3.43 \(\int \frac{a+b \log (c (d+e x)^n)}{(f+g x)^4} \, dx\)

Optimal. Leaf size=141 \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}+\frac{b e^2 n}{3 g (f+g x) (e f-d g)^2}+\frac{b e^3 n \log (d+e x)}{3 g (e f-d g)^3}-\frac{b e^3 n \log (f+g x)}{3 g (e f-d g)^3}+\frac{b e n}{6 g (f+g x)^2 (e f-d g)} \]

[Out]

(b*e*n)/(6*g*(e*f - d*g)*(f + g*x)^2) + (b*e^2*n)/(3*g*(e*f - d*g)^2*(f + g*x)) + (b*e^3*n*Log[d + e*x])/(3*g*
(e*f - d*g)^3) - (a + b*Log[c*(d + e*x)^n])/(3*g*(f + g*x)^3) - (b*e^3*n*Log[f + g*x])/(3*g*(e*f - d*g)^3)

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Rubi [A]  time = 0.0821917, antiderivative size = 141, 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} \[ -\frac{a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}+\frac{b e^2 n}{3 g (f+g x) (e f-d g)^2}+\frac{b e^3 n \log (d+e x)}{3 g (e f-d g)^3}-\frac{b e^3 n \log (f+g x)}{3 g (e f-d g)^3}+\frac{b e n}{6 g (f+g x)^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^4,x]

[Out]

(b*e*n)/(6*g*(e*f - d*g)*(f + g*x)^2) + (b*e^2*n)/(3*g*(e*f - d*g)^2*(f + g*x)) + (b*e^3*n*Log[d + e*x])/(3*g*
(e*f - d*g)^3) - (a + b*Log[c*(d + e*x)^n])/(3*g*(f + g*x)^3) - (b*e^3*n*Log[f + g*x])/(3*g*(e*f - d*g)^3)

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)^4} \, dx &=-\frac{a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}+\frac{(b e n) \int \frac{1}{(d+e x) (f+g x)^3} \, dx}{3 g}\\ &=-\frac{a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}+\frac{(b e n) \int \left (\frac{e^3}{(e f-d g)^3 (d+e x)}-\frac{g}{(e f-d g) (f+g x)^3}-\frac{e g}{(e f-d g)^2 (f+g x)^2}-\frac{e^2 g}{(e f-d g)^3 (f+g x)}\right ) \, dx}{3 g}\\ &=\frac{b e n}{6 g (e f-d g) (f+g x)^2}+\frac{b e^2 n}{3 g (e f-d g)^2 (f+g x)}+\frac{b e^3 n \log (d+e x)}{3 g (e f-d g)^3}-\frac{a+b \log \left (c (d+e x)^n\right )}{3 g (f+g x)^3}-\frac{b e^3 n \log (f+g x)}{3 g (e f-d g)^3}\\ \end{align*}

Mathematica [A]  time = 0.147297, size = 110, normalized size = 0.78 \[ \frac{\frac{b e n (f+g x) \left (2 e^2 (f+g x)^2 \log (d+e x)+(e f-d g) (-d g+3 e f+2 e g x)-2 e^2 (f+g x)^2 \log (f+g x)\right )}{(e f-d g)^3}-2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{6 g (f+g x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^4,x]

[Out]

(-2*(a + b*Log[c*(d + e*x)^n]) + (b*e*n*(f + g*x)*((e*f - d*g)*(3*e*f - d*g + 2*e*g*x) + 2*e^2*(f + g*x)^2*Log
[d + e*x] - 2*e^2*(f + g*x)^2*Log[f + g*x]))/(e*f - d*g)^3)/(6*g*(f + g*x)^3)

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Maple [C]  time = 0.386, size = 950, normalized size = 6.7 \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)^4,x)

[Out]

-1/3*b/g/(g*x+f)^3*ln((e*x+d)^n)+1/6*(-3*b*e^3*f^3*n-2*ln(e*x+d)*b*e^3*f^3*n+2*ln(-g*x-f)*b*e^3*f^3*n-3*I*Pi*b
*d^2*e*f*g^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-2*ln(c)*b*d^3*g^3+2*ln(c)*b*e^3*f^3-b*d^2*e*f*n*g
^2+4*b*d*e^2*f^2*n*g+2*b*d*e^2*g^3*n*x^2-2*b*e^3*f*g^2*n*x^2-b*d^2*e*g^3*n*x-5*b*e^3*f^2*g*n*x-2*a*d^3*g^3-3*I
*Pi*b*d^2*e*f*g^2*csgn(I*c*(e*x+d)^n)^3+3*I*Pi*b*d*e^2*f^2*g*csgn(I*c*(e*x+d)^n)^3+I*Pi*b*d^3*g^3*csgn(I*c)*cs
gn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+2*a*e^3*f^3+6*b*d*e^2*f*g^2*n*x-I*Pi*b*e^3*f^3*csgn(I*c*(e*x+d)^n)^3-I*Pi*
b*e^3*f^3*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+3*I*Pi*b*d^2*e*f*g^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2
+3*I*Pi*b*d^2*e*f*g^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+6*a*d^2*e*f*g^2+2*ln(-g*x-f)*b*e^3*g^3*n*x^3-2*l
n(e*x+d)*b*e^3*g^3*n*x^3+6*ln(c)*b*d^2*e*f*g^2-6*ln(c)*b*d*e^2*f^2*g-I*Pi*b*d^3*g^3*csgn(I*c)*csgn(I*c*(e*x+d)
^n)^2-I*Pi*b*d^3*g^3*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*Pi*b*e^3*f^3*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-6*
a*d*e^2*f^2*g-3*I*Pi*b*d*e^2*f^2*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3*I*Pi*b*d*e^2*f^2*g*csgn(I*(e*x+d)^n)*csgn
(I*c*(e*x+d)^n)^2+3*I*Pi*b*d*e^2*f^2*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*Pi*b*e^3*f^3*csgn(I*(
e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*Pi*b*d^3*g^3*csgn(I*c*(e*x+d)^n)^3+6*ln(-g*x-f)*b*e^3*f*g^2*n*x^2-6*ln(e*x+d
)*b*e^3*f*g^2*n*x^2+6*ln(-g*x-f)*b*e^3*f^2*g*n*x-6*ln(e*x+d)*b*e^3*f^2*g*n*x)/(g*x+f)^3/(d^2*g^2-2*d*e*f*g+e^2
*f^2)/(d*g-e*f)/g

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Maxima [B]  time = 1.19411, size = 406, normalized size = 2.88 \begin{align*} \frac{1}{6} \,{\left (\frac{2 \, e^{2} \log \left (e x + d\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} - \frac{2 \, e^{2} \log \left (g x + f\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} + \frac{2 \, e g x + 3 \, e f - d g}{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 )} b e n - \frac{b \log \left ({\left (e x + d\right )}^{n} c\right )}{3 \,{\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} - \frac{a}{3 \,{\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} 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)^4,x, algorithm="maxima")

[Out]

1/6*(2*e^2*log(e*x + d)/(e^3*f^3*g - 3*d*e^2*f^2*g^2 + 3*d^2*e*f*g^3 - d^3*g^4) - 2*e^2*log(g*x + f)/(e^3*f^3*
g - 3*d*e^2*f^2*g^2 + 3*d^2*e*f*g^3 - d^3*g^4) + (2*e*g*x + 3*e*f - d*g)/(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))*b*e*n - 1/3*
b*log((e*x + d)^n*c)/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*x + f^3*g) - 1/3*a/(g^4*x^3 + 3*f*g^3*x^2 + 3*f^2*g^2*
x + f^3*g)

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Fricas [B]  time = 2.37471, size = 1041, normalized size = 7.38 \begin{align*} -\frac{2 \, a e^{3} f^{3} - 6 \, a d e^{2} f^{2} g + 6 \, a d^{2} e f g^{2} - 2 \, a d^{3} g^{3} - 2 \,{\left (b e^{3} f g^{2} - b d e^{2} g^{3}\right )} n x^{2} -{\left (5 \, b e^{3} f^{2} g - 6 \, b d e^{2} f g^{2} + b d^{2} e g^{3}\right )} n x -{\left (3 \, b e^{3} f^{3} - 4 \, b d e^{2} f^{2} g + b d^{2} e f g^{2}\right )} n - 2 \,{\left (b e^{3} g^{3} n x^{3} + 3 \, b e^{3} f g^{2} n x^{2} + 3 \, b e^{3} f^{2} g n x +{\left (3 \, b d e^{2} f^{2} g - 3 \, b d^{2} e f g^{2} + b d^{3} g^{3}\right )} n\right )} \log \left (e x + d\right ) + 2 \,{\left (b e^{3} g^{3} n x^{3} + 3 \, b e^{3} f g^{2} n x^{2} + 3 \, b e^{3} f^{2} g n x + b e^{3} f^{3} n\right )} \log \left (g x + f\right ) + 2 \,{\left (b e^{3} f^{3} - 3 \, b d e^{2} f^{2} g + 3 \, b d^{2} e f g^{2} - b d^{3} g^{3}\right )} \log \left (c\right )}{6 \,{\left (e^{3} f^{6} g - 3 \, d e^{2} f^{5} g^{2} + 3 \, d^{2} e f^{4} g^{3} - d^{3} f^{3} g^{4} +{\left (e^{3} f^{3} g^{4} - 3 \, d e^{2} f^{2} g^{5} + 3 \, d^{2} e f g^{6} - d^{3} g^{7}\right )} x^{3} + 3 \,{\left (e^{3} f^{4} g^{3} - 3 \, d e^{2} f^{3} g^{4} + 3 \, d^{2} e f^{2} g^{5} - d^{3} f g^{6}\right )} x^{2} + 3 \,{\left (e^{3} f^{5} g^{2} - 3 \, d e^{2} f^{4} g^{3} + 3 \, d^{2} e f^{3} g^{4} - d^{3} f^{2} g^{5}\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)^4,x, algorithm="fricas")

[Out]

-1/6*(2*a*e^3*f^3 - 6*a*d*e^2*f^2*g + 6*a*d^2*e*f*g^2 - 2*a*d^3*g^3 - 2*(b*e^3*f*g^2 - b*d*e^2*g^3)*n*x^2 - (5
*b*e^3*f^2*g - 6*b*d*e^2*f*g^2 + b*d^2*e*g^3)*n*x - (3*b*e^3*f^3 - 4*b*d*e^2*f^2*g + b*d^2*e*f*g^2)*n - 2*(b*e
^3*g^3*n*x^3 + 3*b*e^3*f*g^2*n*x^2 + 3*b*e^3*f^2*g*n*x + (3*b*d*e^2*f^2*g - 3*b*d^2*e*f*g^2 + b*d^3*g^3)*n)*lo
g(e*x + d) + 2*(b*e^3*g^3*n*x^3 + 3*b*e^3*f*g^2*n*x^2 + 3*b*e^3*f^2*g*n*x + b*e^3*f^3*n)*log(g*x + f) + 2*(b*e
^3*f^3 - 3*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2 - b*d^3*g^3)*log(c))/(e^3*f^6*g - 3*d*e^2*f^5*g^2 + 3*d^2*e*f^4*g^3
- d^3*f^3*g^4 + (e^3*f^3*g^4 - 3*d*e^2*f^2*g^5 + 3*d^2*e*f*g^6 - d^3*g^7)*x^3 + 3*(e^3*f^4*g^3 - 3*d*e^2*f^3*
g^4 + 3*d^2*e*f^2*g^5 - d^3*f*g^6)*x^2 + 3*(e^3*f^5*g^2 - 3*d*e^2*f^4*g^3 + 3*d^2*e*f^3*g^4 - d^3*f^2*g^5)*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)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.27562, size = 763, normalized size = 5.41 \begin{align*} \frac{2 \, b g^{3} n x^{3} e^{3} \log \left (g x + f\right ) - 2 \, b g^{3} n x^{3} e^{3} \log \left (x e + d\right ) + 2 \, b d g^{3} n x^{2} e^{2} - b d^{2} g^{3} n x e + 6 \, b f g^{2} n x^{2} e^{3} \log \left (g x + f\right ) - 2 \, b d^{3} g^{3} n \log \left (x e + d\right ) - 6 \, b f g^{2} n x^{2} e^{3} \log \left (x e + d\right ) + 6 \, b d^{2} f g^{2} n e \log \left (x e + d\right ) - 2 \, b f g^{2} n x^{2} e^{3} + 6 \, b d f g^{2} n x e^{2} - b d^{2} f g^{2} n e + 6 \, b f^{2} g n x e^{3} \log \left (g x + f\right ) - 6 \, b f^{2} g n x e^{3} \log \left (x e + d\right ) - 6 \, b d f^{2} g n e^{2} \log \left (x e + d\right ) - 2 \, b d^{3} g^{3} \log \left (c\right ) + 6 \, b d^{2} f g^{2} e \log \left (c\right ) - 2 \, a d^{3} g^{3} - 5 \, b f^{2} g n x e^{3} + 4 \, b d f^{2} g n e^{2} + 6 \, a d^{2} f g^{2} e + 2 \, b f^{3} n e^{3} \log \left (g x + f\right ) - 6 \, b d f^{2} g e^{2} \log \left (c\right ) - 3 \, b f^{3} n e^{3} - 6 \, a d f^{2} g e^{2} + 2 \, b f^{3} e^{3} \log \left (c\right ) + 2 \, a f^{3} e^{3}}{6 \,{\left (d^{3} g^{7} x^{3} - 3 \, d^{2} f g^{6} x^{3} e + 3 \, d^{3} f g^{6} x^{2} + 3 \, d f^{2} g^{5} x^{3} e^{2} - 9 \, d^{2} f^{2} g^{5} x^{2} e + 3 \, d^{3} f^{2} g^{5} x - f^{3} g^{4} x^{3} e^{3} + 9 \, d f^{3} g^{4} x^{2} e^{2} - 9 \, d^{2} f^{3} g^{4} x e + d^{3} f^{3} g^{4} - 3 \, f^{4} g^{3} x^{2} e^{3} + 9 \, d f^{4} g^{3} x e^{2} - 3 \, d^{2} f^{4} g^{3} e - 3 \, f^{5} g^{2} x e^{3} + 3 \, d f^{5} g^{2} e^{2} - f^{6} g e^{3}\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)^4,x, algorithm="giac")

[Out]

1/6*(2*b*g^3*n*x^3*e^3*log(g*x + f) - 2*b*g^3*n*x^3*e^3*log(x*e + d) + 2*b*d*g^3*n*x^2*e^2 - b*d^2*g^3*n*x*e +
6*b*f*g^2*n*x^2*e^3*log(g*x + f) - 2*b*d^3*g^3*n*log(x*e + d) - 6*b*f*g^2*n*x^2*e^3*log(x*e + d) + 6*b*d^2*f*
g^2*n*e*log(x*e + d) - 2*b*f*g^2*n*x^2*e^3 + 6*b*d*f*g^2*n*x*e^2 - b*d^2*f*g^2*n*e + 6*b*f^2*g*n*x*e^3*log(g*x
+ f) - 6*b*f^2*g*n*x*e^3*log(x*e + d) - 6*b*d*f^2*g*n*e^2*log(x*e + d) - 2*b*d^3*g^3*log(c) + 6*b*d^2*f*g^2*e
*log(c) - 2*a*d^3*g^3 - 5*b*f^2*g*n*x*e^3 + 4*b*d*f^2*g*n*e^2 + 6*a*d^2*f*g^2*e + 2*b*f^3*n*e^3*log(g*x + f) -
6*b*d*f^2*g*e^2*log(c) - 3*b*f^3*n*e^3 - 6*a*d*f^2*g*e^2 + 2*b*f^3*e^3*log(c) + 2*a*f^3*e^3)/(d^3*g^7*x^3 - 3
*d^2*f*g^6*x^3*e + 3*d^3*f*g^6*x^2 + 3*d*f^2*g^5*x^3*e^2 - 9*d^2*f^2*g^5*x^2*e + 3*d^3*f^2*g^5*x - f^3*g^4*x^3
*e^3 + 9*d*f^3*g^4*x^2*e^2 - 9*d^2*f^3*g^4*x*e + d^3*f^3*g^4 - 3*f^4*g^3*x^2*e^3 + 9*d*f^4*g^3*x*e^2 - 3*d^2*f
^4*g^3*e - 3*f^5*g^2*x*e^3 + 3*d*f^5*g^2*e^2 - f^6*g*e^3)