3.2.47 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x) (c i+d i x)^2} \, dx\) [147]

Optimal. Leaf size=166 \[ -\frac {A d (a+b x)}{(b c-a d)^2 g i^2 (c+d x)}+\frac {B d n (a+b x)}{(b c-a d)^2 g i^2 (c+d x)}-\frac {B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(b c-a d)^2 g i^2 (c+d x)}+\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{2 B (b c-a d)^2 g i^2 n} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {2561, 2388, 2338, 2332} \begin {gather*} \frac {b \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{2 B g i^2 n (b c-a d)^2}-\frac {A d (a+b x)}{g i^2 (c+d x) (b c-a d)^2}-\frac {B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g i^2 (c+d x) (b c-a d)^2}+\frac {B d n (a+b x)}{g i^2 (c+d x) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2338

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 2388

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

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(147 c+147 d x)^2 (a g+b g x)} \, dx &=\int \left (\frac {b^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g (a+b x)}-\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d) g (c+d x)^2}-\frac {b d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g (c+d x)}\right ) \, dx\\ &=\frac {b^2 \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a+b x} \, dx}{21609 (b c-a d)^2 g}-\frac {(b d) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c+d x} \, dx}{21609 (b c-a d)^2 g}-\frac {d \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c+d x)^2} \, dx}{21609 (b c-a d) g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {(b B n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (a+b x)}{a+b x} \, dx}{21609 (b c-a d)^2 g}+\frac {(b B n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{21609 (b c-a d)^2 g}-\frac {(B n) \int \frac {b c-a d}{(a+b x) (c+d x)^2} \, dx}{21609 (b c-a d) g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {(B n) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{21609 g}-\frac {(b B n) \int \left (\frac {b \log (a+b x)}{a+b x}-\frac {d \log (a+b x)}{c+d x}\right ) \, dx}{21609 (b c-a d)^2 g}+\frac {(b B n) \int \left (\frac {b \log (c+d x)}{a+b x}-\frac {d \log (c+d x)}{c+d x}\right ) \, dx}{21609 (b c-a d)^2 g}\\ &=\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {(B n) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{21609 g}-\frac {\left (b^2 B n\right ) \int \frac {\log (a+b x)}{a+b x} \, dx}{21609 (b c-a d)^2 g}+\frac {\left (b^2 B n\right ) \int \frac {\log (c+d x)}{a+b x} \, dx}{21609 (b c-a d)^2 g}+\frac {(b B d n) \int \frac {\log (a+b x)}{c+d x} \, dx}{21609 (b c-a d)^2 g}-\frac {(b B d n) \int \frac {\log (c+d x)}{c+d x} \, dx}{21609 (b c-a d)^2 g}\\ &=-\frac {B n}{21609 (b c-a d) g (c+d x)}-\frac {b B n \log (a+b x)}{21609 (b c-a d)^2 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}+\frac {b B n \log (c+d x)}{21609 (b c-a d)^2 g}+\frac {b B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}+\frac {b B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{21609 (b c-a d)^2 g}-\frac {(b B n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{21609 (b c-a d)^2 g}-\frac {(b B n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{21609 (b c-a d)^2 g}-\frac {\left (b^2 B n\right ) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{21609 (b c-a d)^2 g}-\frac {(b B d n) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{21609 (b c-a d)^2 g}\\ &=-\frac {B n}{21609 (b c-a d) g (c+d x)}-\frac {b B n \log (a+b x)}{21609 (b c-a d)^2 g}-\frac {b B n \log ^2(a+b x)}{43218 (b c-a d)^2 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}+\frac {b B n \log (c+d x)}{21609 (b c-a d)^2 g}+\frac {b B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {b B n \log ^2(c+d x)}{43218 (b c-a d)^2 g}+\frac {b B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{21609 (b c-a d)^2 g}-\frac {(b B n) \text {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{21609 (b c-a d)^2 g}-\frac {(b B n) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{21609 (b c-a d)^2 g}\\ &=-\frac {B n}{21609 (b c-a d) g (c+d x)}-\frac {b B n \log (a+b x)}{21609 (b c-a d)^2 g}-\frac {b B n \log ^2(a+b x)}{43218 (b c-a d)^2 g}+\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{21609 (b c-a d) g (c+d x)}+\frac {b \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21609 (b c-a d)^2 g}+\frac {b B n \log (c+d x)}{21609 (b c-a d)^2 g}+\frac {b B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21609 (b c-a d)^2 g}-\frac {b B n \log ^2(c+d x)}{43218 (b c-a d)^2 g}+\frac {b B n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{21609 (b c-a d)^2 g}+\frac {b B n \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{21609 (b c-a d)^2 g}+\frac {b B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{21609 (b c-a d)^2 g}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.21, size = 304, normalized size = 1.83 \begin {gather*} \frac {2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 b (c+d x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 b (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)-2 B n (b c-a d+b (c+d x) \log (a+b x)-b (c+d x) \log (c+d x))-b B n (c+d x) \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 n (c+d x) \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 )}{2 (b c-a d)^2 g i^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{\left (b g x +a g \right ) \left (d i x +c i \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (153) = 306\).
time = 0.29, size = 385, normalized size = 2.32 \begin {gather*} -B {\left (\frac {b \log \left (b x + a\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g} - \frac {b \log \left (d x + c\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g} + \frac {1}{{\left (b c d - a d^{2}\right )} g x + {\left (b c^{2} - a c d\right )} g}\right )} \log \left ({\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n} e\right ) + \frac {{\left ({\left (b d x + b c\right )} \log \left (b x + a\right )^{2} + {\left (b d x + b c\right )} \log \left (d x + c\right )^{2} + 2 \, b c - 2 \, a d + 2 \, {\left (b d x + b c\right )} \log \left (b x + a\right ) - 2 \, {\left (b d x + b c + {\left (b d x + b c\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} B n}{2 \, {\left (b^{2} c^{3} g - 2 \, a b c^{2} d g + a^{2} c d^{2} g + {\left (b^{2} c^{2} d g - 2 \, a b c d^{2} g + a^{2} d^{3} g\right )} x\right )}} - A {\left (\frac {b \log \left (b x + a\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g} - \frac {b \log \left (d x + c\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} g} + \frac {1}{{\left (b c d - a d^{2}\right )} g x + {\left (b c^{2} - a c d\right )} g}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*(b*log(b*x + a)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*g) - b*log(d*x + c)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*g) +
1/((b*c*d - a*d^2)*g*x + (b*c^2 - a*c*d)*g))*log((b*x/(d*x + c) + a/(d*x + c))^n*e) + 1/2*((b*d*x + b*c)*log(b
*x + a)^2 + (b*d*x + b*c)*log(d*x + c)^2 + 2*b*c - 2*a*d + 2*(b*d*x + b*c)*log(b*x + a) - 2*(b*d*x + b*c + (b*
d*x + b*c)*log(b*x + a))*log(d*x + c))*B*n/(b^2*c^3*g - 2*a*b*c^2*d*g + a^2*c*d^2*g + (b^2*c^2*d*g - 2*a*b*c*d
^2*g + a^2*d^3*g)*x) - A*(b*log(b*x + a)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*g) - b*log(d*x + c)/((b^2*c^2 - 2*a*
b*c*d + a^2*d^2)*g) + 1/((b*c*d - a*d^2)*g*x + (b*c^2 - a*c*d)*g))

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 159, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (A + B\right )} b c - 2 \, {\left (A + B\right )} a d + {\left (B b d n x + B b c n\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} - 2 \, {\left (B b c - B a d\right )} n - 2 \, {\left (B a d n - {\left (A + B\right )} b c + {\left (B b d n - {\left (A + B\right )} b d\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, {\left ({\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} g x + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} g\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(A + B)*b*c - 2*(A + B)*a*d + (B*b*d*n*x + B*b*c*n)*log((b*x + a)/(d*x + c))^2 - 2*(B*b*c - B*a*d)*n -
2*(B*a*d*n - (A + B)*b*c + (B*b*d*n - (A + B)*b*d)*x)*log((b*x + a)/(d*x + c)))/((b^2*c^2*d - 2*a*b*c*d^2 + a
^2*d^3)*g*x + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*g)

________________________________________________________________________________________

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))**n))/(b*g*x+a*g)/(d*i*x+c*i)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 4.07, size = 180, normalized size = 1.08 \begin {gather*} -\frac {1}{2} \, {\left (\frac {B b n \log \left (\frac {b x + a}{d x + c}\right )^{2}}{b c g - a d g} - \frac {2 \, {\left (b x + a\right )} B d n \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b c g - a d g\right )} {\left (d x + c\right )}} + \frac {2 \, {\left (A b + B b\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b c g - a d g} + \frac {2 \, {\left (B d n - A d - B d\right )} {\left (b x + a\right )}}{{\left (b c g - a d g\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.82, size = 241, normalized size = 1.45 \begin {gather*} \frac {B\,n}{g\,i^2\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )}-\frac {A}{g\,i^2\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{g\,i^2\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )}+\frac {B\,b\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{2\,g\,i^2\,n\,{\left (a\,d-b\,c\right )}^2}-\frac {A\,b\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{g\,i^2\,{\left (a\,d-b\,c\right )}^2}+\frac {B\,b\,n\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{g\,i^2\,{\left (a\,d-b\,c\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________