3.148 \(\int (a+b x)^3 (A+B \log (e (a+b x)^n (c+d x)^{-n})) \, dx\)

Optimal. Leaf size=142 \[ \frac{(a+b x)^4 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{4 b}-\frac{B n x (b c-a d)^3}{4 d^3}+\frac{B n (a+b x)^2 (b c-a d)^2}{8 b d^2}+\frac{B n (b c-a d)^4 \log (c+d x)}{4 b d^4}-\frac{B n (a+b x)^3 (b c-a d)}{12 b d} \]

[Out]

-(B*(b*c - a*d)^3*n*x)/(4*d^3) + (B*(b*c - a*d)^2*n*(a + b*x)^2)/(8*b*d^2) - (B*(b*c - a*d)*n*(a + b*x)^3)/(12
*b*d) + (B*(b*c - a*d)^4*n*Log[c + d*x])/(4*b*d^4) + ((a + b*x)^4*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]))/(4
*b)

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Rubi [A]  time = 0.135311, antiderivative size = 154, normalized size of antiderivative = 1.08, 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, 43} \[ \frac{A (a+b x)^4}{4 b}-\frac{B n x (b c-a d)^3}{4 d^3}+\frac{B n (a+b x)^2 (b c-a d)^2}{8 b d^2}+\frac{B n (b c-a d)^4 \log (c+d x)}{4 b d^4}+\frac{B (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{4 b}-\frac{B n (a+b x)^3 (b c-a d)}{12 b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]),x]

[Out]

-(B*(b*c - a*d)^3*n*x)/(4*d^3) + (B*(b*c - a*d)^2*n*(a + b*x)^2)/(8*b*d^2) - (B*(b*c - a*d)*n*(a + b*x)^3)/(12
*b*d) + (A*(a + b*x)^4)/(4*b) + (B*(b*c - a*d)^4*n*Log[c + d*x])/(4*b*d^4) + (B*(a + b*x)^4*Log[(e*(a + b*x)^n
)/(c + d*x)^n])/(4*b)

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (a+b x)^3 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx &=\int \left (A (a+b x)^3+B (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx\\ &=\frac{A (a+b x)^4}{4 b}+B \int (a+b x)^3 \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \, dx\\ &=\frac{A (a+b x)^4}{4 b}+\frac{B (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{4 b}-\frac{(B (b c-a d) n) \int \frac{(a+b x)^3}{c+d x} \, dx}{4 b}\\ &=\frac{A (a+b x)^4}{4 b}+\frac{B (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{4 b}-\frac{(B (b c-a d) n) \int \left (\frac{b (b c-a d)^2}{d^3}-\frac{b (b c-a d) (a+b x)}{d^2}+\frac{b (a+b x)^2}{d}+\frac{(-b c+a d)^3}{d^3 (c+d x)}\right ) \, dx}{4 b}\\ &=-\frac{B (b c-a d)^3 n x}{4 d^3}+\frac{B (b c-a d)^2 n (a+b x)^2}{8 b d^2}-\frac{B (b c-a d) n (a+b x)^3}{12 b d}+\frac{A (a+b x)^4}{4 b}+\frac{B (b c-a d)^4 n \log (c+d x)}{4 b d^4}+\frac{B (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.472615, size = 273, normalized size = 1.92 \[ \frac{b d x \left (9 a^2 b d^2 (4 A d x-4 B c n+B d n x)+6 a^3 d^3 (4 A+3 B n)+2 a b^2 d \left (12 A d^2 x^2+B n \left (12 c^2-6 c d x+d^2 x^2\right )\right )+b^3 \left (6 A d^3 x^3+B c n \left (-6 c^2+3 c d x-2 d^2 x^2\right )\right )\right )+6 B n \left (6 a^2 b^2 c^2 d^2-4 a^3 b c d^3+4 a^4 d^4-4 a b^3 c^3 d+b^4 c^4\right ) \log (c+d x)+6 B d^4 \left (6 a^2 b^2 x^2+4 a^3 b x+4 a^4+4 a b^3 x^3+b^4 x^4\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )-18 a^4 B d^4 n \log (a+b x)}{24 b d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]),x]

[Out]

(b*d*x*(6*a^3*d^3*(4*A + 3*B*n) + 9*a^2*b*d^2*(-4*B*c*n + 4*A*d*x + B*d*n*x) + b^3*(6*A*d^3*x^3 + B*c*n*(-6*c^
2 + 3*c*d*x - 2*d^2*x^2)) + 2*a*b^2*d*(12*A*d^2*x^2 + B*n*(12*c^2 - 6*c*d*x + d^2*x^2))) - 18*a^4*B*d^4*n*Log[
a + b*x] + 6*B*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 4*a^4*d^4)*n*Log[c + d*x] + 6*B*
d^4*(4*a^4 + 4*a^3*b*x + 6*a^2*b^2*x^2 + 4*a*b^3*x^3 + b^4*x^4)*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(24*b*d^4)

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Maple [C]  time = 0.564, size = 1840, normalized size = 13. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n))),x)

[Out]

3/2*b/d^2*B*ln(d*x+c)*a^2*c^2*n-b^2/d^3*B*ln(d*x+c)*a*c^3*n+1/8*I*b^3*B*Pi*x^4*csgn(I*e)*csgn(I*e/((d*x+c)^n)*
(b*x+a)^n)^2+1/8*I*b^3*B*Pi*x^4*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/8*I*b^3*B*Pi*x^4*csgn(I*
(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/8*I*b^3*B*Pi*x^4*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^
n)*(b*x+a)^n)^2+1/4*B*a^4*n/b*ln(-b*x-a)+B*a^3*x*ln((b*x+a)^n)-1/4*(b*x+a)^4*B/b*ln((d*x+c)^n)+1/4*b^3*A*x^4+1
/4*b^3*B*x^4*ln((b*x+a)^n)+1/4*b^3*B*ln(e)*x^4+B*ln(e)*a^3*x-1/2*I*b^2*B*Pi*a*x^3*csgn(I*(b*x+a)^n/((d*x+c)^n)
)^3-1/2*I*b^2*B*Pi*a*x^3*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-3/4*I*b*B*Pi*a^2*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))^
3-3/4*I*b*B*Pi*a^2*x^2*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3+1/2*I*B*Pi*a^3*x*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+
a)^n)^2+b^2*A*a*x^3+3/2*b*A*a^2*x^2+A*a^3*x+b^2*B*a*x^3*ln((b*x+a)^n)+b^2*B*ln(e)*a*x^3+3/2*b*B*a^2*x^2*ln((b*
x+a)^n)+3/2*b*B*ln(e)*a^2*x^2+1/4/b*B*ln(d*x+c)*a^4*n+1/12*b^2*B*a*n*x^3-1/12*b^3/d*B*c*n*x^3-1/2*I*B*Pi*a^3*x
*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/8*I*b^3*B*Pi*x^4*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/8*I*b^3*B*Pi*x^4*csgn(
I*e/((d*x+c)^n)*(b*x+a)^n)^3-1/8*I*b^3*B*Pi*x^4*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(
b*x+a)^n)-1/8*I*b^3*B*Pi*x^4*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))+1/2*I*b^2*B*P
i*a*x^3*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I*b^2*B*Pi*a*x^3*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/
((d*x+c)^n))^2+1/2*I*b^2*B*Pi*a*x^3*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*b^2*B*Pi*a*x^3*csg
n(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I*b^2*B*Pi*a*x^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)-1/2*I*b^2*B*Pi*a*x^3*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*
(b*x+a)^n/((d*x+c)^n))-3/4*I*b*B*Pi*a^2*x^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)-3/4*I*b*B*Pi*a^2*x^2*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-1/2*b^2/d*B*a*c
*n*x^2+3/4*I*b*B*Pi*a^2*x^2*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+3/4*I*b*B*Pi*a^2*x^2*csgn(I/((d*x+c)^n
))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+3/4*I*b*B*Pi*a^2*x^2*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+3/4*
I*b*B*Pi*a^2*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I*B*Pi*a^3*x*csgn(I*e)*cs
gn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)-1/2*I*B*Pi*a^3*x*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)
^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-3/2*b/d*B*a^2*c*n*x+b^2/d^2*B*a*c^2*n*x+3/8*b*B*a^2*n*x^2+1/8*b^3/d^2*B*c^2
*n*x^2+3/4*B*a^3*n*x-1/4*b^3/d^3*B*c^3*n*x-1/d*B*ln(d*x+c)*a^3*c*n+1/4*b^3/d^4*B*ln(d*x+c)*c^4*n-1/2*I*B*Pi*a^
3*x*csgn(I*(b*x+a)^n/((d*x+c)^n))^3+1/2*I*B*Pi*a^3*x*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I
*B*Pi*a^3*x*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*a^3*x*csgn(I*(b*x+a)^n/((d*x+c)^n))*c
sgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2

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Maxima [B]  time = 1.24692, size = 630, normalized size = 4.44 \begin{align*} \frac{1}{4} \, B b^{3} x^{4} \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + \frac{1}{4} \, A b^{3} x^{4} + B a b^{2} x^{3} \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A a b^{2} x^{3} + \frac{3}{2} \, B a^{2} b x^{2} \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + \frac{3}{2} \, A a^{2} b x^{2} + B a^{3} x \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A a^{3} x + \frac{{\left (\frac{a e n \log \left (b x + a\right )}{b} - \frac{c e n \log \left (d x + c\right )}{d}\right )} B a^{3}}{e} - \frac{3 \,{\left (\frac{a^{2} e n \log \left (b x + a\right )}{b^{2}} - \frac{c^{2} e n \log \left (d x + c\right )}{d^{2}} + \frac{{\left (b c e n - a d e n\right )} x}{b d}\right )} B a^{2} b}{2 \, e} + \frac{{\left (\frac{2 \, a^{3} e n \log \left (b x + a\right )}{b^{3}} - \frac{2 \, c^{3} e n \log \left (d x + c\right )}{d^{3}} - \frac{{\left (b^{2} c d e n - a b d^{2} e n\right )} x^{2} - 2 \,{\left (b^{2} c^{2} e n - a^{2} d^{2} e n\right )} x}{b^{2} d^{2}}\right )} B a b^{2}}{2 \, e} - \frac{{\left (\frac{6 \, a^{4} e n \log \left (b x + a\right )}{b^{4}} - \frac{6 \, c^{4} e n \log \left (d x + c\right )}{d^{4}} + \frac{2 \,{\left (b^{3} c d^{2} e n - a b^{2} d^{3} e n\right )} x^{3} - 3 \,{\left (b^{3} c^{2} d e n - a^{2} b d^{3} e n\right )} x^{2} + 6 \,{\left (b^{3} c^{3} e n - a^{3} d^{3} e n\right )} x}{b^{3} d^{3}}\right )} B b^{3}}{24 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="maxima")

[Out]

1/4*B*b^3*x^4*log((b*x + a)^n*e/(d*x + c)^n) + 1/4*A*b^3*x^4 + B*a*b^2*x^3*log((b*x + a)^n*e/(d*x + c)^n) + A*
a*b^2*x^3 + 3/2*B*a^2*b*x^2*log((b*x + a)^n*e/(d*x + c)^n) + 3/2*A*a^2*b*x^2 + B*a^3*x*log((b*x + a)^n*e/(d*x
+ c)^n) + A*a^3*x + (a*e*n*log(b*x + a)/b - c*e*n*log(d*x + c)/d)*B*a^3/e - 3/2*(a^2*e*n*log(b*x + a)/b^2 - c^
2*e*n*log(d*x + c)/d^2 + (b*c*e*n - a*d*e*n)*x/(b*d))*B*a^2*b/e + 1/2*(2*a^3*e*n*log(b*x + a)/b^3 - 2*c^3*e*n*
log(d*x + c)/d^3 - ((b^2*c*d*e*n - a*b*d^2*e*n)*x^2 - 2*(b^2*c^2*e*n - a^2*d^2*e*n)*x)/(b^2*d^2))*B*a*b^2/e -
1/24*(6*a^4*e*n*log(b*x + a)/b^4 - 6*c^4*e*n*log(d*x + c)/d^4 + (2*(b^3*c*d^2*e*n - a*b^2*d^3*e*n)*x^3 - 3*(b^
3*c^2*d*e*n - a^2*b*d^3*e*n)*x^2 + 6*(b^3*c^3*e*n - a^3*d^3*e*n)*x)/(b^3*d^3))*B*b^3/e

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Fricas [B]  time = 1.06623, size = 871, normalized size = 6.13 \begin{align*} \frac{6 \, A b^{4} d^{4} x^{4} + 2 \,{\left (12 \, A a b^{3} d^{4} -{\left (B b^{4} c d^{3} - B a b^{3} d^{4}\right )} n\right )} x^{3} + 3 \,{\left (12 \, A a^{2} b^{2} d^{4} +{\left (B b^{4} c^{2} d^{2} - 4 \, B a b^{3} c d^{3} + 3 \, B a^{2} b^{2} d^{4}\right )} n\right )} x^{2} + 6 \,{\left (4 \, A a^{3} b d^{4} -{\left (B b^{4} c^{3} d - 4 \, B a b^{3} c^{2} d^{2} + 6 \, B a^{2} b^{2} c d^{3} - 3 \, B a^{3} b d^{4}\right )} n\right )} x + 6 \,{\left (B b^{4} d^{4} n x^{4} + 4 \, B a b^{3} d^{4} n x^{3} + 6 \, B a^{2} b^{2} d^{4} n x^{2} + 4 \, B a^{3} b d^{4} n x + B a^{4} d^{4} n\right )} \log \left (b x + a\right ) - 6 \,{\left (B b^{4} d^{4} n x^{4} + 4 \, B a b^{3} d^{4} n x^{3} + 6 \, B a^{2} b^{2} d^{4} n x^{2} + 4 \, B a^{3} b d^{4} n x -{\left (B b^{4} c^{4} - 4 \, B a b^{3} c^{3} d + 6 \, B a^{2} b^{2} c^{2} d^{2} - 4 \, B a^{3} b c d^{3}\right )} n\right )} \log \left (d x + c\right ) + 6 \,{\left (B b^{4} d^{4} x^{4} + 4 \, B a b^{3} d^{4} x^{3} + 6 \, B a^{2} b^{2} d^{4} x^{2} + 4 \, B a^{3} b d^{4} x\right )} \log \left (e\right )}{24 \, b d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="fricas")

[Out]

1/24*(6*A*b^4*d^4*x^4 + 2*(12*A*a*b^3*d^4 - (B*b^4*c*d^3 - B*a*b^3*d^4)*n)*x^3 + 3*(12*A*a^2*b^2*d^4 + (B*b^4*
c^2*d^2 - 4*B*a*b^3*c*d^3 + 3*B*a^2*b^2*d^4)*n)*x^2 + 6*(4*A*a^3*b*d^4 - (B*b^4*c^3*d - 4*B*a*b^3*c^2*d^2 + 6*
B*a^2*b^2*c*d^3 - 3*B*a^3*b*d^4)*n)*x + 6*(B*b^4*d^4*n*x^4 + 4*B*a*b^3*d^4*n*x^3 + 6*B*a^2*b^2*d^4*n*x^2 + 4*B
*a^3*b*d^4*n*x + B*a^4*d^4*n)*log(b*x + a) - 6*(B*b^4*d^4*n*x^4 + 4*B*a*b^3*d^4*n*x^3 + 6*B*a^2*b^2*d^4*n*x^2
+ 4*B*a^3*b*d^4*n*x - (B*b^4*c^4 - 4*B*a*b^3*c^3*d + 6*B*a^2*b^2*c^2*d^2 - 4*B*a^3*b*c*d^3)*n)*log(d*x + c) +
6*(B*b^4*d^4*x^4 + 4*B*a*b^3*d^4*x^3 + 6*B*a^2*b^2*d^4*x^2 + 4*B*a^3*b*d^4*x)*log(e))/(b*d^4)

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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((b*x+a)**3*(A+B*ln(e*(b*x+a)**n/((d*x+c)**n))),x)

[Out]

Timed out

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Giac [B]  time = 5.42466, size = 479, normalized size = 3.37 \begin{align*} \frac{B a^{4} n \log \left (b x + a\right )}{4 \, b} + \frac{1}{4} \,{\left (A b^{3} + B b^{3}\right )} x^{4} - \frac{{\left (B b^{3} c n - B a b^{2} d n - 12 \, A a b^{2} d - 12 \, B a b^{2} d\right )} x^{3}}{12 \, d} + \frac{1}{4} \,{\left (B b^{3} n x^{4} + 4 \, B a b^{2} n x^{3} + 6 \, B a^{2} b n x^{2} + 4 \, B a^{3} n x\right )} \log \left (b x + a\right ) - \frac{1}{4} \,{\left (B b^{3} n x^{4} + 4 \, B a b^{2} n x^{3} + 6 \, B a^{2} b n x^{2} + 4 \, B a^{3} n x\right )} \log \left (d x + c\right ) + \frac{{\left (B b^{3} c^{2} n - 4 \, B a b^{2} c d n + 3 \, B a^{2} b d^{2} n + 12 \, A a^{2} b d^{2} + 12 \, B a^{2} b d^{2}\right )} x^{2}}{8 \, d^{2}} - \frac{{\left (B b^{3} c^{3} n - 4 \, B a b^{2} c^{2} d n + 6 \, B a^{2} b c d^{2} n - 3 \, B a^{3} d^{3} n - 4 \, A a^{3} d^{3} - 4 \, B a^{3} d^{3}\right )} x}{4 \, d^{3}} + \frac{{\left (B b^{3} c^{4} n - 4 \, B a b^{2} c^{3} d n + 6 \, B a^{2} b c^{2} d^{2} n - 4 \, B a^{3} c d^{3} n\right )} \log \left (d x + c\right )}{4 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(A+B*log(e*(b*x+a)^n/((d*x+c)^n))),x, algorithm="giac")

[Out]

1/4*B*a^4*n*log(b*x + a)/b + 1/4*(A*b^3 + B*b^3)*x^4 - 1/12*(B*b^3*c*n - B*a*b^2*d*n - 12*A*a*b^2*d - 12*B*a*b
^2*d)*x^3/d + 1/4*(B*b^3*n*x^4 + 4*B*a*b^2*n*x^3 + 6*B*a^2*b*n*x^2 + 4*B*a^3*n*x)*log(b*x + a) - 1/4*(B*b^3*n*
x^4 + 4*B*a*b^2*n*x^3 + 6*B*a^2*b*n*x^2 + 4*B*a^3*n*x)*log(d*x + c) + 1/8*(B*b^3*c^2*n - 4*B*a*b^2*c*d*n + 3*B
*a^2*b*d^2*n + 12*A*a^2*b*d^2 + 12*B*a^2*b*d^2)*x^2/d^2 - 1/4*(B*b^3*c^3*n - 4*B*a*b^2*c^2*d*n + 6*B*a^2*b*c*d
^2*n - 3*B*a^3*d^3*n - 4*A*a^3*d^3 - 4*B*a^3*d^3)*x/d^3 + 1/4*(B*b^3*c^4*n - 4*B*a*b^2*c^3*d*n + 6*B*a^2*b*c^2
*d^2*n - 4*B*a^3*c*d^3*n)*log(d*x + c)/d^4