3.152 \(\int \frac{A+B \log (e (a+b x)^n (c+d x)^{-n})}{(a+b x)^2} \, dx\)

Optimal. Leaf size=97 \[ -\frac{B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{b (a+b x)}-\frac{B d n \log (a+b x)}{b (b c-a d)}+\frac{B d n \log (c+d x)}{b (b c-a d)}-\frac{B n}{b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0835585, antiderivative size = 72, normalized size of antiderivative = 0.74, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {6742, 2490, 32} \[ -\frac{A}{b (a+b x)}-\frac{B (c+d x) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x) (b c-a d)}-\frac{B n}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

Rule 2490

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0895557, size = 89, normalized size = 0.92 \[ \frac{-(b c-a d) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A+B n\right )+B d n (a+b x) \log (c+d x)-B d n (a+b x) \log (a+b x)}{b (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.4, size = 823, normalized size = 8.5 \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.18632, size = 157, normalized size = 1.62 \begin{align*} -\frac{{\left (\frac{d e n \log \left (b x + a\right )}{b^{2} c - a b d} - \frac{d e n \log \left (d x + c\right )}{b^{2} c - a b d} + \frac{e n}{b^{2} x + a b}\right )} B}{e} - \frac{B \log \left (\frac{{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{b^{2} x + a b} - \frac{A}{b^{2} x + a b} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.06594, size = 242, normalized size = 2.49 \begin{align*} -\frac{A b c - A a d +{\left (B b c - B a d\right )} n +{\left (B b d n x + B b c n\right )} \log \left (b x + a\right ) -{\left (B b d n x + B b c n\right )} \log \left (d x + c\right ) +{\left (B b c - B a d\right )} \log \left (e\right )}{a b^{2} c - a^{2} b d +{\left (b^{3} c - a b^{2} d\right )} x} \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)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.26185, size = 146, normalized size = 1.51 \begin{align*} -\frac{B d n \log \left (b x + a\right )}{b^{2} c - a b d} + \frac{B d n \log \left (d x + c\right )}{b^{2} c - a b d} - \frac{B n \log \left (b x + a\right )}{b^{2} x + a b} + \frac{B n \log \left (d x + c\right )}{b^{2} x + a b} - \frac{B n + A + B}{b^{2} x + a b} \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)^2,x, algorithm="giac")

[Out]

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