\(\int (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2 \, dx\) [275]

Optimal result

Integrand size = 23, antiderivative size = 129 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}+\frac {8 B^2 (b c-a d) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]

[Out]

(b*x+a)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2/b+4*B*(-a*d+b*c)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))*ln((-a*d+b*c)/b/(d*
x+c))/b/d+8*B^2*(-a*d+b*c)*polylog(2,d*(b*x+a)/b/(d*x+c))/b/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2536, 2542, 2458, 2378, 2370, 2352} \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\frac {4 B (b c-a d) \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b d}+\frac {(a+b x) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{b}+\frac {8 B^2 (b c-a d) \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]

[In]

Int[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2,x]

[Out]

((a + b*x)*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2)/b + (4*B*(b*c - a*d)*(A + B*Log[(e*(a + b*x)^2)/(c + d*
x)^2])*Log[(b*c - a*d)/(b*(c + d*x))])/(b*d) + (8*B^2*(b*c - a*d)*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))])/(b*
d)

Rule 2352

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

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2378

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

Rule 2458

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 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rule 2542

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

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}-\frac {(4 B (b c-a d)) \int \frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{c+d x} \, dx}{b} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}-\frac {\left (8 B^2 (b c-a d)^2\right ) \int \frac {\log \left (\frac {b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{b d} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}-\frac {\left (8 B^2 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {b c-a d}{b x}\right )}{x \left (\frac {-b c+a d}{d}+\frac {b x}{d}\right )} \, dx,x,c+d x\right )}{b d^2} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}+\frac {\left (8 B^2 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {(b c-a d) x}{b}\right )}{\left (\frac {-b c+a d}{d}+\frac {b}{d x}\right ) x} \, dx,x,\frac {1}{c+d x}\right )}{b d^2} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}+\frac {\left (8 B^2 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {(b c-a d) x}{b}\right )}{\frac {b}{d}+\frac {(-b c+a d) x}{d}} \, dx,x,\frac {1}{c+d x}\right )}{b d^2} \\ & = \frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{b}+\frac {4 B (b c-a d) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{b d}+\frac {8 B^2 (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.71 \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=x \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2+\frac {4 B \left (a d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )-b c \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \log (c+d x)-a B d \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+b B c \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b d} \]

[In]

Integrate[(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2,x]

[Out]

x*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2 + (4*B*(a*d*Log[a + b*x]*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])
- b*c*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])*Log[c + d*x] - a*B*d*(Log[a + b*x]*(Log[a + b*x] - 2*Log[(b*(c
+ d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]) + b*B*c*((2*Log[(d*(a + b*x))/(-(b*c) + a
*d)] - Log[c + d*x])*Log[c + d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])))/(b*d)

Maple [F]

\[\int {\left (A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )}^{2}d x\]

[In]

int((A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2,x)

[Out]

int((A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2,x)

Fricas [F]

\[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

integral(B^2*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2))^2 + 2*A*B*log((b^2*e*x^2 + 2*a*b*e
*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2)) + A^2, x)

Sympy [F(-1)]

Timed out. \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\text {Timed out} \]

[In]

integrate((A+B*ln(e*(b*x+a)**2/(d*x+c)**2))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

2*(x*log((b*x + a)^2*e/(d*x + c)^2) + 2*(a*e*log(b*x + a)/b - c*e*log(d*x + c)/d)/e)*A*B + A^2*x + B^2*(4*(b*d
*x*log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^2 - (b*d*x*log(e) + 2*(b*d*x + a*d)*log(b*x + a))*log(d*x + c))
/(b*d) + integrate(((log(e)^2 + 4*log(e))*b^2*d*x^2 + a*b*c*log(e)^2 + (b^2*c*log(e)^2 + (log(e)^2 + 4*log(e))
*a*b*d)*x + 4*(b^2*d*x^2*log(e) + a*b*c*log(e) + 2*a^2*d + (a*b*d*(log(e) + 4) + b^2*c*(log(e) - 2))*x)*log(b*
x + a))/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x))

Giac [F]

\[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int { {\left (B \log \left (\frac {{\left (b x + a\right )}^{2} e}{{\left (d x + c\right )}^{2}}\right ) + A\right )}^{2} \,d x } \]

[In]

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

[Out]

integrate((B*log((b*x + a)^2*e/(d*x + c)^2) + A)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \, dx=\int {\left (A+B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\right )}^2 \,d x \]

[In]

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2,x)

[Out]

int((A + B*log((e*(a + b*x)^2)/(c + d*x)^2))^2, x)