\(\int \frac {A+B \log (e (a+b x)^n (c+d x)^{-n})}{(a+b x)^3} \, dx\) [153]

Optimal result

Integrand size = 31, antiderivative size = 137 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=-\frac {B n}{4 b (a+b x)^2}+\frac {B d n}{2 b (b c-a d) (a+b x)}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2}-\frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{2 b (a+b x)^2} \]

[Out]

-1/4*B*n/b/(b*x+a)^2+1/2*B*d*n/b/(-a*d+b*c)/(b*x+a)+1/2*B*d^2*n*ln(b*x+a)/b/(-a*d+b*c)^2-1/2*B*d^2*n*ln(d*x+c)
/b/(-a*d+b*c)^2+1/2*(-A-B*ln(e*(b*x+a)^n/((d*x+c)^n)))/b/(b*x+a)^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2548, 46} \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{2 b (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2}+\frac {B d n}{2 b (a+b x) (b c-a d)}-\frac {B n}{4 b (a+b x)^2} \]

[In]

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

[Out]

-1/4*(B*n)/(b*(a + b*x)^2) + (B*d*n)/(2*b*(b*c - a*d)*(a + b*x)) + (B*d^2*n*Log[a + b*x])/(2*b*(b*c - a*d)^2)
- (B*d^2*n*Log[c + d*x])/(2*b*(b*c - a*d)^2) - (A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(2*b*(a + b*x)^2)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2548

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

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=-\frac {\frac {2 A}{(a+b x)^2}+B n \left (\frac {1+\frac {2 d (a+b x)}{-b c+a d}}{(a+b x)^2}-\frac {2 d^2 \log (a+b x)}{(b c-a d)^2}+\frac {2 d^2 \log (c+d x)}{(b c-a d)^2}\right )+\frac {2 B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^2}}{4 b} \]

[In]

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

[Out]

-1/4*((2*A)/(a + b*x)^2 + B*n*((1 + (2*d*(a + b*x))/(-(b*c) + a*d))/(a + b*x)^2 - (2*d^2*Log[a + b*x])/(b*c -
a*d)^2 + (2*d^2*Log[c + d*x])/(b*c - a*d)^2) + (2*B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(a + b*x)^2)/b

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(333\) vs. \(2(130)=260\).

Time = 19.18 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.44

method	result	size
parallelrisch	\(-\frac {-4 B a c \,d^{2} n \,b^{4}+2 B \ln \left (d x +c \right ) x^{2} b^{5} d^{3} n -2 B \ln \left (b x +a \right ) a^{2} b^{3} d^{3} n +2 B \ln \left (d x +c \right ) a^{2} b^{3} d^{3} n +2 B x a \,b^{4} d^{3} n -2 B x \,b^{5} c \,d^{2} n -4 B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) a \,b^{4} c \,d^{2}+2 A \,a^{2} b^{3} d^{3}+2 A \,b^{5} c^{2} d -4 A a \,b^{4} c \,d^{2}-2 B \ln \left (b x +a \right ) x^{2} b^{5} d^{3} n +3 B \,a^{2} b^{3} d^{3} n +2 B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) a^{2} b^{3} d^{3}+2 B \ln \left (e \left (b x +a \right )^{n} \left (d x +c \right )^{-n}\right ) b^{5} c^{2} d +B \,b^{5} c^{2} n d -4 B \ln \left (b x +a \right ) x a \,b^{4} d^{3} n +4 B \ln \left (d x +c \right ) x a \,b^{4} d^{3} n}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b x +a \right )^{2} b^{4} d}\)	\(334\)
risch	\(\text {Expression too large to display}\)	\(1379\)

[In]

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

[Out]

-1/4*(-4*B*a*c*d^2*n*b^4+2*B*ln(d*x+c)*x^2*b^5*d^3*n-2*B*ln(b*x+a)*a^2*b^3*d^3*n+2*B*ln(d*x+c)*a^2*b^3*d^3*n+2
*B*x*a*b^4*d^3*n-2*B*x*b^5*c*d^2*n-4*B*ln(e*(b*x+a)^n/((d*x+c)^n))*a*b^4*c*d^2+2*A*a^2*b^3*d^3+2*A*b^5*c^2*d-4
*A*a*b^4*c*d^2-2*B*ln(b*x+a)*x^2*b^5*d^3*n+3*B*a^2*b^3*d^3*n+2*B*ln(e*(b*x+a)^n/((d*x+c)^n))*a^2*b^3*d^3+2*B*l
n(e*(b*x+a)^n/((d*x+c)^n))*b^5*c^2*d+B*b^5*c^2*n*d-4*B*ln(b*x+a)*x*a*b^4*d^3*n+4*B*ln(d*x+c)*x*a*b^4*d^3*n)/(a
^2*d^2-2*a*b*c*d+b^2*c^2)/(b*x+a)^2/b^4/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (127) = 254\).

Time = 0.27 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=-\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x + {\left (B b^{2} c^{2} - 4 \, B a b c d + 3 \, B a^{2} d^{2}\right )} n - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (b x + a\right ) + 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (d x + c\right ) + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \left (e\right )}{4 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.68 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=\frac {{\left (\frac {2 \, d^{2} e n \log \left (b x + a\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} - \frac {2 \, d^{2} e n \log \left (d x + c\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} + \frac {2 \, b d e n x - b c e n + 3 \, a d e n}{a^{2} b^{2} c - a^{3} b d + {\left (b^{4} c - a b^{3} d\right )} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} x}\right )} B}{4 \, e} - \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {A}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.77 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=\frac {B d^{2} n \log \left (b x + a\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {B d^{2} n \log \left (d x + c\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {B n \log \left (b x + a\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {B n \log \left (d x + c\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {2 \, B b d n x - B b c n + 3 \, B a d n - 2 \, B b c \log \left (e\right ) + 2 \, B a d \log \left (e\right ) - 2 \, A b c + 2 \, A a d}{4 \, {\left (b^{4} c x^{2} - a b^{3} d x^{2} + 2 \, a b^{3} c x - 2 \, a^{2} b^{2} d x + a^{2} b^{2} c - a^{3} b d\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.40 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^3} \, dx=-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d\,n-B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2}-\frac {B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}{2\,b\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}-\frac {B\,d^2\,n\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2-2\,a^2\,b\,d^2}{2\,b\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,{\left (a\,d-b\,c\right )}^2} \]

[In]

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

[Out]

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