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

Optimal. Leaf size=149 \[ \frac{(f x+g)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f}+\frac{b n x (d f-e g)^3}{4 e^3}-\frac{b n (f x+g)^2 (d f-e g)^2}{8 e^2 f}-\frac{b n (d f-e g)^4 \log (d+e x)}{4 e^4 f}+\frac{b n (f x+g)^3 (d f-e g)}{12 e f}-\frac{b n (f x+g)^4}{16 f} \]

[Out]

(b*(d*f - e*g)^3*n*x)/(4*e^3) - (b*(d*f - e*g)^2*n*(g + f*x)^2)/(8*e^2*f) + (b*(d*f - e*g)*n*(g + f*x)^3)/(12*
e*f) - (b*n*(g + f*x)^4)/(16*f) - (b*(d*f - e*g)^4*n*Log[d + e*x])/(4*e^4*f) + ((g + f*x)^4*(a + b*Log[c*(d +
e*x)^n]))/(4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.12266, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2412, 2395, 43} \[ \frac{(f x+g)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f}+\frac{b n x (d f-e g)^3}{4 e^3}-\frac{b n (f x+g)^2 (d f-e g)^2}{8 e^2 f}-\frac{b n (d f-e g)^4 \log (d+e x)}{4 e^4 f}+\frac{b n (f x+g)^3 (d f-e g)}{12 e f}-\frac{b n (f x+g)^4}{16 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(d*f - e*g)^3*n*x)/(4*e^3) - (b*(d*f - e*g)^2*n*(g + f*x)^2)/(8*e^2*f) + (b*(d*f - e*g)*n*(g + f*x)^3)/(12*
e*f) - (b*n*(g + f*x)^4)/(16*f) - (b*(d*f - e*g)^4*n*Log[d + e*x])/(4*e^4*f) + ((g + f*x)^4*(a + b*Log[c*(d +
e*x)^n]))/(4*f)

Rule 2412

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol]
:> Int[(g + f*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q}, x] && EqQ[m,
q] && IntegerQ[q]

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (f+\frac{g}{x}\right )^3 x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=\int (g+f x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx\\ &=\frac{(g+f x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f}-\frac{(b e n) \int \frac{(g+f x)^4}{d+e x} \, dx}{4 f}\\ &=\frac{(g+f x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f}-\frac{(b e n) \int \left (\frac{f (-d f+e g)^3}{e^4}+\frac{(-d f+e g)^4}{e^4 (d+e x)}+\frac{f (-d f+e g)^2 (g+f x)}{e^3}+\frac{f (-d f+e g) (g+f x)^2}{e^2}+\frac{f (g+f x)^3}{e}\right ) \, dx}{4 f}\\ &=\frac{b (d f-e g)^3 n x}{4 e^3}-\frac{b (d f-e g)^2 n (g+f x)^2}{8 e^2 f}+\frac{b (d f-e g) n (g+f x)^3}{12 e f}-\frac{b n (g+f x)^4}{16 f}-\frac{b (d f-e g)^4 n \log (d+e x)}{4 e^4 f}+\frac{(g+f x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.208142, size = 226, normalized size = 1.52 \[ \frac{e x \left (12 a e^3 \left (4 f^2 g x^2+f^3 x^3+6 f g^2 x+4 g^3\right )+b n \left (-6 d^2 e f^2 (f x+8 g)+12 d^3 f^3+4 d e^2 f \left (f^2 x^2+6 f g x+18 g^2\right )+e^3 \left (-\left (16 f^2 g x^2+3 f^3 x^3+36 f g^2 x+48 g^3\right )\right )\right )\right )+12 b e^3 \left (4 d g^3+e x \left (4 f^2 g x^2+f^3 x^3+6 f g^2 x+4 g^3\right )\right ) \log \left (c (d+e x)^n\right )-12 b d^2 f n \left (d^2 f^2-4 d e f g+6 e^2 g^2\right ) \log (d+e x)}{48 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(12*a*e^3*(4*g^3 + 6*f*g^2*x + 4*f^2*g*x^2 + f^3*x^3) + b*n*(12*d^3*f^3 - 6*d^2*e*f^2*(8*g + f*x) + 4*d*e
^2*f*(18*g^2 + 6*f*g*x + f^2*x^2) - e^3*(48*g^3 + 36*f*g^2*x + 16*f^2*g*x^2 + 3*f^3*x^3))) - 12*b*d^2*f*(d^2*f
^2 - 4*d*e*f*g + 6*e^2*g^2)*n*Log[d + e*x] + 12*b*e^3*(4*d*g^3 + e*x*(4*g^3 + 6*f*g^2*x + 4*f^2*g*x^2 + f^3*x^
3))*Log[c*(d + e*x)^n])/(48*e^4)

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Maple [C]  time = 0.481, size = 836, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f+g/x)^3*x^3*(a+b*ln(c*(e*x+d)^n)),x)

[Out]

1/4*f^3*a*x^4+1/4*f^3*ln(c)*b*x^4+ln(c)*b*g^3*x+1/4*(f*x+g)^4*b/f*ln((e*x+d)^n)+3/2*f*a*g^2*x^2-1/16*f^3*b*n*x
^4+f^2*a*g*x^3+a*g^3*x-1/4/f*ln(e*x+d)*b*g^4*n+f^2*ln(c)*b*g*x^3+3/2*f*ln(c)*b*g^2*x^2+3/4*I*f*Pi*b*g^2*x^2*cs
gn(I*c)*csgn(I*c*(e*x+d)^n)^2+3/4*I*f*Pi*b*g^2*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/2*I*f^2*Pi*b*g*x^
3*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/8*I*f^3*Pi*b*x^4*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*f^2
*Pi*b*g*x^3*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*Pi*b*g^3*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x
+d)^n)+3/2/e*f*b*d*g^2*n*x+1/2/e*f^2*b*d*g*n*x^2-1/e^2*f^2*b*d^2*g*n*x+1/e^3*f^2*ln(e*x+d)*b*d^3*g*n-3/2/e^2*f
*ln(e*x+d)*b*d^2*g^2*n+1/8*I*f^3*Pi*b*x^4*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-3/4*f*b*g^2*n*x^2+1/12/e*f^3*b*d*n*x
^3-1/3*f^2*b*g*n*x^3-1/8/e^2*f^3*b*d^2*n*x^2-3/4*I*f*Pi*b*g^2*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)
^n)-1/2*I*f^2*Pi*b*g*x^3*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4/e^3*f^3*b*d^3*n*x-b*g^3*n*x+1/e*l
n(e*x+d)*b*d*g^3*n-1/4/e^4*f^3*ln(e*x+d)*b*d^4*n-1/2*I*Pi*b*g^3*x*csgn(I*c*(e*x+d)^n)^3-1/8*I*f^3*Pi*b*x^4*csg
n(I*c*(e*x+d)^n)^3-3/4*I*f*Pi*b*g^2*x^2*csgn(I*c*(e*x+d)^n)^3+1/8*I*f^3*Pi*b*x^4*csgn(I*(e*x+d)^n)*csgn(I*c*(e
*x+d)^n)^2-1/2*I*f^2*Pi*b*g*x^3*csgn(I*c*(e*x+d)^n)^3+1/2*I*Pi*b*g^3*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2
+1/2*I*Pi*b*g^3*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2

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Maxima [B]  time = 1.14845, size = 383, normalized size = 2.57 \begin{align*} \frac{1}{4} \, b f^{3} x^{4} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac{1}{4} \, a f^{3} x^{4} + b f^{2} g x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + a f^{2} g x^{3} - b e g^{3} n{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} - \frac{1}{48} \, b e f^{3} n{\left (\frac{12 \, d^{4} \log \left (e x + d\right )}{e^{5}} + \frac{3 \, e^{3} x^{4} - 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} - 12 \, d^{3} x}{e^{4}}\right )} + \frac{1}{6} \, b e f^{2} g n{\left (\frac{6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac{2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} - \frac{3}{4} \, b e f g^{2} n{\left (\frac{2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac{e x^{2} - 2 \, d x}{e^{2}}\right )} + \frac{3}{2} \, b f g^{2} x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac{3}{2} \, a f g^{2} x^{2} + b g^{3} x \log \left ({\left (e x + d\right )}^{n} c\right ) + a g^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*f^3*x^4*log((e*x + d)^n*c) + 1/4*a*f^3*x^4 + b*f^2*g*x^3*log((e*x + d)^n*c) + a*f^2*g*x^3 - b*e*g^3*n*(x
/e - d*log(e*x + d)/e^2) - 1/48*b*e*f^3*n*(12*d^4*log(e*x + d)/e^5 + (3*e^3*x^4 - 4*d*e^2*x^3 + 6*d^2*e*x^2 -
12*d^3*x)/e^4) + 1/6*b*e*f^2*g*n*(6*d^3*log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^2*x)/e^3) - 3/4*b*e*f*
g^2*n*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 3/2*b*f*g^2*x^2*log((e*x + d)^n*c) + 3/2*a*f*g^2*x^2 +
b*g^3*x*log((e*x + d)^n*c) + a*g^3*x

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Fricas [B]  time = 1.95985, size = 713, normalized size = 4.79 \begin{align*} -\frac{3 \,{\left (b e^{4} f^{3} n - 4 \, a e^{4} f^{3}\right )} x^{4} - 4 \,{\left (12 \, a e^{4} f^{2} g +{\left (b d e^{3} f^{3} - 4 \, b e^{4} f^{2} g\right )} n\right )} x^{3} - 6 \,{\left (12 \, a e^{4} f g^{2} -{\left (b d^{2} e^{2} f^{3} - 4 \, b d e^{3} f^{2} g + 6 \, b e^{4} f g^{2}\right )} n\right )} x^{2} - 12 \,{\left (4 \, a e^{4} g^{3} +{\left (b d^{3} e f^{3} - 4 \, b d^{2} e^{2} f^{2} g + 6 \, b d e^{3} f g^{2} - 4 \, b e^{4} g^{3}\right )} n\right )} x - 12 \,{\left (b e^{4} f^{3} n x^{4} + 4 \, b e^{4} f^{2} g n x^{3} + 6 \, b e^{4} f g^{2} n x^{2} + 4 \, b e^{4} g^{3} n x -{\left (b d^{4} f^{3} - 4 \, b d^{3} e f^{2} g + 6 \, b d^{2} e^{2} f g^{2} - 4 \, b d e^{3} g^{3}\right )} n\right )} \log \left (e x + d\right ) - 12 \,{\left (b e^{4} f^{3} x^{4} + 4 \, b e^{4} f^{2} g x^{3} + 6 \, b e^{4} f g^{2} x^{2} + 4 \, b e^{4} g^{3} x\right )} \log \left (c\right )}{48 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(b*e^4*f^3*n - 4*a*e^4*f^3)*x^4 - 4*(12*a*e^4*f^2*g + (b*d*e^3*f^3 - 4*b*e^4*f^2*g)*n)*x^3 - 6*(12*a*
e^4*f*g^2 - (b*d^2*e^2*f^3 - 4*b*d*e^3*f^2*g + 6*b*e^4*f*g^2)*n)*x^2 - 12*(4*a*e^4*g^3 + (b*d^3*e*f^3 - 4*b*d^
2*e^2*f^2*g + 6*b*d*e^3*f*g^2 - 4*b*e^4*g^3)*n)*x - 12*(b*e^4*f^3*n*x^4 + 4*b*e^4*f^2*g*n*x^3 + 6*b*e^4*f*g^2*
n*x^2 + 4*b*e^4*g^3*n*x - (b*d^4*f^3 - 4*b*d^3*e*f^2*g + 6*b*d^2*e^2*f*g^2 - 4*b*d*e^3*g^3)*n)*log(e*x + d) -
12*(b*e^4*f^3*x^4 + 4*b*e^4*f^2*g*x^3 + 6*b*e^4*f*g^2*x^2 + 4*b*e^4*g^3*x)*log(c))/e^4

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Sympy [A]  time = 42.5615, size = 450, normalized size = 3.02 \begin{align*} \begin{cases} \frac{a f^{3} x^{4}}{4} + a f^{2} g x^{3} + \frac{3 a f g^{2} x^{2}}{2} + a g^{3} x - \frac{b d^{4} f^{3} n \log{\left (d + e x \right )}}{4 e^{4}} + \frac{b d^{3} f^{3} n x}{4 e^{3}} + \frac{b d^{3} f^{2} g n \log{\left (d + e x \right )}}{e^{3}} - \frac{b d^{2} f^{3} n x^{2}}{8 e^{2}} - \frac{b d^{2} f^{2} g n x}{e^{2}} - \frac{3 b d^{2} f g^{2} n \log{\left (d + e x \right )}}{2 e^{2}} + \frac{b d f^{3} n x^{3}}{12 e} + \frac{b d f^{2} g n x^{2}}{2 e} + \frac{3 b d f g^{2} n x}{2 e} + \frac{b d g^{3} n \log{\left (d + e x \right )}}{e} + \frac{b f^{3} n x^{4} \log{\left (d + e x \right )}}{4} - \frac{b f^{3} n x^{4}}{16} + \frac{b f^{3} x^{4} \log{\left (c \right )}}{4} + b f^{2} g n x^{3} \log{\left (d + e x \right )} - \frac{b f^{2} g n x^{3}}{3} + b f^{2} g x^{3} \log{\left (c \right )} + \frac{3 b f g^{2} n x^{2} \log{\left (d + e x \right )}}{2} - \frac{3 b f g^{2} n x^{2}}{4} + \frac{3 b f g^{2} x^{2} \log{\left (c \right )}}{2} + b g^{3} n x \log{\left (d + e x \right )} - b g^{3} n x + b g^{3} x \log{\left (c \right )} & \text{for}\: e \neq 0 \\\left (a + b \log{\left (c d^{n} \right )}\right ) \left (\frac{f^{3} x^{4}}{4} + f^{2} g x^{3} + \frac{3 f g^{2} x^{2}}{2} + g^{3} x\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g/x)**3*x**3*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Piecewise((a*f**3*x**4/4 + a*f**2*g*x**3 + 3*a*f*g**2*x**2/2 + a*g**3*x - b*d**4*f**3*n*log(d + e*x)/(4*e**4)
+ b*d**3*f**3*n*x/(4*e**3) + b*d**3*f**2*g*n*log(d + e*x)/e**3 - b*d**2*f**3*n*x**2/(8*e**2) - b*d**2*f**2*g*n
*x/e**2 - 3*b*d**2*f*g**2*n*log(d + e*x)/(2*e**2) + b*d*f**3*n*x**3/(12*e) + b*d*f**2*g*n*x**2/(2*e) + 3*b*d*f
*g**2*n*x/(2*e) + b*d*g**3*n*log(d + e*x)/e + b*f**3*n*x**4*log(d + e*x)/4 - b*f**3*n*x**4/16 + b*f**3*x**4*lo
g(c)/4 + b*f**2*g*n*x**3*log(d + e*x) - b*f**2*g*n*x**3/3 + b*f**2*g*x**3*log(c) + 3*b*f*g**2*n*x**2*log(d + e
*x)/2 - 3*b*f*g**2*n*x**2/4 + 3*b*f*g**2*x**2*log(c)/2 + b*g**3*n*x*log(d + e*x) - b*g**3*n*x + b*g**3*x*log(c
), Ne(e, 0)), ((a + b*log(c*d**n))*(f**3*x**4/4 + f**2*g*x**3 + 3*f*g**2*x**2/2 + g**3*x), True))

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Giac [B]  time = 1.33416, size = 1053, normalized size = 7.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x*e + d)^4*b*f^3*n*e^(-4)*log(x*e + d) - (x*e + d)^3*b*d*f^3*n*e^(-4)*log(x*e + d) + 3/2*(x*e + d)^2*b*d^
2*f^3*n*e^(-4)*log(x*e + d) - (x*e + d)*b*d^3*f^3*n*e^(-4)*log(x*e + d) - 1/16*(x*e + d)^4*b*f^3*n*e^(-4) + 1/
3*(x*e + d)^3*b*d*f^3*n*e^(-4) - 3/4*(x*e + d)^2*b*d^2*f^3*n*e^(-4) + (x*e + d)*b*d^3*f^3*n*e^(-4) + (x*e + d)
^3*b*f^2*g*n*e^(-3)*log(x*e + d) - 3*(x*e + d)^2*b*d*f^2*g*n*e^(-3)*log(x*e + d) + 3*(x*e + d)*b*d^2*f^2*g*n*e
^(-3)*log(x*e + d) + 1/4*(x*e + d)^4*b*f^3*e^(-4)*log(c) - (x*e + d)^3*b*d*f^3*e^(-4)*log(c) + 3/2*(x*e + d)^2
*b*d^2*f^3*e^(-4)*log(c) - (x*e + d)*b*d^3*f^3*e^(-4)*log(c) - 1/3*(x*e + d)^3*b*f^2*g*n*e^(-3) + 3/2*(x*e + d
)^2*b*d*f^2*g*n*e^(-3) - 3*(x*e + d)*b*d^2*f^2*g*n*e^(-3) + 1/4*(x*e + d)^4*a*f^3*e^(-4) - (x*e + d)^3*a*d*f^3
*e^(-4) + 3/2*(x*e + d)^2*a*d^2*f^3*e^(-4) - (x*e + d)*a*d^3*f^3*e^(-4) + 3/2*(x*e + d)^2*b*f*g^2*n*e^(-2)*log
(x*e + d) - 3*(x*e + d)*b*d*f*g^2*n*e^(-2)*log(x*e + d) + (x*e + d)^3*b*f^2*g*e^(-3)*log(c) - 3*(x*e + d)^2*b*
d*f^2*g*e^(-3)*log(c) + 3*(x*e + d)*b*d^2*f^2*g*e^(-3)*log(c) - 3/4*(x*e + d)^2*b*f*g^2*n*e^(-2) + 3*(x*e + d)
*b*d*f*g^2*n*e^(-2) + (x*e + d)^3*a*f^2*g*e^(-3) - 3*(x*e + d)^2*a*d*f^2*g*e^(-3) + 3*(x*e + d)*a*d^2*f^2*g*e^
(-3) + (x*e + d)*b*g^3*n*e^(-1)*log(x*e + d) + 3/2*(x*e + d)^2*b*f*g^2*e^(-2)*log(c) - 3*(x*e + d)*b*d*f*g^2*e
^(-2)*log(c) - (x*e + d)*b*g^3*n*e^(-1) + 3/2*(x*e + d)^2*a*f*g^2*e^(-2) - 3*(x*e + d)*a*d*f*g^2*e^(-2) + (x*e
+ d)*b*g^3*e^(-1)*log(c) + (x*e + d)*a*g^3*e^(-1)