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

Optimal. Leaf size=120 \[ -\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{h (g+h x)}+\frac {B n (b c-a d) \log (g+h x)}{(b g-a h) (d g-c h)}+\frac {b B n \log (a+b x)}{h (b g-a h)}-\frac {B d n \log (c+d x)}{h (d g-c h)} \]

[Out]

b*B*n*ln(b*x+a)/h/(-a*h+b*g)-B*d*n*ln(d*x+c)/h/(-c*h+d*g)+(-A-B*ln(e*(b*x+a)^n/((d*x+c)^n)))/h/(h*x+g)+B*(-a*d
+b*c)*n*ln(h*x+g)/(-a*h+b*g)/(-c*h+d*g)

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Rubi [A]  time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6742, 2490, 36, 31} \[ \frac {B (a+b x) \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x) (b g-a h)}-\frac {B n (b c-a d) \log (c+d x)}{(b g-a h) (d g-c h)}+\frac {B n (b c-a d) \log (g+h x)}{(b g-a h) (d g-c h)}-\frac {A}{h (g+h x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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

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Mathematica [A]  time = 0.20, size = 117, normalized size = 0.98 \[ \frac {-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{g+h x}+\frac {B n (b \log (a+b x) (d g-c h)+\log (c+d x) (a d h-b d g)+h (b c-a d) \log (g+h x))}{(b g-a h) (d g-c h)}-\frac {A}{g+h x}}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 11.67, size = 250, normalized size = 2.08 \[ -\frac {A b d g^{2} + A a c h^{2} - {\left (A b c + A a d\right )} g h - {\left ({\left (B b d g h - B b c h^{2}\right )} n x + {\left (B a d g h - B a c h^{2}\right )} n\right )} \log \left (b x + a\right ) + {\left ({\left (B b d g h - B a d h^{2}\right )} n x + {\left (B b c g h - B a c h^{2}\right )} n\right )} \log \left (d x + c\right ) - {\left ({\left (B b c - B a d\right )} h^{2} n x + {\left (B b c - B a d\right )} g h n\right )} \log \left (h x + g\right ) + {\left (B b d g^{2} + B a c h^{2} - {\left (B b c + B a d\right )} g h\right )} \log \relax (e)}{b d g^{3} h + a c g h^{3} - {\left (b c + a d\right )} g^{2} h^{2} + {\left (b d g^{2} h^{2} + a c h^{4} - {\left (b c + a d\right )} g h^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(h*x+g)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.38, size = 166, normalized size = 1.38 \[ \frac {B b^{2} n \log \left ({\left | -b x - a \right |}\right )}{b^{2} g h - a b h^{2}} - \frac {B d^{2} n \log \left ({\left | d x + c \right |}\right )}{d^{2} g h - c d h^{2}} - \frac {B n \log \left (b x + a\right )}{h^{2} x + g h} + \frac {B n \log \left (d x + c\right )}{h^{2} x + g h} + \frac {{\left (B b c n - B a d n\right )} \log \left (h x + g\right )}{b d g^{2} - b c g h - a d g h + a c h^{2}} - \frac {A + B}{h^{2} x + g h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(h*x+g)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.53, size = 1796, normalized size = 14.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(b*x+a)^n/((d*x+c)^n)))/(h*x+g)^2,x)

[Out]

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

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maxima [A]  time = 0.76, size = 151, normalized size = 1.26 \[ \frac {{\left (\frac {b e n \log \left (b x + a\right )}{b g h - a h^{2}} - \frac {d e n \log \left (d x + c\right )}{d g h - c h^{2}} - \frac {{\left (b c e n - a d e n\right )} \log \left (h x + g\right )}{{\left (d g h - c h^{2}\right )} a - {\left (d g^{2} - c g h\right )} b}\right )} B}{e} - \frac {B \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{h^{2} x + g h} - \frac {A}{h^{2} x + g h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))/(h*x+g)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.72, size = 141, normalized size = 1.18 \[ \frac {B\,d\,n\,\ln \left (c+d\,x\right )}{c\,h^2-d\,g\,h}-\frac {\ln \left (g+h\,x\right )\,\left (B\,a\,d\,n-B\,b\,c\,n\right )}{a\,c\,h^2+b\,d\,g^2-a\,d\,g\,h-b\,c\,g\,h}-\frac {B\,\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}{h\,\left (g+h\,x\right )}-\frac {B\,b\,n\,\ln \left (a+b\,x\right )}{a\,h^2-b\,g\,h}-\frac {A}{x\,h^2+g\,h} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log((e*(a + b*x)^n)/(c + d*x)^n))/(g + h*x)^2,x)

[Out]

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

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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)))/(h*x+g)**2,x)

[Out]

Timed out

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