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

Optimal. Leaf size=171 \[ \frac{(a+b x)^5 \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )}{5 b}+\frac{B n x (b c-a d)^4}{5 d^4}-\frac{B n (a+b x)^2 (b c-a d)^3}{10 b d^3}+\frac{B n (a+b x)^3 (b c-a d)^2}{15 b d^2}-\frac{B n (b c-a d)^5 \log (c+d x)}{5 b d^5}-\frac{B n (a+b x)^4 (b c-a d)}{20 b d} \]

[Out]

(B*(b*c - a*d)^4*n*x)/(5*d^4) - (B*(b*c - a*d)^3*n*(a + b*x)^2)/(10*b*d^3) + (B*(b*c - a*d)^2*n*(a + b*x)^3)/(
15*b*d^2) - (B*(b*c - a*d)*n*(a + b*x)^4)/(20*b*d) - (B*(b*c - a*d)^5*n*Log[c + d*x])/(5*b*d^5) + ((a + b*x)^5
*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n]))/(5*b)

________________________________________________________________________________________

Rubi [A]  time = 0.183445, antiderivative size = 183, 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, 43} \[ \frac{A (a+b x)^5}{5 b}+\frac{B n x (b c-a d)^4}{5 d^4}-\frac{B n (a+b x)^2 (b c-a d)^3}{10 b d^3}+\frac{B n (a+b x)^3 (b c-a d)^2}{15 b d^2}-\frac{B n (b c-a d)^5 \log (c+d x)}{5 b d^5}+\frac{B (a+b x)^5 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{5 b}-\frac{B n (a+b x)^4 (b c-a d)}{20 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(b*c - a*d)^4*n*x)/(5*d^4) - (B*(b*c - a*d)^3*n*(a + b*x)^2)/(10*b*d^3) + (B*(b*c - a*d)^2*n*(a + b*x)^3)/(
15*b*d^2) - (B*(b*c - a*d)*n*(a + b*x)^4)/(20*b*d) + (A*(a + b*x)^5)/(5*b) - (B*(b*c - a*d)^5*n*Log[c + d*x])/
(5*b*d^5) + (B*(a + b*x)^5*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(5*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)^4 \left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx &=\int \left (A (a+b x)^4+B (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right ) \, dx\\ &=\frac{A (a+b x)^5}{5 b}+B \int (a+b x)^4 \log \left (e (a+b x)^n (c+d x)^{-n}\right ) \, dx\\ &=\frac{A (a+b x)^5}{5 b}+\frac{B (a+b x)^5 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{5 b}-\frac{(B (b c-a d) n) \int \frac{(a+b x)^4}{c+d x} \, dx}{5 b}\\ &=\frac{A (a+b x)^5}{5 b}+\frac{B (a+b x)^5 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{5 b}-\frac{(B (b c-a d) n) \int \left (-\frac{b (b c-a d)^3}{d^4}+\frac{b (b c-a d)^2 (a+b x)}{d^3}-\frac{b (b c-a d) (a+b x)^2}{d^2}+\frac{b (a+b x)^3}{d}+\frac{(-b c+a d)^4}{d^4 (c+d x)}\right ) \, dx}{5 b}\\ &=\frac{B (b c-a d)^4 n x}{5 d^4}-\frac{B (b c-a d)^3 n (a+b x)^2}{10 b d^3}+\frac{B (b c-a d)^2 n (a+b x)^3}{15 b d^2}-\frac{B (b c-a d) n (a+b x)^4}{20 b d}+\frac{A (a+b x)^5}{5 b}-\frac{B (b c-a d)^5 n \log (c+d x)}{5 b d^5}+\frac{B (a+b x)^5 \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{5 b}\\ \end{align*}

Mathematica [B]  time = 0.784216, size = 364, normalized size = 2.13 \[ \frac{b d x \left (4 a^2 b^2 d^2 \left (30 A d^2 x^2+B n \left (30 c^2-15 c d x+4 d^2 x^2\right )\right )+12 a^3 b d^3 (10 A d x-10 B c n+3 B d n x)+12 a^4 d^4 (5 A+4 B n)+a b^3 d \left (60 A d^3 x^3+B n \left (30 c^2 d x-60 c^3-20 c d^2 x^2+3 d^3 x^3\right )\right )+b^4 \left (12 A d^4 x^4+B c n \left (-6 c^2 d x+12 c^3+4 c d^2 x^2-3 d^3 x^3\right )\right )\right )-12 B n \left (10 a^2 b^3 c^3 d^2-10 a^3 b^2 c^2 d^3+5 a^4 b c d^4-5 a^5 d^5-5 a b^4 c^4 d+b^5 c^5\right ) \log (c+d x)+12 B d^5 \left (10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a^4 b x+5 a^5+5 a b^4 x^4+b^5 x^5\right ) \log \left (e (a+b x)^n (c+d x)^{-n}\right )-48 a^5 B d^5 n \log (a+b x)}{60 b d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*(12*a^4*d^4*(5*A + 4*B*n) + 12*a^3*b*d^3*(-10*B*c*n + 10*A*d*x + 3*B*d*n*x) + 4*a^2*b^2*d^2*(30*A*d^2*x
^2 + B*n*(30*c^2 - 15*c*d*x + 4*d^2*x^2)) + b^4*(12*A*d^4*x^4 + B*c*n*(12*c^3 - 6*c^2*d*x + 4*c*d^2*x^2 - 3*d^
3*x^3)) + a*b^3*d*(60*A*d^3*x^3 + B*n*(-60*c^3 + 30*c^2*d*x - 20*c*d^2*x^2 + 3*d^3*x^3))) - 48*a^5*B*d^5*n*Log
[a + b*x] - 12*B*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - 5*a^5*d^
5)*n*Log[c + d*x] + 12*B*d^5*(5*a^5 + 5*a^4*b*x + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a*b^4*x^4 + b^5*x^5)*Log
[(e*(a + b*x)^n)/(c + d*x)^n])/(60*b*d^5)

________________________________________________________________________________________

Maple [C]  time = 0.833, size = 2374, normalized size = 13.9 \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)^4*(A+B*ln(e*(b*x+a)^n/((d*x+c)^n))),x)

[Out]

1/5*B*a^5*n/b*ln(-b*x-a)+1/5*b^4*A*x^5+1/5*b^4*B*ln(e)*x^5+1/5*b^4*B*x^5*ln((b*x+a)^n)+B*ln(e)*a^4*x-1/2*I*b^3
*B*Pi*a*x^4*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-I*b^2*B*Pi*a^2*x^3*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-I*b^2*B*Pi*a^
2*x^3*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^3-I*b*B*Pi*a^3*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-I*b*B*Pi*a^3*x^2*csgn
(I*e/((d*x+c)^n)*(b*x+a)^n)^3+1/2*I*B*Pi*a^4*x*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*
a^4*x*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*B*Pi*a^4*x*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a
)^n)^2+1/2*I*B*Pi*a^4*x*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+B*a^4*x*ln((b*x+a)^n)-
1/5*(b*x+a)^5*B/b*ln((d*x+c)^n)-1/3*b^3/d*B*a*c*n*x^3-b^2/d*B*a^2*c*n*x^2+1/2*b^3/d^2*B*a*c^2*n*x^2-2*b/d*B*a^
3*c*n*x+b^3*A*a*x^4+2*b^2*A*a^2*x^3+2*b*A*a^3*x^2+A*a^4*x+1/5/b*B*ln(d*x+c)*a^5*n+b^3*B*ln(e)*a*x^4+b^3*B*a*x^
4*ln((b*x+a)^n)+2*b^2*B*ln(e)*a^2*x^3+2*b^2*B*a^2*x^3*ln((b*x+a)^n)+2*b*B*ln(e)*a^3*x^2+2*b*B*a^3*x^2*ln((b*x+
a)^n)+3/5*b*B*a^3*n*x^2-1/10*b^4/d^3*B*c^3*n*x^2+4/5*B*a^4*n*x+1/5*b^4/d^4*B*c^4*n*x-1/d*B*ln(d*x+c)*a^4*c*n-1
/5*b^4/d^5*B*ln(d*x+c)*c^5*n-1/10*I*b^4*B*Pi*x^5*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/10*I*b^4*B*Pi*x^5*csgn(I*e/
((d*x+c)^n)*(b*x+a)^n)^3-1/2*I*B*Pi*a^4*x*csgn(I*(b*x+a)^n/((d*x+c)^n))^3-1/2*I*B*Pi*a^4*x*csgn(I*e/((d*x+c)^n
)*(b*x+a)^n)^3+I*b*B*Pi*a^3*x^2*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*b*B*Pi*a^3*x^2*csgn(I*e)*c
sgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+I*b*B*Pi*a^3*x^2*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n
)^2-1/10*I*b^4*B*Pi*x^5*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-1/10*I*b^4*B*Pi*x^
5*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^3*B*Pi*a*x^4*csgn(I/((d*x+c)
^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/2*I*b^3*B*Pi*a*x^4*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/
2*I*b^3*B*Pi*a*x^4*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+1/2*I*b^3*B*Pi*a*x^4*csgn(I*(b*x+a)^n/((d*x+c)^
n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2-1/2*I*B*Pi*a^4*x*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*Pi*a^4*x*csgn(I*e)*csgn(I*(b*x+a)^n/((d*x+c)^n))*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)+1/20*b^3
*B*a*n*x^4-1/20*b^4/d*B*c*n*x^4+4/15*b^2*B*a^2*n*x^3+1/15*b^4/d^2*B*c^2*n*x^3+I*b^2*B*Pi*a^2*x^3*csgn(I/((d*x+
c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*b^2*B*Pi*a^2*x^3*csgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+I*
b^2*B*Pi*a^2*x^3*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+I*b^2*B*Pi*a^2*x^3*csgn(I*(b*x+a)^n/((d*x+c)^n))*
csgn(I*e/((d*x+c)^n)*(b*x+a)^n)^2+I*b*B*Pi*a^3*x^2*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+2*b^2/d
^2*B*a^2*c^2*n*x-b^3/d^3*B*a*c^3*n*x+2*b/d^2*B*ln(d*x+c)*a^3*c^2*n-2*b^2/d^3*B*ln(d*x+c)*a^2*c^3*n+b^3/d^4*B*l
n(d*x+c)*a*c^4*n+1/10*I*b^4*B*Pi*x^5*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/10*I*b^4*B*Pi*x^5*c
sgn(I*(b*x+a)^n)*csgn(I*(b*x+a)^n/((d*x+c)^n))^2+1/10*I*b^4*B*Pi*x^5*csgn(I*e)*csgn(I*e/((d*x+c)^n)*(b*x+a)^n)
^2+1/10*I*b^4*B*Pi*x^5*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^3*B*Pi*a*x^4*cs
gn(I*(b*x+a)^n/((d*x+c)^n))^3-1/2*I*b^3*B*Pi*a*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^3*B*Pi*a*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)-I*b^2*B*
Pi*a^2*x^3*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-I*b^2*B*Pi*a^2*x^3*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)-I*b*B*Pi*a^3*x^2*csgn(I*(b*x+a)^n)*csgn(I/((d*x+c)
^n))*csgn(I*(b*x+a)^n/((d*x+c)^n))-I*b*B*Pi*a^3*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)

________________________________________________________________________________________

Maxima [B]  time = 1.29535, size = 906, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*b^4*x^5*log((b*x + a)^n*e/(d*x + c)^n) + 1/5*A*b^4*x^5 + B*a*b^3*x^4*log((b*x + a)^n*e/(d*x + c)^n) + A*
a*b^3*x^4 + 2*B*a^2*b^2*x^3*log((b*x + a)^n*e/(d*x + c)^n) + 2*A*a^2*b^2*x^3 + 2*B*a^3*b*x^2*log((b*x + a)^n*e
/(d*x + c)^n) + 2*A*a^3*b*x^2 + B*a^4*x*log((b*x + a)^n*e/(d*x + c)^n) + A*a^4*x + (a*e*n*log(b*x + a)/b - c*e
*n*log(d*x + c)/d)*B*a^4/e - 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^3*b/e + (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^2*b^2/e - 1/6*(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*a*b^3/e + 1/60*(12*a^5*e*n*log(b*x + a)/b^5 - 12*c^5*e*n*log(d*x + c)/d^5 - (3*(b
^4*c*d^3*e*n - a*b^3*d^4*e*n)*x^4 - 4*(b^4*c^2*d^2*e*n - a^2*b^2*d^4*e*n)*x^3 + 6*(b^4*c^3*d*e*n - a^3*b*d^4*e
*n)*x^2 - 12*(b^4*c^4*e*n - a^4*d^4*e*n)*x)/(b^4*d^4))*B*b^4/e

________________________________________________________________________________________

Fricas [B]  time = 1.0928, size = 1188, normalized size = 6.95 \begin{align*} \frac{12 \, A b^{5} d^{5} x^{5} + 3 \,{\left (20 \, A a b^{4} d^{5} -{\left (B b^{5} c d^{4} - B a b^{4} d^{5}\right )} n\right )} x^{4} + 4 \,{\left (30 \, A a^{2} b^{3} d^{5} +{\left (B b^{5} c^{2} d^{3} - 5 \, B a b^{4} c d^{4} + 4 \, B a^{2} b^{3} d^{5}\right )} n\right )} x^{3} + 6 \,{\left (20 \, A a^{3} b^{2} d^{5} -{\left (B b^{5} c^{3} d^{2} - 5 \, B a b^{4} c^{2} d^{3} + 10 \, B a^{2} b^{3} c d^{4} - 6 \, B a^{3} b^{2} d^{5}\right )} n\right )} x^{2} + 12 \,{\left (5 \, A a^{4} b d^{5} +{\left (B b^{5} c^{4} d - 5 \, B a b^{4} c^{3} d^{2} + 10 \, B a^{2} b^{3} c^{2} d^{3} - 10 \, B a^{3} b^{2} c d^{4} + 4 \, B a^{4} b d^{5}\right )} n\right )} x + 12 \,{\left (B b^{5} d^{5} n x^{5} + 5 \, B a b^{4} d^{5} n x^{4} + 10 \, B a^{2} b^{3} d^{5} n x^{3} + 10 \, B a^{3} b^{2} d^{5} n x^{2} + 5 \, B a^{4} b d^{5} n x + B a^{5} d^{5} n\right )} \log \left (b x + a\right ) - 12 \,{\left (B b^{5} d^{5} n x^{5} + 5 \, B a b^{4} d^{5} n x^{4} + 10 \, B a^{2} b^{3} d^{5} n x^{3} + 10 \, B a^{3} b^{2} d^{5} n x^{2} + 5 \, B a^{4} b d^{5} n x +{\left (B b^{5} c^{5} - 5 \, B a b^{4} c^{4} d + 10 \, B a^{2} b^{3} c^{3} d^{2} - 10 \, B a^{3} b^{2} c^{2} d^{3} + 5 \, B a^{4} b c d^{4}\right )} n\right )} \log \left (d x + c\right ) + 12 \,{\left (B b^{5} d^{5} x^{5} + 5 \, B a b^{4} d^{5} x^{4} + 10 \, B a^{2} b^{3} d^{5} x^{3} + 10 \, B a^{3} b^{2} d^{5} x^{2} + 5 \, B a^{4} b d^{5} x\right )} \log \left (e\right )}{60 \, b d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*A*b^5*d^5*x^5 + 3*(20*A*a*b^4*d^5 - (B*b^5*c*d^4 - B*a*b^4*d^5)*n)*x^4 + 4*(30*A*a^2*b^3*d^5 + (B*b^5
*c^2*d^3 - 5*B*a*b^4*c*d^4 + 4*B*a^2*b^3*d^5)*n)*x^3 + 6*(20*A*a^3*b^2*d^5 - (B*b^5*c^3*d^2 - 5*B*a*b^4*c^2*d^
3 + 10*B*a^2*b^3*c*d^4 - 6*B*a^3*b^2*d^5)*n)*x^2 + 12*(5*A*a^4*b*d^5 + (B*b^5*c^4*d - 5*B*a*b^4*c^3*d^2 + 10*B
*a^2*b^3*c^2*d^3 - 10*B*a^3*b^2*c*d^4 + 4*B*a^4*b*d^5)*n)*x + 12*(B*b^5*d^5*n*x^5 + 5*B*a*b^4*d^5*n*x^4 + 10*B
*a^2*b^3*d^5*n*x^3 + 10*B*a^3*b^2*d^5*n*x^2 + 5*B*a^4*b*d^5*n*x + B*a^5*d^5*n)*log(b*x + a) - 12*(B*b^5*d^5*n*
x^5 + 5*B*a*b^4*d^5*n*x^4 + 10*B*a^2*b^3*d^5*n*x^3 + 10*B*a^3*b^2*d^5*n*x^2 + 5*B*a^4*b*d^5*n*x + (B*b^5*c^5 -
5*B*a*b^4*c^4*d + 10*B*a^2*b^3*c^3*d^2 - 10*B*a^3*b^2*c^2*d^3 + 5*B*a^4*b*c*d^4)*n)*log(d*x + c) + 12*(B*b^5*
d^5*x^5 + 5*B*a*b^4*d^5*x^4 + 10*B*a^2*b^3*d^5*x^3 + 10*B*a^3*b^2*d^5*x^2 + 5*B*a^4*b*d^5*x)*log(e))/(b*d^5)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 13.9964, size = 671, normalized size = 3.92 \begin{align*} \frac{B a^{5} n \log \left (b x + a\right )}{5 \, b} + \frac{1}{5} \,{\left (A b^{4} + B b^{4}\right )} x^{5} - \frac{{\left (B b^{4} c n - B a b^{3} d n - 20 \, A a b^{3} d - 20 \, B a b^{3} d\right )} x^{4}}{20 \, d} + \frac{{\left (B b^{4} c^{2} n - 5 \, B a b^{3} c d n + 4 \, B a^{2} b^{2} d^{2} n + 30 \, A a^{2} b^{2} d^{2} + 30 \, B a^{2} b^{2} d^{2}\right )} x^{3}}{15 \, d^{2}} + \frac{1}{5} \,{\left (B b^{4} n x^{5} + 5 \, B a b^{3} n x^{4} + 10 \, B a^{2} b^{2} n x^{3} + 10 \, B a^{3} b n x^{2} + 5 \, B a^{4} n x\right )} \log \left (b x + a\right ) - \frac{1}{5} \,{\left (B b^{4} n x^{5} + 5 \, B a b^{3} n x^{4} + 10 \, B a^{2} b^{2} n x^{3} + 10 \, B a^{3} b n x^{2} + 5 \, B a^{4} n x\right )} \log \left (d x + c\right ) - \frac{{\left (B b^{4} c^{3} n - 5 \, B a b^{3} c^{2} d n + 10 \, B a^{2} b^{2} c d^{2} n - 6 \, B a^{3} b d^{3} n - 20 \, A a^{3} b d^{3} - 20 \, B a^{3} b d^{3}\right )} x^{2}}{10 \, d^{3}} + \frac{{\left (B b^{4} c^{4} n - 5 \, B a b^{3} c^{3} d n + 10 \, B a^{2} b^{2} c^{2} d^{2} n - 10 \, B a^{3} b c d^{3} n + 4 \, B a^{4} d^{4} n + 5 \, A a^{4} d^{4} + 5 \, B a^{4} d^{4}\right )} x}{5 \, d^{4}} - \frac{{\left (B b^{4} c^{5} n - 5 \, B a b^{3} c^{4} d n + 10 \, B a^{2} b^{2} c^{3} d^{2} n - 10 \, B a^{3} b c^{2} d^{3} n + 5 \, B a^{4} c d^{4} n\right )} \log \left (-d x - c\right )}{5 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*a^5*n*log(b*x + a)/b + 1/5*(A*b^4 + B*b^4)*x^5 - 1/20*(B*b^4*c*n - B*a*b^3*d*n - 20*A*a*b^3*d - 20*B*a*b
^3*d)*x^4/d + 1/15*(B*b^4*c^2*n - 5*B*a*b^3*c*d*n + 4*B*a^2*b^2*d^2*n + 30*A*a^2*b^2*d^2 + 30*B*a^2*b^2*d^2)*x
^3/d^2 + 1/5*(B*b^4*n*x^5 + 5*B*a*b^3*n*x^4 + 10*B*a^2*b^2*n*x^3 + 10*B*a^3*b*n*x^2 + 5*B*a^4*n*x)*log(b*x + a
) - 1/5*(B*b^4*n*x^5 + 5*B*a*b^3*n*x^4 + 10*B*a^2*b^2*n*x^3 + 10*B*a^3*b*n*x^2 + 5*B*a^4*n*x)*log(d*x + c) - 1
/10*(B*b^4*c^3*n - 5*B*a*b^3*c^2*d*n + 10*B*a^2*b^2*c*d^2*n - 6*B*a^3*b*d^3*n - 20*A*a^3*b*d^3 - 20*B*a^3*b*d^
3)*x^2/d^3 + 1/5*(B*b^4*c^4*n - 5*B*a*b^3*c^3*d*n + 10*B*a^2*b^2*c^2*d^2*n - 10*B*a^3*b*c*d^3*n + 4*B*a^4*d^4*
n + 5*A*a^4*d^4 + 5*B*a^4*d^4)*x/d^4 - 1/5*(B*b^4*c^5*n - 5*B*a*b^3*c^4*d*n + 10*B*a^2*b^2*c^3*d^2*n - 10*B*a^
3*b*c^2*d^3*n + 5*B*a^4*c*d^4*n)*log(-d*x - c)/d^5