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

Optimal. Leaf size=202 \[ \frac {B g (b c-a d)^2 \log \left (1-\frac {d (a+b x)}{b (c+d x)}\right ) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )}{b d^2}+\frac {B g (c+d x) (b c-a d) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )}{d^2}+\frac {g (a+b x)^2 \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )^2}{2 b}-\frac {B^2 g (b c-a d)^2 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{b d^2}+\frac {B^2 g (b c-a d)^2 \log (a+b x)}{b d^2} \]

[Out]

B^2*(-a*d+b*c)^2*g*ln(b*x+a)/b/d^2+B*(-a*d+b*c)*g*(d*x+c)*(A+B*ln(e*(d*x+c)/(b*x+a)))/d^2+1/2*g*(b*x+a)^2*(A+B
*ln(e*(d*x+c)/(b*x+a)))^2/b+B*(-a*d+b*c)^2*g*(A+B*ln(e*(d*x+c)/(b*x+a)))*ln(1-d*(b*x+a)/b/(d*x+c))/b/d^2-B^2*(
-a*d+b*c)^2*g*polylog(2,d*(b*x+a)/b/(d*x+c))/b/d^2

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Rubi [A]  time = 0.42, antiderivative size = 284, normalized size of antiderivative = 1.41, number of steps used = 16, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2525, 12, 2528, 2486, 31, 2524, 2418, 2394, 2393, 2391, 2390, 2301} \[ -\frac {B^2 g (b c-a d)^2 \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{b d^2}-\frac {B g (b c-a d)^2 \log (c+d x) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )}{b d^2}+\frac {g (a+b x)^2 \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )^2}{2 b}+\frac {A B g x (b c-a d)}{d}+\frac {B^2 g (b c-a d)^2 \log ^2(c+d x)}{2 b d^2}+\frac {B^2 g (b c-a d)^2 \log (c+d x)}{b d^2}-\frac {B^2 g (b c-a d)^2 \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{b d^2}+\frac {B^2 g (a+b x) (b c-a d) \log \left (\frac {e (c+d x)}{a+b x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 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 2525

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

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

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Mathematica [A]  time = 0.18, size = 203, normalized size = 1.00 \[ \frac {g \left (\frac {B (b c-a d) \left (-2 (b c-a d) \log (c+d x) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )+2 b B (c+d x) \log \left (\frac {e (c+d x)}{a+b x}\right )-B (b c-a d) \left (2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac {d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )+2 B (b c-a d) \log (a+b x)+2 A b d x\right )}{d^2}+(a+b x)^2 \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (A^{2} b g x + A^{2} a g + {\left (B^{2} b g x + B^{2} a g\right )} \log \left (\frac {d e x + c e}{b x + a}\right )^{2} + 2 \, {\left (A B b g x + A B a g\right )} \log \left (\frac {d e x + c e}{b x + a}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int \left (b g x +a g \right ) \left (B \ln \left (\frac {\left (d x +c \right ) e}{b x +a}\right )+A \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.97, size = 619, normalized size = 3.06 \[ \frac {1}{2} \, A^{2} b g x^{2} + 2 \, {\left (x \log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right ) - \frac {a \log \left (b x + a\right )}{b} + \frac {c \log \left (d x + c\right )}{d}\right )} A B a g + {\left (x^{2} \log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right ) + \frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} A B b g + A^{2} a g x - \frac {{\left ({\left (g \log \relax (e) - g\right )} b c^{2} - {\left (2 \, g \log \relax (e) - g\right )} a c d\right )} B^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b^{2} c^{2} g - 2 \, a b c d g + a^{2} d^{2} g\right )} {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} B^{2}}{b d^{2}} + \frac {B^{2} b^{2} d^{2} g x^{2} \log \relax (e)^{2} + 2 \, {\left (b^{2} c d g \log \relax (e) + {\left (g \log \relax (e)^{2} - g \log \relax (e)\right )} a b d^{2}\right )} B^{2} x + {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x + B^{2} a^{2} d^{2} g\right )} \log \left (b x + a\right )^{2} + {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x - {\left (b^{2} c^{2} g - 2 \, a b c d g\right )} B^{2}\right )} \log \left (d x + c\right )^{2} - 2 \, {\left (B^{2} b^{2} d^{2} g x^{2} \log \relax (e) + {\left ({\left (2 \, g \log \relax (e) - g\right )} a b d^{2} + b^{2} c d g\right )} B^{2} x + {\left ({\left (g \log \relax (e) - g\right )} a^{2} d^{2} + a b c d g\right )} B^{2}\right )} \log \left (b x + a\right ) + 2 \, {\left (B^{2} b^{2} d^{2} g x^{2} \log \relax (e) + {\left ({\left (2 \, g \log \relax (e) - g\right )} a b d^{2} + b^{2} c d g\right )} B^{2} x - {\left (B^{2} b^{2} d^{2} g x^{2} + 2 \, B^{2} a b d^{2} g x + B^{2} a^{2} d^{2} g\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A^2*b*g*x^2 + 2*(x*log(d*e*x/(b*x + a) + c*e/(b*x + a)) - a*log(b*x + a)/b + c*log(d*x + c)/d)*A*B*a*g + (
x^2*log(d*e*x/(b*x + a) + c*e/(b*x + a)) + a^2*log(b*x + a)/b^2 - c^2*log(d*x + c)/d^2 + (b*c - a*d)*x/(b*d))*
A*B*b*g + A^2*a*g*x - ((g*log(e) - g)*b*c^2 - (2*g*log(e) - g)*a*c*d)*B^2*log(d*x + c)/d^2 + (b^2*c^2*g - 2*a*
b*c*d*g + a^2*d^2*g)*(log(b*x + a)*log((b*d*x + a*d)/(b*c - a*d) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))*B^2
/(b*d^2) + 1/2*(B^2*b^2*d^2*g*x^2*log(e)^2 + 2*(b^2*c*d*g*log(e) + (g*log(e)^2 - g*log(e))*a*b*d^2)*B^2*x + (B
^2*b^2*d^2*g*x^2 + 2*B^2*a*b*d^2*g*x + B^2*a^2*d^2*g)*log(b*x + a)^2 + (B^2*b^2*d^2*g*x^2 + 2*B^2*a*b*d^2*g*x
- (b^2*c^2*g - 2*a*b*c*d*g)*B^2)*log(d*x + c)^2 - 2*(B^2*b^2*d^2*g*x^2*log(e) + ((2*g*log(e) - g)*a*b*d^2 + b^
2*c*d*g)*B^2*x + ((g*log(e) - g)*a^2*d^2 + a*b*c*d*g)*B^2)*log(b*x + a) + 2*(B^2*b^2*d^2*g*x^2*log(e) + ((2*g*
log(e) - g)*a*b*d^2 + b^2*c*d*g)*B^2*x - (B^2*b^2*d^2*g*x^2 + 2*B^2*a*b*d^2*g*x + B^2*a^2*d^2*g)*log(b*x + a))
*log(d*x + c))/(b*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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