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

Optimal. Leaf size=137 \[ -\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{2 b (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2}+\frac {B d n}{2 b (a+b x) (b c-a d)}-\frac {B n}{4 b (a+b x)^2} \]

[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

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Rubi [A]  time = 0.15, antiderivative size = 149, normalized size of antiderivative = 1.09, 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} \[ -\frac {A}{2 b (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2}-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{2 b (a+b x)^2}+\frac {B d n}{2 b (a+b x) (b c-a d)}-\frac {B n}{4 b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 121, normalized size = 0.88 \[ -\frac {\frac {2 A}{(a+b x)^2}+B n \left (-\frac {2 d^2 \log (a+b x)}{(b c-a d)^2}+\frac {2 d^2 \log (c+d x)}{(b c-a d)^2}+\frac {\frac {2 d (a+b x)}{a d-b c}+1}{(a+b x)^2}\right )+\frac {2 B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x)^2}}{4 b} \]

Antiderivative was successfully verified.

[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

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fricas [B]  time = 0.56, size = 296, normalized size = 2.16 \[ -\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x + {\left (B b^{2} c^{2} - 4 \, B a b c d + 3 \, B a^{2} d^{2}\right )} n - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (b x + a\right ) + 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (d x + c\right ) + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \relax (e)}{4 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x\right )}} \]

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)^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)

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giac [A]  time = 0.22, size = 239, normalized size = 1.74 \[ \frac {B d^{2} n \log \left (b x + a\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {B d^{2} n \log \left (d x + c\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {B n \log \left (b x + a\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {B n \log \left (d x + c\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {2 \, B b d n x - B b c n + 3 \, B a d n - 2 \, A b c - 2 \, B b c + 2 \, A a d + 2 \, B a d}{4 \, {\left (b^{4} c x^{2} - a b^{3} d x^{2} + 2 \, a b^{3} c x - 2 \, a^{2} b^{2} d x + a^{2} b^{2} c - a^{3} b d\right )}} \]

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)^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*A*b*c - 2*B*b*c + 2*A*a*d + 2*B*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)

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maple [C]  time = 0.49, size = 1379, normalized size = 10.07 \[ \text {result too large to display} \]

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)^3,x)

[Out]

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

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maxima [A]  time = 1.38, size = 230, normalized size = 1.68 \[ \frac {{\left (\frac {2 \, d^{2} e n \log \left (b x + a\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} - \frac {2 \, d^{2} e n \log \left (d x + c\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} + \frac {2 \, b d e n x - b c e n + 3 \, a d e n}{a^{2} b^{2} c - a^{3} b d + {\left (b^{4} c - a b^{3} d\right )} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} x}\right )} B}{4 \, e} - \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {A}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

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)^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)

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mupad [B]  time = 4.66, size = 192, normalized size = 1.40 \[ -\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d\,n-B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2}-\frac {B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}{2\,b\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}-\frac {B\,d^2\,n\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2-2\,a^2\,b\,d^2}{2\,b\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,{\left (a\,d-b\,c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3,x)

[Out]

Timed out

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