3.1.58 \(\int (c i+d i x) (A+B \log (\frac {e (a+b x)}{c+d x}))^2 \, dx\) [58]

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

[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 = 0.15, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2552, 2356, 2389, 2379, 2438, 2351, 31} \begin {gather*} -\frac {B^2 i (b c-a d)^2 \text {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^2 d}+\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 \log (c+d x)}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*(b*c - a*d)*i*(a + b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/b^2) + (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*(b*c - a*d)^2*i*(A + B*Log[(e*(a +
b*x))/(c + d*x)])*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/(b^2*d) - (B^2*(b*c - a*d)^2*i*PolyLog[2, (b*(c + d*x)
)/(d*(a + b*x))])/(b^2*d)

Rule 31

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

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

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2438

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 2552

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x],
x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && EqQ[n + mn, 0] && IGtQ[n, 0] && NeQ
[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

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 ) \text {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 ) \text {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 = 0.14, size = 205, normalized size = 1.01 \begin {gather*} \frac {i \left ((c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )^2-\frac {B (b c-a d) \left ((-b B c+a B d) \log ^2(a+b x)+2 \left (A b d x+B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )+(-b B c+a B d) \log (c+d x)\right )+2 (b c-a d) \log (a+b x) \left (A+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 )\right )+2 B (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{-b c+a d}\right )\right )}{b^2}\right )}{2 d} \end {gather*}

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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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (d i x +c i \right ) \left (A +B \ln \left (\frac {e \left (b x +a \right )}{d x +c}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (200) = 400\).
time = 0.36, size = 534, normalized size = 2.63 \begin {gather*} \frac {1}{2} i \, A^{2} d x^{2} + 2 i \, {\left (x \log \left (\frac {b x e}{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 x e}{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 x - \frac {i \, B^{2} a c \log \left (d x + c\right )}{b} - \frac {{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\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 {i \, B^{2} b^{2} d^{2} x^{2} + 2 i \, B^{2} a b d^{2} x + {\left (i \, B^{2} b^{2} d^{2} x^{2} + 2 i \, B^{2} b^{2} c d x + {\left (2 i \, a b c d - i \, a^{2} d^{2}\right )} B^{2}\right )} \log \left (b x + a\right )^{2} + {\left (i \, B^{2} b^{2} d^{2} x^{2} + 2 i \, B^{2} b^{2} c d x + i \, B^{2} b^{2} c^{2}\right )} \log \left (d x + c\right )^{2} - 2 \, {\left (-i \, B^{2} b^{2} d^{2} x^{2} - i \, B^{2} a b c d + {\left (-i \, b^{2} c d - i \, a b d^{2}\right )} B^{2} x\right )} \log \left (b x + a\right ) - 2 \, {\left (i \, B^{2} b^{2} d^{2} x^{2} + {\left (i \, b^{2} c d + i \, a b d^{2}\right )} B^{2} x + {\left (i \, B^{2} b^{2} d^{2} x^{2} + 2 i \, B^{2} b^{2} c d x + {\left (2 i \, a b c d - i \, a^{2} d^{2}\right )} B^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b^{2} d} \end {gather*}

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

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

[Out]

1/2*(I*B^2*d*x^2 + 2*I*B^2*c*x)*log((b*x + a)*e/(d*x + c))^2 + integral((I*A^2*b*d^2*x^3 + I*A^2*a*c^2 + (2*I*
A^2*b*c*d + I*A^2*a*d^2)*x^2 + (I*A^2*b*c^2 + 2*I*A^2*a*c*d)*x + (2*I*A*B*b*d^2*x^3 + 2*I*A*B*a*c^2 + ((4*I*A*
B - I*B^2)*b*c*d + (2*I*A*B + I*B^2)*a*d^2)*x^2 - 2*((-I*A*B + I*B^2)*b*c^2 + (-2*I*A*B - I*B^2)*a*c*d)*x)*log
((b*x + a)*e/(d*x + c)))/(b*d*x^2 + a*c + (b*c + a*d)*x), 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((d*i*x+c*i)*(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((d*i*x+c*i)*(A+B*log(e*(b*x+a)/(d*x+c)))^2,x, algorithm="giac")

[Out]

integrate((I*d*x + I*c)*(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 = -0.00 \begin {gather*} \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 \end {gather*}

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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