3.42 \(\int \frac{a+b \log (c (d+e x)^n)}{(f+g x)^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{2 g (f+g x)^2}+\frac{b e^2 n \log (d+e x)}{2 g (e f-d g)^2}-\frac{b e^2 n \log (f+g x)}{2 g (e f-d g)^2}+\frac{b e n}{2 g (f+g x) (e f-d g)} \]

[Out]

(b*e*n)/(2*g*(e*f - d*g)*(f + g*x)) + (b*e^2*n*Log[d + e*x])/(2*g*(e*f - d*g)^2) - (a + b*Log[c*(d + e*x)^n])/
(2*g*(f + g*x)^2) - (b*e^2*n*Log[f + g*x])/(2*g*(e*f - d*g)^2)

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Rubi [A]  time = 0.0622702, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2395, 44} \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{2 g (f+g x)^2}+\frac{b e^2 n \log (d+e x)}{2 g (e f-d g)^2}-\frac{b e^2 n \log (f+g x)}{2 g (e f-d g)^2}+\frac{b e n}{2 g (f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*n)/(2*g*(e*f - d*g)*(f + g*x)) + (b*e^2*n*Log[d + e*x])/(2*g*(e*f - d*g)^2) - (a + b*Log[c*(d + e*x)^n])/
(2*g*(f + g*x)^2) - (b*e^2*n*Log[f + g*x])/(2*g*(e*f - d*g)^2)

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && 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])

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c (d+e x)^n\right )}{(f+g x)^3} \, dx &=-\frac{a+b \log \left (c (d+e x)^n\right )}{2 g (f+g x)^2}+\frac{(b e n) \int \frac{1}{(d+e x) (f+g x)^2} \, dx}{2 g}\\ &=-\frac{a+b \log \left (c (d+e x)^n\right )}{2 g (f+g x)^2}+\frac{(b e n) \int \left (\frac{e^2}{(e f-d g)^2 (d+e x)}-\frac{g}{(e f-d g) (f+g x)^2}-\frac{e g}{(e f-d g)^2 (f+g x)}\right ) \, dx}{2 g}\\ &=\frac{b e n}{2 g (e f-d g) (f+g x)}+\frac{b e^2 n \log (d+e x)}{2 g (e f-d g)^2}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 g (f+g x)^2}-\frac{b e^2 n \log (f+g x)}{2 g (e f-d g)^2}\\ \end{align*}

Mathematica [A]  time = 0.103032, size = 83, normalized size = 0.74 \[ -\frac{a+b \log \left (c (d+e x)^n\right )-\frac{b e n (f+g x) (e (f+g x) \log (d+e x)-d g-e (f+g x) \log (f+g x)+e f)}{(e f-d g)^2}}{2 g (f+g x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*Log[c*(d + e*x)^n] - (b*e*n*(f + g*x)*(e*f - d*g + e*(f + g*x)*Log[d + e*x] - e*(f + g*x)*Log[f + g*x]
))/(e*f - d*g)^2)/(2*g*(f + g*x)^2)

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Maple [C]  time = 0.374, size = 633, normalized size = 5.7 \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(c*(e*x+d)^n))/(g*x+f)^3,x)

[Out]

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

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Maxima [A]  time = 1.1057, size = 225, normalized size = 2.01 \begin{align*} \frac{1}{2} \, b e n{\left (\frac{e \log \left (e x + d\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} - \frac{e \log \left (g x + f\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} + \frac{1}{e f^{2} g - d f g^{2} +{\left (e f g^{2} - d g^{3}\right )} x}\right )} - \frac{b \log \left ({\left (e x + d\right )}^{n} c\right )}{2 \,{\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} - \frac{a}{2 \,{\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*e*n*(e*log(e*x + d)/(e^2*f^2*g - 2*d*e*f*g^2 + d^2*g^3) - e*log(g*x + f)/(e^2*f^2*g - 2*d*e*f*g^2 + d^2*
g^3) + 1/(e*f^2*g - d*f*g^2 + (e*f*g^2 - d*g^3)*x)) - 1/2*b*log((e*x + d)^n*c)/(g^3*x^2 + 2*f*g^2*x + f^2*g) -
1/2*a/(g^3*x^2 + 2*f*g^2*x + f^2*g)

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Fricas [B]  time = 2.54747, size = 579, normalized size = 5.17 \begin{align*} -\frac{a e^{2} f^{2} - 2 \, a d e f g + a d^{2} g^{2} -{\left (b e^{2} f g - b d e g^{2}\right )} n x -{\left (b e^{2} f^{2} - b d e f g\right )} n -{\left (b e^{2} g^{2} n x^{2} + 2 \, b e^{2} f g n x +{\left (2 \, b d e f g - b d^{2} g^{2}\right )} n\right )} \log \left (e x + d\right ) +{\left (b e^{2} g^{2} n x^{2} + 2 \, b e^{2} f g n x + b e^{2} f^{2} n\right )} \log \left (g x + f\right ) +{\left (b e^{2} f^{2} - 2 \, b d e f g + b d^{2} g^{2}\right )} \log \left (c\right )}{2 \,{\left (e^{2} f^{4} g - 2 \, d e f^{3} g^{2} + d^{2} f^{2} g^{3} +{\left (e^{2} f^{2} g^{3} - 2 \, d e f g^{4} + d^{2} g^{5}\right )} x^{2} + 2 \,{\left (e^{2} f^{3} g^{2} - 2 \, d e f^{2} g^{3} + d^{2} f g^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.22982, size = 408, normalized size = 3.64 \begin{align*} -\frac{b g^{2} n x^{2} e^{2} \log \left (g x + f\right ) - b g^{2} n x^{2} e^{2} \log \left (x e + d\right ) + b d g^{2} n x e + 2 \, b f g n x e^{2} \log \left (g x + f\right ) + b d^{2} g^{2} n \log \left (x e + d\right ) - 2 \, b f g n x e^{2} \log \left (x e + d\right ) - 2 \, b d f g n e \log \left (x e + d\right ) - b f g n x e^{2} + b d f g n e + b f^{2} n e^{2} \log \left (g x + f\right ) + b d^{2} g^{2} \log \left (c\right ) - 2 \, b d f g e \log \left (c\right ) + a d^{2} g^{2} - b f^{2} n e^{2} - 2 \, a d f g e + b f^{2} e^{2} \log \left (c\right ) + a f^{2} e^{2}}{2 \,{\left (d^{2} g^{5} x^{2} - 2 \, d f g^{4} x^{2} e + 2 \, d^{2} f g^{4} x + f^{2} g^{3} x^{2} e^{2} - 4 \, d f^{2} g^{3} x e + d^{2} f^{2} g^{3} + 2 \, f^{3} g^{2} x e^{2} - 2 \, d f^{3} g^{2} e + f^{4} g e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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