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

Optimal. Leaf size=203 \[ \frac {B i (b c-a d)^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^2 d}-\frac {B i (a+b x) (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^2}+\frac {i (c+d x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{2 d}-\frac {B^2 i (b c-a d)^2 \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b^2 d}+\frac {B^2 i (b c-a d)^2 \log (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 283, normalized size of antiderivative = 1.39, 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, 2390, 2301, 2394, 2393, 2391} \[ -\frac {B^2 i (b c-a d)^2 \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 d}-\frac {B i (b c-a d)^2 \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^2 d}+\frac {i (c+d x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2}{2 d}-\frac {A B i x (b c-a d)}{b}-\frac {B^2 i (a+b x) (b c-a d) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}+\frac {B^2 i (b c-a d)^2 \log ^2(a+b x)}{2 b^2 d}+\frac {B^2 i (b c-a d)^2 \log (c+d x)}{b^2 d}-\frac {B^2 i (b c-a d)^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*B*(b*c - a*d)*i*x)/b) + (B^2*(b*c - a*d)^2*i*Log[a + b*x]^2)/(2*b^2*d) - (B^2*(b*c - a*d)*i*(a + b*x)*Log
[(e*(a + b*x))/(c + d*x)])/b^2 - (B*(b*c - a*d)^2*i*Log[a + b*x]*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b^2*d)
+ (i*(c + d*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)])^2)/(2*d) + (B^2*(b*c - a*d)^2*i*Log[c + d*x])/(b^2*d) -
(B^2*(b*c - a*d)^2*i*Log[a + b*x]*Log[(b*(c + d*x))/(b*c - a*d)])/(b^2*d) - (B^2*(b*c - a*d)^2*i*PolyLog[2, -
((d*(a + b*x))/(b*c - a*d))])/(b^2*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 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 (58 c+58 d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2 \, dx &=\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}-\frac {B \int \frac {3364 (b c-a d) (c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{a+b x} \, dx}{58 d}\\ &=\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}-\frac {(58 B (b c-a d)) \int \frac {(c+d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{a+b x} \, dx}{d}\\ &=\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}-\frac {(58 B (b c-a d)) \int \left (\frac {d \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b}+\frac {(b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b (a+b x)}\right ) \, dx}{d}\\ &=\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}-\frac {(58 B (b c-a d)) \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx}{b}-\frac {\left (58 B (b c-a d)^2\right ) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{a+b x} \, dx}{b d}\\ &=-\frac {58 A B (b c-a d) x}{b}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}-\frac {\left (58 B^2 (b c-a d)\right ) \int \log \left (\frac {e (a+b x)}{c+d x}\right ) \, dx}{b}+\frac {\left (58 B^2 (b c-a d)^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^2 d}\\ &=-\frac {58 A B (b c-a d) x}{b}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {\left (58 B^2 (b c-a d)^2\right ) \int \frac {1}{c+d x} \, dx}{b^2}+\frac {\left (58 B^2 (b c-a d)^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^2 d e}\\ &=-\frac {58 A B (b c-a d) x}{b}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {58 B^2 (b c-a d)^2 \log (c+d x)}{b^2 d}+\frac {\left (58 B^2 (b c-a d)^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^2 d e}\\ &=-\frac {58 A B (b c-a d) x}{b}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {58 B^2 (b c-a d)^2 \log (c+d x)}{b^2 d}-\frac {\left (58 B^2 (b c-a d)^2\right ) \int \frac {\log (a+b x)}{c+d x} \, dx}{b^2}+\frac {\left (58 B^2 (b c-a d)^2\right ) \int \frac {\log (a+b x)}{a+b x} \, dx}{b d}\\ &=-\frac {58 A B (b c-a d) x}{b}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {58 B^2 (b c-a d)^2 \log (c+d x)}{b^2 d}-\frac {58 B^2 (b c-a d)^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 d}+\frac {\left (58 B^2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{b^2 d}+\frac {\left (58 B^2 (b c-a d)^2\right ) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{b d}\\ &=-\frac {58 A B (b c-a d) x}{b}+\frac {29 B^2 (b c-a d)^2 \log ^2(a+b x)}{b^2 d}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {58 B^2 (b c-a d)^2 \log (c+d x)}{b^2 d}-\frac {58 B^2 (b c-a d)^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 d}+\frac {\left (58 B^2 (b c-a d)^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^2 d}\\ &=-\frac {58 A B (b c-a d) x}{b}+\frac {29 B^2 (b c-a d)^2 \log ^2(a+b x)}{b^2 d}-\frac {58 B^2 (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2}-\frac {58 B (b c-a d)^2 \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 d}+\frac {29 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2}{d}+\frac {58 B^2 (b c-a d)^2 \log (c+d x)}{b^2 d}-\frac {58 B^2 (b c-a d)^2 \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 d}-\frac {58 B^2 (b c-a d)^2 \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 205, normalized size = 1.01 \[ \frac {i \left ((c+d x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )^2-\frac {B (b c-a d) \left (2 (b c-a d) \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+B \log \left (\frac {b (c+d x)}{b c-a d}\right )+A\right )+2 \left (B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )+\log (c+d x) (a B d-b B c)+A b d x\right )+2 B (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )+\log ^2(a+b x) (a B d-b B c)\right )}{b^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (A^{2} d i x + A^{2} c i + {\left (B^{2} d i x + B^{2} c i\right )} \log \left (\frac {b e x + a e}{d x + c}\right )^{2} + 2 \, {\left (A B d i x + A B c i\right )} \log \left (\frac {b e x + a e}{d x + c}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [F]  time = 1.58, size = 0, normalized size = 0.00 \[ \int \left (d i x +c i \right ) \left (B \ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )+A \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.97, size = 633, normalized size = 3.12 \[ \frac {1}{2} \, A^{2} d i x^{2} + 2 \, {\left (x \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right ) + \frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} A B c i + {\left (x^{2} \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\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 d i + A^{2} c i x - \frac {{\left ({\left (i \log \relax (e) - i\right )} b c^{2} + a c d i\right )} B^{2} \log \left (d x + c\right )}{b d} - \frac {{\left (b^{2} c^{2} i - 2 \, a b c d i + a^{2} d^{2} i\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^{2} d} + \frac {B^{2} b^{2} d^{2} i x^{2} \log \relax (e)^{2} + 2 \, {\left (a b d^{2} i \log \relax (e) + {\left (i \log \relax (e)^{2} - i \log \relax (e)\right )} b^{2} c d\right )} B^{2} x + {\left (B^{2} b^{2} d^{2} i x^{2} + 2 \, B^{2} b^{2} c d i x + {\left (2 \, a b c d i - a^{2} d^{2} i\right )} B^{2}\right )} \log \left (b x + a\right )^{2} + {\left (B^{2} b^{2} d^{2} i x^{2} + 2 \, B^{2} b^{2} c d i x + B^{2} b^{2} c^{2} i\right )} \log \left (d x + c\right )^{2} + 2 \, {\left (B^{2} b^{2} d^{2} i x^{2} \log \relax (e) + {\left ({\left (2 \, i \log \relax (e) - i\right )} b^{2} c d + a b d^{2} i\right )} B^{2} x + {\left ({\left (2 \, i \log \relax (e) - i\right )} a b c d - {\left (i \log \relax (e) - i\right )} a^{2} d^{2}\right )} B^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (B^{2} b^{2} d^{2} i x^{2} \log \relax (e) + {\left ({\left (2 \, i \log \relax (e) - i\right )} b^{2} c d + a b d^{2} i\right )} B^{2} x + {\left (B^{2} b^{2} d^{2} i x^{2} + 2 \, B^{2} b^{2} c d i x + {\left (2 \, a b c d i - a^{2} d^{2} i\right )} B^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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