3.3.43 \(\int (A+B \log (\frac {e (a+b x)}{c+d x}))^2 \, dx\) [243]

Optimal. Leaf size=125 \[ \frac {2 B (b c-a d) \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b d}+\frac {(a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{b}+\frac {2 B^2 (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2536, 2544, 2458, 2378, 2370, 2352} \begin {gather*} \frac {2 B^2 (b c-a d) \text {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{b d}+\frac {2 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)}{c+d x}\right )+A\right )}{b d}+\frac {(a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(b*c - a*d)*Log[(b*c - a*d)/(b*(c + d*x))]*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b*d) + ((a + b*x)*(A +
B*Log[(e*(a + b*x))/(c + d*x)])^2)/b + (2*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 2544

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)/(b*(c + d*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)/(b*(c + d*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[d*f - c*g, 0]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 214, normalized size = 1.71 \begin {gather*} x \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2+\frac {B \left (2 a d \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 b c \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\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 \text {Li}_2\left (\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 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (A +B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((a*e*log(b*x + a)/b - c*e*log(d*x + c)/d)*e^(-1) + x*log((b*x + a)*e/(d*x + c)))*A*B + A^2*x + B^2*((b*d*x*
log(b*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^2 - 2*(b*d*x + (b*d*x + a*d)*log(b*x + a))*log(d*x + c))/(b*d) + i
ntegrate((3*b^2*d*x^2 + a*b*c + (b^2*c + 3*a*b*d)*x + 2*(b^2*d*x^2 + 3*a*b*d*x + a*b*c + a^2*d)*log(b*x + a))/
(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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