3.454 \(\int \frac{(f+g x) (a+b \log (c x^n))}{(d+e x)^3} \, dx\)
Optimal. Leaf size=115 \[ -\frac{(f+g x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 (d+e x)^2 (e f-d g)}-\frac{b n (d g+e f) \log (d+e x)}{2 d^2 e^2}+\frac{b f^2 n \log (x)}{2 d^2 (e f-d g)}+\frac{b n (e f-d g)}{2 d e^2 (d+e x)} \]
[Out]
(b*(e*f - d*g)*n)/(2*d*e^2*(d + e*x)) + (b*f^2*n*Log[x])/(2*d^2*(e*f - d*g)) - ((f + g*x)^2*(a + b*Log[c*x^n])
)/(2*(e*f - d*g)*(d + e*x)^2) - (b*(e*f + d*g)*n*Log[d + e*x])/(2*d^2*e^2)
________________________________________________________________________________________
Rubi [A] time = 0.145514, antiderivative size = 151, normalized size of antiderivative =
1.31, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules
used = {2357, 2319, 44, 2314, 31} \[ -\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 (d+e x)^2}+\frac{g x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}+\frac{b n \log (x) (e f-d g)}{2 d^2 e^2}-\frac{b n (e f-d g) \log (d+e x)}{2 d^2 e^2}+\frac{b n (e f-d g)}{2 d e^2 (d+e x)}-\frac{b g n \log (d+e x)}{d e^2} \]
Antiderivative was successfully verified.
[In]
Int[((f + g*x)*(a + b*Log[c*x^n]))/(d + e*x)^3,x]
[Out]
(b*(e*f - d*g)*n)/(2*d*e^2*(d + e*x)) + (b*(e*f - d*g)*n*Log[x])/(2*d^2*e^2) - ((e*f - d*g)*(a + b*Log[c*x^n])
)/(2*e^2*(d + e*x)^2) + (g*x*(a + b*Log[c*x^n]))/(d*e*(d + e*x)) - (b*g*n*Log[d + e*x])/(d*e^2) - (b*(e*f - d*
g)*n*Log[d + e*x])/(2*d^2*e^2)
Rule 2357
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]
Rule 2319
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] && !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))
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 2314
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \frac{(f+g x) \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx &=\int \left (\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)^3}+\frac{g \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)^2}\right ) \, dx\\ &=\frac{g \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e}+\frac{(e f-d g) \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e}\\ &=-\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 (d+e x)^2}+\frac{g x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}-\frac{(b g n) \int \frac{1}{d+e x} \, dx}{d e}+\frac{(b (e f-d g) n) \int \frac{1}{x (d+e x)^2} \, dx}{2 e^2}\\ &=-\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 (d+e x)^2}+\frac{g x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}-\frac{b g n \log (d+e x)}{d e^2}+\frac{(b (e f-d g) n) \int \left (\frac{1}{d^2 x}-\frac{e}{d (d+e x)^2}-\frac{e}{d^2 (d+e x)}\right ) \, dx}{2 e^2}\\ &=\frac{b (e f-d g) n}{2 d e^2 (d+e x)}+\frac{b (e f-d g) n \log (x)}{2 d^2 e^2}-\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 (d+e x)^2}+\frac{g x \left (a+b \log \left (c x^n\right )\right )}{d e (d+e x)}-\frac{b g n \log (d+e x)}{d e^2}-\frac{b (e f-d g) n \log (d+e x)}{2 d^2 e^2}\\ \end{align*}
Mathematica [A] time = 0.171129, size = 108, normalized size = 0.94 \[ \frac{-\frac{(e f-d g) \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac{2 g \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac{b n (e f-d g) \left (\frac{d}{d+e x}-\log (d+e x)+\log (x)\right )}{d^2}+\frac{2 b g n (\log (x)-\log (d+e x))}{d}}{2 e^2} \]
Antiderivative was successfully verified.
[In]
Integrate[((f + g*x)*(a + b*Log[c*x^n]))/(d + e*x)^3,x]
[Out]
(-(((e*f - d*g)*(a + b*Log[c*x^n]))/(d + e*x)^2) - (2*g*(a + b*Log[c*x^n]))/(d + e*x) + (2*b*g*n*(Log[x] - Log
[d + e*x]))/d + (b*(e*f - d*g)*n*(d/(d + e*x) + Log[x] - Log[d + e*x]))/d^2)/(2*e^2)
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Maple [C] time = 0.147, size = 624, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*x+f)*(a+b*ln(c*x^n))/(e*x+d)^3,x)
[Out]
-1/2*b*(2*e*g*x+d*g+e*f)/(e*x+d)^2/e^2*ln(x^n)+1/4*(I*Pi*b*d^2*e*f*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+I*Pi*b*
d^3*g*csgn(I*c*x^n)^3-2*a*d^3*g-2*I*Pi*b*d^2*e*g*x*csgn(I*c*x^n)^2*csgn(I*c)-2*I*Pi*b*d^2*e*g*x*csgn(I*x^n)*cs
gn(I*c*x^n)^2-2*b*d^3*g*n-2*a*d^2*e*f+2*ln(-x)*b*e^3*f*n*x^2+2*ln(-x)*b*d^2*e*f*n-2*ln(e*x+d)*b*e^3*f*n*x^2-2*
ln(e*x+d)*b*d^2*e*f*n-4*ln(c)*b*d^2*e*g*x+2*I*Pi*b*d^2*e*g*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-4*a*d^2*e*g*x
+2*b*d^2*e*f*n+2*ln(-x)*b*d^3*g*n-2*ln(e*x+d)*b*d^3*g*n-2*ln(c)*b*d^2*e*f-2*b*d^2*e*g*n*x+2*b*d*e^2*f*n*x-2*ln
(c)*b*d^3*g-I*Pi*b*d^3*g*csgn(I*c*x^n)^2*csgn(I*c)+I*Pi*b*d^2*e*f*csgn(I*c*x^n)^3-I*Pi*b*d^2*e*f*csgn(I*c*x^n)
^2*csgn(I*c)+I*Pi*b*d^3*g*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+2*I*Pi*b*d^2*e*g*x*csgn(I*c*x^n)^3-I*Pi*b*d^2*e*
f*csgn(I*x^n)*csgn(I*c*x^n)^2+2*ln(-x)*b*d*e^2*g*n*x^2+4*ln(-x)*b*d^2*e*g*n*x+4*ln(-x)*b*d*e^2*f*n*x-2*ln(e*x+
d)*b*d*e^2*g*n*x^2-4*ln(e*x+d)*b*d^2*e*g*n*x-4*ln(e*x+d)*b*d*e^2*f*n*x-I*Pi*b*d^3*g*csgn(I*x^n)*csgn(I*c*x^n)^
2)/e^2/d^2/(e*x+d)^2
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Maxima [B] time = 1.2056, size = 294, normalized size = 2.56 \begin{align*} \frac{1}{2} \, b f n{\left (\frac{1}{d e^{2} x + d^{2} e} - \frac{\log \left (e x + d\right )}{d^{2} e} + \frac{\log \left (x\right )}{d^{2} e}\right )} - \frac{1}{2} \, b g n{\left (\frac{1}{e^{3} x + d e^{2}} + \frac{\log \left (e x + d\right )}{d e^{2}} - \frac{\log \left (x\right )}{d e^{2}}\right )} - \frac{{\left (2 \, e x + d\right )} b g \log \left (c x^{n}\right )}{2 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} - \frac{{\left (2 \, e x + d\right )} a g}{2 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} - \frac{b f \log \left (c x^{n}\right )}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac{a f}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="maxima")
[Out]
1/2*b*f*n*(1/(d*e^2*x + d^2*e) - log(e*x + d)/(d^2*e) + log(x)/(d^2*e)) - 1/2*b*g*n*(1/(e^3*x + d*e^2) + log(e
*x + d)/(d*e^2) - log(x)/(d*e^2)) - 1/2*(2*e*x + d)*b*g*log(c*x^n)/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) - 1/2*(2*e*
x + d)*a*g/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) - 1/2*b*f*log(c*x^n)/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 1/2*a*f/(e^3*x
^2 + 2*d*e^2*x + d^2*e)
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Fricas [B] time = 1.32886, size = 466, normalized size = 4.05 \begin{align*} -\frac{a d^{2} e f + a d^{3} g -{\left (b d^{2} e f - b d^{3} g\right )} n +{\left (2 \, a d^{2} e g -{\left (b d e^{2} f - b d^{2} e g\right )} n\right )} x +{\left ({\left (b e^{3} f + b d e^{2} g\right )} n x^{2} + 2 \,{\left (b d e^{2} f + b d^{2} e g\right )} n x +{\left (b d^{2} e f + b d^{3} g\right )} n\right )} \log \left (e x + d\right ) +{\left (2 \, b d^{2} e g x + b d^{2} e f + b d^{3} g\right )} \log \left (c\right ) -{\left (2 \, b d e^{2} f n x +{\left (b e^{3} f + b d e^{2} g\right )} n x^{2}\right )} \log \left (x\right )}{2 \,{\left (d^{2} e^{4} x^{2} + 2 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="fricas")
[Out]
-1/2*(a*d^2*e*f + a*d^3*g - (b*d^2*e*f - b*d^3*g)*n + (2*a*d^2*e*g - (b*d*e^2*f - b*d^2*e*g)*n)*x + ((b*e^3*f
+ b*d*e^2*g)*n*x^2 + 2*(b*d*e^2*f + b*d^2*e*g)*n*x + (b*d^2*e*f + b*d^3*g)*n)*log(e*x + d) + (2*b*d^2*e*g*x +
b*d^2*e*f + b*d^3*g)*log(c) - (2*b*d*e^2*f*n*x + (b*e^3*f + b*d*e^2*g)*n*x^2)*log(x))/(d^2*e^4*x^2 + 2*d^3*e^3
*x + d^4*e^2)
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Sympy [A] time = 5.41824, size = 1103, normalized size = 9.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(a+b*ln(c*x**n))/(e*x+d)**3,x)
[Out]
Piecewise((zoo*(-a*f/(2*x**2) - a*g/x - b*f*n*log(x)/(2*x**2) - b*f*n/(4*x**2) - b*f*log(c)/(2*x**2) - b*g*n*l
og(x)/x - b*g*n/x - b*g*log(c)/x), Eq(d, 0) & Eq(e, 0)), ((-a*f/(2*x**2) - a*g/x - b*f*n*log(x)/(2*x**2) - b*f
*n/(4*x**2) - b*f*log(c)/(2*x**2) - b*g*n*log(x)/x - b*g*n/x - b*g*log(c)/x)/e**3, Eq(d, 0)), ((a*f*x + a*g*x*
*2/2 + b*f*n*x*log(x) - b*f*n*x + b*f*x*log(c) + b*g*n*x**2*log(x)/2 - b*g*n*x**2/4 + b*g*x**2*log(c)/2)/d**3,
Eq(e, 0)), (2*a*d*e**2*f*x/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) + a*d*e**2*g*x**2/(2*d**4*e**2 +
4*d**3*e**3*x + 2*d**2*e**4*x**2) + a*e**3*f*x**2/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) - b*d**3*g*
n*log(d/e + x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) - b*d**2*e*f*n*log(d/e + x)/(2*d**4*e**2 + 4*d
**3*e**3*x + 2*d**2*e**4*x**2) - 2*b*d**2*e*g*n*x*log(d/e + x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2
) + b*d**2*e*g*n*x/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) + 2*b*d*e**2*f*n*x*log(x)/(2*d**4*e**2 + 4
*d**3*e**3*x + 2*d**2*e**4*x**2) - 2*b*d*e**2*f*n*x*log(d/e + x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x*
*2) - b*d*e**2*f*n*x/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) + 2*b*d*e**2*f*x*log(c)/(2*d**4*e**2 + 4
*d**3*e**3*x + 2*d**2*e**4*x**2) + b*d*e**2*g*n*x**2*log(x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) -
b*d*e**2*g*n*x**2*log(d/e + x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) + b*d*e**2*g*n*x**2/(2*d**4*e
**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) + b*d*e**2*g*x**2*log(c)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x*
*2) + b*e**3*f*n*x**2*log(x)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) - b*e**3*f*n*x**2*log(d/e + x)/(
2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2) - b*e**3*f*n*x**2/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x
**2) + b*e**3*f*x**2*log(c)/(2*d**4*e**2 + 4*d**3*e**3*x + 2*d**2*e**4*x**2), True))
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Giac [B] time = 1.32223, size = 340, normalized size = 2.96 \begin{align*} -\frac{b d g n x^{2} e^{2} \log \left (x e + d\right ) + 2 \, b d^{2} g n x e \log \left (x e + d\right ) - b d g n x^{2} e^{2} \log \left (x\right ) + b d^{2} g n x e + b d^{3} g n \log \left (x e + d\right ) + b f n x^{2} e^{3} \log \left (x e + d\right ) + 2 \, b d f n x e^{2} \log \left (x e + d\right ) + b d^{2} f n e \log \left (x e + d\right ) + 2 \, b d^{2} g x e \log \left (c\right ) - b f n x^{2} e^{3} \log \left (x\right ) - 2 \, b d f n x e^{2} \log \left (x\right ) + b d^{3} g n - b d f n x e^{2} - b d^{2} f n e + 2 \, a d^{2} g x e + b d^{3} g \log \left (c\right ) + b d^{2} f e \log \left (c\right ) + a d^{3} g + a d^{2} f e}{2 \,{\left (d^{2} x^{2} e^{4} + 2 \, d^{3} x e^{3} + d^{4} e^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(a+b*log(c*x^n))/(e*x+d)^3,x, algorithm="giac")
[Out]
-1/2*(b*d*g*n*x^2*e^2*log(x*e + d) + 2*b*d^2*g*n*x*e*log(x*e + d) - b*d*g*n*x^2*e^2*log(x) + b*d^2*g*n*x*e + b
*d^3*g*n*log(x*e + d) + b*f*n*x^2*e^3*log(x*e + d) + 2*b*d*f*n*x*e^2*log(x*e + d) + b*d^2*f*n*e*log(x*e + d) +
2*b*d^2*g*x*e*log(c) - b*f*n*x^2*e^3*log(x) - 2*b*d*f*n*x*e^2*log(x) + b*d^3*g*n - b*d*f*n*x*e^2 - b*d^2*f*n*
e + 2*a*d^2*g*x*e + b*d^3*g*log(c) + b*d^2*f*e*log(c) + a*d^3*g + a*d^2*f*e)/(d^2*x^2*e^4 + 2*d^3*x*e^3 + d^4*
e^2)