3.154 \(\int \frac{A+B \log (e (a+b x)^n (c+d x)^{-n})}{(a+b x)^4} \, dx\)

Optimal. Leaf size=166 \[ -\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{3 b (a+b x)^3}-\frac{B d^2 n}{3 b (a+b x) (b c-a d)^2}-\frac{B d^3 n \log (a+b x)}{3 b (b c-a d)^3}+\frac{B d^3 n \log (c+d x)}{3 b (b c-a d)^3}+\frac{B d n}{6 b (a+b x)^2 (b c-a d)}-\frac{B n}{9 b (a+b x)^3} \]

[Out]

-(B*n)/(9*b*(a + b*x)^3) + (B*d*n)/(6*b*(b*c - a*d)*(a + b*x)^2) - (B*d^2*n)/(3*b*(b*c - a*d)^2*(a + b*x)) - (
B*d^3*n*Log[a + b*x])/(3*b*(b*c - a*d)^3) + (B*d^3*n*Log[c + d*x])/(3*b*(b*c - a*d)^3) - (A + B*Log[(e*(a + b*
x)^n)/(c + d*x)^n])/(3*b*(a + b*x)^3)

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Rubi [A]  time = 0.169881, antiderivative size = 178, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {6742, 2492, 44} \[ -\frac{A}{3 b (a+b x)^3}-\frac{B d^2 n}{3 b (a+b x) (b c-a d)^2}-\frac{B d^3 n \log (a+b x)}{3 b (b c-a d)^3}+\frac{B d^3 n \log (c+d x)}{3 b (b c-a d)^3}-\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b (a+b x)^3}+\frac{B d n}{6 b (a+b x)^2 (b c-a d)}-\frac{B n}{9 b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(a + b*x)^4,x]

[Out]

-A/(3*b*(a + b*x)^3) - (B*n)/(9*b*(a + b*x)^3) + (B*d*n)/(6*b*(b*c - a*d)*(a + b*x)^2) - (B*d^2*n)/(3*b*(b*c -
a*d)^2*(a + b*x)) - (B*d^3*n*Log[a + b*x])/(3*b*(b*c - a*d)^3) + (B*d^3*n*Log[c + d*x])/(3*b*(b*c - a*d)^3) -
(B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(3*b*(a + b*x)^3)

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2492

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^4} \, dx &=\int \left (\frac{A}{(a+b x)^4}+\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^4}\right ) \, dx\\ &=-\frac{A}{3 b (a+b x)^3}+B \int \frac{\log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^4} \, dx\\ &=-\frac{A}{3 b (a+b x)^3}-\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b (a+b x)^3}+\frac{(B (b c-a d) n) \int \frac{1}{(a+b x)^4 (c+d x)} \, dx}{3 b}\\ &=-\frac{A}{3 b (a+b x)^3}-\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b (a+b x)^3}+\frac{(B (b c-a d) n) \int \left (\frac{b}{(b c-a d) (a+b x)^4}-\frac{b d}{(b c-a d)^2 (a+b x)^3}+\frac{b d^2}{(b c-a d)^3 (a+b x)^2}-\frac{b d^3}{(b c-a d)^4 (a+b x)}+\frac{d^4}{(b c-a d)^4 (c+d x)}\right ) \, dx}{3 b}\\ &=-\frac{A}{3 b (a+b x)^3}-\frac{B n}{9 b (a+b x)^3}+\frac{B d n}{6 b (b c-a d) (a+b x)^2}-\frac{B d^2 n}{3 b (b c-a d)^2 (a+b x)}-\frac{B d^3 n \log (a+b x)}{3 b (b c-a d)^3}+\frac{B d^3 n \log (c+d x)}{3 b (b c-a d)^3}-\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{3 b (a+b x)^3}\\ \end{align*}

Mathematica [A]  time = 0.400264, size = 143, normalized size = 0.86 \[ -\frac{\frac{6 A}{(a+b x)^3}+B n \left (\frac{\frac{6 d^2 (a+b x)^2}{(b c-a d)^2}+\frac{3 d (a+b x)}{a d-b c}+2}{(a+b x)^3}+\frac{6 d^3 \log (a+b x)}{(b c-a d)^3}-\frac{6 d^3 \log (c+d x)}{(b c-a d)^3}\right )+\frac{6 B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3}}{18 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(a + b*x)^4,x]

[Out]

-((6*A)/(a + b*x)^3 + B*n*((2 + (3*d*(a + b*x))/(-(b*c) + a*d) + (6*d^2*(a + b*x)^2)/(b*c - a*d)^2)/(a + b*x)^
3 + (6*d^3*Log[a + b*x])/(b*c - a*d)^3 - (6*d^3*Log[c + d*x])/(b*c - a*d)^3) + (6*B*Log[(e*(a + b*x)^n)/(c + d
*x)^n])/(a + b*x)^3)/(18*b)

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Maple [C]  time = 0.475, size = 1976, normalized size = 11.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/(b*x+a)^4,x)

[Out]

1/3*B/b/(b*x+a)^3*ln((d*x+c)^n)-1/18*(-6*B*a^3*d^3*ln((b*x+a)^n)+6*B*b^3*c^3*ln((b*x+a)^n)+18*A*a^2*b*c*d^2-18
*A*a*b^2*c^2*d-11*B*a^3*d^3*n-6*B*ln(d*x+c)*a^3*d^3*n+6*B*a^3*n*ln(-b*x-a)*d^3+6*A*b^3*c^3-6*A*a^3*d^3+18*B*ln
(e)*a^2*b*c*d^2-18*B*ln(e)*a*b^2*c^2*d-6*B*ln(e)*a^3*d^3+6*B*ln(e)*b^3*c^3-6*B*ln(d*x+c)*b^3*d^3*n*x^3+6*B*ln(
-b*x-a)*b^3*d^3*n*x^3+9*I*B*Pi*a^2*b*c*d^2*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-9*I
*B*Pi*a*b^2*c^2*d*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-9*I*B*Pi*a*b^2*c^2*d*csgn(I*(b*x+a)^n)*csgn(I*(b
*x+a)^n/((d*x+c)^n))^2-9*I*B*Pi*a*b^2*c^2*d*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-9*I*B*Pi*a*b^2
*c^2*d*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+2*B*b^3*c^3*n-9*B*a*c^2*d*n*b^2+18*B*a*
b^2*c*d^2*n*x-9*I*B*Pi*a^2*b*c*d^2*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-9*I*B*Pi*a^2*b*c*d^2*csgn(I*e/((d*x+c)^n)*(
b*x+a)^n)^3+9*I*B*Pi*a*b^2*c^2*d*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+9*I*B*Pi*a*b^2*c^2*d*csgn(I*e/((d*x+c)^n)*(b*
x+a)^n)^3+3*I*B*Pi*a^3*d^3*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+3*I*B*Pi*a^
3*d^3*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-3*I*B*Pi*b^3*c^3*csgn(I*e)*csgn(I*(b
*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-3*I*B*Pi*b^3*c^3*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*cs
gn(I*(b*x+a)^n/((d*x+c)^n))+9*I*B*Pi*a^2*b*c*d^2*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+9*I*B*Pi*a^2*b*c*
d^2*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+9*I*B*Pi*a^2*b*c*d^2*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^
n/((d*x+c)^n))^2-18*B*ln(d*x+c)*a*b^2*d^3*n*x^2+18*B*ln(-b*x-a)*a*b^2*d^3*n*x^2-18*B*ln(d*x+c)*a^2*b*d^3*n*x+1
8*B*ln(-b*x-a)*a^2*b*d^3*n*x-3*I*B*Pi*b^3*c^3*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-3*I*B*Pi*b^3*c^3*csgn(I*e/((d*x+
c)^n)*(b*x+a)^n)^3+3*I*B*Pi*b^3*c^3*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-15*B*a^2*b
*d^3*n*x-3*B*b^3*c^2*d*n*x-3*I*B*Pi*a^3*d^3*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+18*B*a^2*b*c*d^2*n+18*
B*a^2*b*c*d^2*ln((b*x+a)^n)-18*B*a*b^2*c^2*d*ln((b*x+a)^n)-6*B*a*b^2*d^3*n*x^2+6*B*b^3*c*d^2*n*x^2-9*I*B*Pi*a^
2*b*c*d^2*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-9*I*B*Pi*a^2*b*c*d^2*csgn(I*
(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+3*I*B*Pi*a^3*d^3*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+
3*I*B*Pi*a^3*d^3*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-3*I*B*Pi*a^3*d^3*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c
)^n))^2-3*I*B*Pi*a^3*d^3*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2-3*I*B*Pi*a^3*d^3*csgn(I*(b*x+a)^n
/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+3*I*B*Pi*b^3*c^3*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+3
*I*B*Pi*b^3*c^3*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+3*I*B*Pi*b^3*c^3*csgn(I/((d*x+c)^n))*csgn(I*
(b*x+a)^n/((d*x+c)^n))^2+9*I*B*Pi*a*b^2*c^2*d*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*
x+a)^n)+9*I*B*Pi*a*b^2*c^2*d*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n)))/(b*x+a)^3/(a
^2*d^2-2*a*b*c*d+b^2*c^2)/(-a*d+b*c)/b

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Maxima [B]  time = 1.25239, size = 540, normalized size = 3.25 \begin{align*} -\frac{{\left (\frac{6 \, d^{3} e n \log \left (b x + a\right )}{b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}} - \frac{6 \, d^{3} e n \log \left (d x + c\right )}{b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}} + \frac{6 \, b^{2} d^{2} e n x^{2} + 2 \, b^{2} c^{2} e n - 7 \, a b c d e n + 11 \, a^{2} d^{2} e n - 3 \,{\left (b^{2} c d e n - 5 \, a b d^{2} e n\right )} x}{a^{3} b^{3} c^{2} - 2 \, a^{4} b^{2} c d + a^{5} b d^{2} +{\left (b^{6} c^{2} - 2 \, a b^{5} c d + a^{2} b^{4} d^{2}\right )} x^{3} + 3 \,{\left (a b^{5} c^{2} - 2 \, a^{2} b^{4} c d + a^{3} b^{3} d^{2}\right )} x^{2} + 3 \,{\left (a^{2} b^{4} c^{2} - 2 \, a^{3} b^{3} c d + a^{4} b^{2} d^{2}\right )} x}\right )} B}{18 \, e} - \frac{B \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{3 \,{\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac{A}{3 \,{\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/18*(6*d^3*e*n*log(b*x + a)/(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) - 6*d^3*e*n*log(d*x + c)
/(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) + (6*b^2*d^2*e*n*x^2 + 2*b^2*c^2*e*n - 7*a*b*c*d*e*n
+ 11*a^2*d^2*e*n - 3*(b^2*c*d*e*n - 5*a*b*d^2*e*n)*x)/(a^3*b^3*c^2 - 2*a^4*b^2*c*d + a^5*b*d^2 + (b^6*c^2 - 2*
a*b^5*c*d + a^2*b^4*d^2)*x^3 + 3*(a*b^5*c^2 - 2*a^2*b^4*c*d + a^3*b^3*d^2)*x^2 + 3*(a^2*b^4*c^2 - 2*a^3*b^3*c*
d + a^4*b^2*d^2)*x))*B/e - 1/3*B*log((b*x + a)^n*e/(d*x + c)^n)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)
- 1/3*A/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)

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Fricas [B]  time = 1.13249, size = 1123, normalized size = 6.77 \begin{align*} -\frac{6 \, A b^{3} c^{3} - 18 \, A a b^{2} c^{2} d + 18 \, A a^{2} b c d^{2} - 6 \, A a^{3} d^{3} + 6 \,{\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} n x^{2} - 3 \,{\left (B b^{3} c^{2} d - 6 \, B a b^{2} c d^{2} + 5 \, B a^{2} b d^{3}\right )} n x +{\left (2 \, B b^{3} c^{3} - 9 \, B a b^{2} c^{2} d + 18 \, B a^{2} b c d^{2} - 11 \, B a^{3} d^{3}\right )} n + 6 \,{\left (B b^{3} d^{3} n x^{3} + 3 \, B a b^{2} d^{3} n x^{2} + 3 \, B a^{2} b d^{3} n x +{\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2}\right )} n\right )} \log \left (b x + a\right ) - 6 \,{\left (B b^{3} d^{3} n x^{3} + 3 \, B a b^{2} d^{3} n x^{2} + 3 \, B a^{2} b d^{3} n x +{\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2}\right )} n\right )} \log \left (d x + c\right ) + 6 \,{\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2} - B a^{3} d^{3}\right )} \log \left (e\right )}{18 \,{\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3} +{\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{3} + 3 \,{\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{2} + 3 \,{\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/18*(6*A*b^3*c^3 - 18*A*a*b^2*c^2*d + 18*A*a^2*b*c*d^2 - 6*A*a^3*d^3 + 6*(B*b^3*c*d^2 - B*a*b^2*d^3)*n*x^2 -
3*(B*b^3*c^2*d - 6*B*a*b^2*c*d^2 + 5*B*a^2*b*d^3)*n*x + (2*B*b^3*c^3 - 9*B*a*b^2*c^2*d + 18*B*a^2*b*c*d^2 - 1
1*B*a^3*d^3)*n + 6*(B*b^3*d^3*n*x^3 + 3*B*a*b^2*d^3*n*x^2 + 3*B*a^2*b*d^3*n*x + (B*b^3*c^3 - 3*B*a*b^2*c^2*d +
3*B*a^2*b*c*d^2)*n)*log(b*x + a) - 6*(B*b^3*d^3*n*x^3 + 3*B*a*b^2*d^3*n*x^2 + 3*B*a^2*b*d^3*n*x + (B*b^3*c^3
- 3*B*a*b^2*c^2*d + 3*B*a^2*b*c*d^2)*n)*log(d*x + c) + 6*(B*b^3*c^3 - 3*B*a*b^2*c^2*d + 3*B*a^2*b*c*d^2 - B*a^
3*d^3)*log(e))/(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3 + (b^7*c^3 - 3*a*b^6*c^2*d + 3*a^2
*b^5*c*d^2 - a^3*b^4*d^3)*x^3 + 3*(a*b^6*c^3 - 3*a^2*b^5*c^2*d + 3*a^3*b^4*c*d^2 - a^4*b^3*d^3)*x^2 + 3*(a^2*b
^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*(b*x+a)**n/((d*x+c)**n)))/(b*x+a)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.29256, size = 605, normalized size = 3.64 \begin{align*} -\frac{B d^{3} n \log \left (b x + a\right )}{3 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}} + \frac{B d^{3} n \log \left (d x + c\right )}{3 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}} - \frac{B n \log \left (b x + a\right )}{3 \,{\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac{B n \log \left (d x + c\right )}{3 \,{\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac{6 \, B b^{2} d^{2} n x^{2} - 3 \, B b^{2} c d n x + 15 \, B a b d^{2} n x + 2 \, B b^{2} c^{2} n - 7 \, B a b c d n + 11 \, B a^{2} d^{2} n + 6 \, A b^{2} c^{2} + 6 \, B b^{2} c^{2} - 12 \, A a b c d - 12 \, B a b c d + 6 \, A a^{2} d^{2} + 6 \, B a^{2} d^{2}}{18 \,{\left (b^{6} c^{2} x^{3} - 2 \, a b^{5} c d x^{3} + a^{2} b^{4} d^{2} x^{3} + 3 \, a b^{5} c^{2} x^{2} - 6 \, a^{2} b^{4} c d x^{2} + 3 \, a^{3} b^{3} d^{2} x^{2} + 3 \, a^{2} b^{4} c^{2} x - 6 \, a^{3} b^{3} c d x + 3 \, a^{4} b^{2} d^{2} x + a^{3} b^{3} c^{2} - 2 \, a^{4} b^{2} c d + a^{5} b d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(b*x+a)^4,x, algorithm="giac")

[Out]

-1/3*B*d^3*n*log(b*x + a)/(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) + 1/3*B*d^3*n*log(d*x + c)/(
b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) - 1/3*B*n*log(b*x + a)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b
^2*x + a^3*b) + 1/3*B*n*log(d*x + c)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) - 1/18*(6*B*b^2*d^2*n*x^2 -
3*B*b^2*c*d*n*x + 15*B*a*b*d^2*n*x + 2*B*b^2*c^2*n - 7*B*a*b*c*d*n + 11*B*a^2*d^2*n + 6*A*b^2*c^2 + 6*B*b^2*c
^2 - 12*A*a*b*c*d - 12*B*a*b*c*d + 6*A*a^2*d^2 + 6*B*a^2*d^2)/(b^6*c^2*x^3 - 2*a*b^5*c*d*x^3 + a^2*b^4*d^2*x^3
+ 3*a*b^5*c^2*x^2 - 6*a^2*b^4*c*d*x^2 + 3*a^3*b^3*d^2*x^2 + 3*a^2*b^4*c^2*x - 6*a^3*b^3*c*d*x + 3*a^4*b^2*d^2
*x + a^3*b^3*c^2 - 2*a^4*b^2*c*d + a^5*b*d^2)