[v, x] &&  !BinomialMatchQ[v, x]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int \log ^2\left (\frac {c (b+a x)}{x}\right ) \, dx &=\int \log ^2\left (a c+\frac {b c}{x}\right ) \, dx\\ &=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}+\frac {(2 b) \int \frac {\log \left (a c+\frac {b c}{x}\right )}{x} \, dx}{a}\\ &=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log (a c+b c x)}{x} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (c \left (a+\frac {b}{x}\right )\right ) \log \left (-\frac {b}{a x}\right )}{a}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a c+b c x} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {(b+a x) \log ^2\left (a c+\frac {b c}{x}\right )}{a}-\frac {2 b \log \left (c \left (a+\frac {b}{x}\right )\right ) \log \left (-\frac {b}{a x}\right )}{a}-\frac {2 b \text {Li}_2\left (1+\frac {b}{a x}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.94 \[ \frac {\log \left (\frac {c (a x+b)}{x}\right ) \left ((a x+b) \log \left (\frac {c (a x+b)}{x}\right )-2 b \log \left (-\frac {b}{a x}\right )\right )-2 b \text {Li}_2\left (\frac {b}{a x}+1\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(c*(b + a*x))/x]^2,x]

[Out]

(Log[(c*(b + a*x))/x]*(-2*b*Log[-(b/(a*x))] + (b + a*x)*Log[(c*(b + a*x))/x]) - 2*b*PolyLog[2, 1 + b/(a*x)])/a

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\log \left (\frac {a c x + b c}{x}\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a*x+b)/x)^2,x, algorithm="fricas")

[Out]

integral(log((a*c*x + b*c)/x)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a*x+b)/x)^2,x, algorithm="giac")

[Out]

integrate(log((a*x + b)*c/x)^2, x)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \ln \left (\frac {\left (a x +b \right ) c}{x}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a*x+b)/x)^2,x)

[Out]

int(ln(c*(a*x+b)/x)^2,x)

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maxima [A]  time = 0.66, size = 113, normalized size = 1.69 \[ x \log \left (\frac {{\left (a x + b\right )} c}{x}\right )^{2} + \frac {2 \, b \log \left (a x + b\right ) \log \left (\frac {{\left (a x + b\right )} c}{x}\right )}{a} + \frac {{\left (\frac {c \log \left (a x + b\right )^{2}}{a} - \frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} c}{a}\right )} b - \frac {2 \, {\left (c \log \left (a x + b\right ) - c \log \relax (x)\right )} b \log \left (a x + b\right )}{a}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a*x+b)/x)^2,x, algorithm="maxima")

[Out]

x*log((a*x + b)*c/x)^2 + 2*b*log(a*x + b)*log((a*x + b)*c/x)/a + ((c*log(a*x + b)^2/a - 2*(log(a*x/b + 1)*log(
x) + dilog(-a*x/b))*c/a)*b - 2*(c*log(a*x + b) - c*log(x))*b*log(a*x + b)/a)/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\ln \left (\frac {c\,\left (b+a\,x\right )}{x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*(b + a*x))/x)^2,x)

[Out]

int(log((c*(b + a*x))/x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 b \int \frac {\log {\left (a c + \frac {b c}{x} \right )}}{a x + b}\, dx + x \log {\left (\frac {c \left (a x + b\right )}{x} \right )}^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(a*x+b)/x)**2,x)

[Out]

2*b*Integral(log(a*c + b*c/x)/(a*x + b), x) + x*log(c*(a*x + b)/x)**2

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