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

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

[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 = 0.64, antiderivative size = 246, normalized size of antiderivative = 1.97, number of steps used = 22, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2523, 12, 2528, 2524, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac {2 a B^2 \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b}+\frac {2 B^2 c \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{d}+\frac {2 a B \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b}-\frac {2 B c \log (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d}+x \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2+\frac {2 B^2 c \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{d}+\frac {2 a B^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b}-\frac {a B^2 \log ^2(a+b x)}{b}-\frac {B^2 c \log ^2(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, 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 2391

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

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2524

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

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.21, size = 214, normalized size = 1.71 \[ \frac {B \left (2 a d \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )-2 b c \log (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )-a B d \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )\right )+b B c \left (2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac {d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )\right )}{b d}+x \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2 \]

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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fricas [F]  time = 1.17, size = 0, normalized size = 0.00 \[ {\rm integral}\left (B^{2} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, A B \log \left (\frac {b e x + a e}{d x + c}\right ) + A^{2}, x\right ) \]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F]  time = 1.52, size = 0, normalized size = 0.00 \[ \int \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, {\left (x \log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right ) + \frac {\frac {a e \log \left (b x + a\right )}{b} - \frac {c e \log \left (d x + c\right )}{d}}{e}\right )} A B + A^{2} x + B^{2} {\left (\frac {b d x \log \left (b x + a\right )^{2} + {\left (b d x + b c\right )} \log \left (d x + c\right )^{2} - 2 \, {\left (b d x \log \relax (e) + {\left (b d x + a d\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{b d} + \int \frac {{\left (\log \relax (e)^{2} + 2 \, \log \relax (e)\right )} b^{2} d x^{2} + a b c \log \relax (e)^{2} + {\left (b^{2} c \log \relax (e)^{2} + {\left (\log \relax (e)^{2} + 2 \, \log \relax (e)\right )} a b d\right )} x + 2 \, {\left (b^{2} d x^{2} \log \relax (e) + a b c \log \relax (e) + a^{2} d + {\left (a b d {\left (\log \relax (e) + 2\right )} + b^{2} c {\left (\log \relax (e) - 1\right )}\right )} x\right )} \log \left (b x + a\right )}{b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x}\,{d x}\right )} \]

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*(x*log((b*x + a)*e/(d*x + c)) + (a*e*log(b*x + a)/b - c*e*log(d*x + c)/d)/e)*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*log(e) + (b*d*x + a*d)*log(b*x + a))*log(d*x + c))/(b*d) +
integrate(((log(e)^2 + 2*log(e))*b^2*d*x^2 + a*b*c*log(e)^2 + (b^2*c*log(e)^2 + (log(e)^2 + 2*log(e))*a*b*d)*
x + 2*(b^2*d*x^2*log(e) + a*b*c*log(e) + a^2*d + (a*b*d*(log(e) + 2) + b^2*c*(log(e) - 1))*x)*log(b*x + a))/(b
^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x), x))

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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