3.304 \(\int (f+\frac {g}{x})^2 x^2 (a+b \log (c (d+e x)^n)) \, dx\)

Optimal. Leaf size=120 \[ \frac {(f x+g)^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 f}+\frac {b n (d f-e g)^3 \log (d+e x)}{3 e^3 f}-\frac {b n x (d f-e g)^2}{3 e^2}+\frac {b n (f x+g)^2 (d f-e g)}{6 e f}-\frac {b n (f x+g)^3}{9 f} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 120, normalized size of antiderivative = 1.00, 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} \[ \frac {(f x+g)^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 f}-\frac {b n x (d f-e g)^2}{3 e^2}+\frac {b n (d f-e g)^3 \log (d+e x)}{3 e^3 f}+\frac {b n (f x+g)^2 (d f-e g)}{6 e f}-\frac {b n (f x+g)^3}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(d*f - e*g)^2*n*x)/(3*e^2) + (b*(d*f - e*g)*n*(g + f*x)^2)/(6*e*f) - (b*n*(g + f*x)^3)/(9*f) + (b*(d*f - e
*g)^3*n*Log[d + e*x])/(3*e^3*f) + ((g + f*x)^3*(a + b*Log[c*(d + e*x)^n]))/(3*f)

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])

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 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]

Rubi steps

\begin {align*} \int \left (f+\frac {g}{x}\right )^2 x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=\int (g+f x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx\\ &=\frac {(g+f x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 f}-\frac {(b e n) \int \frac {(g+f x)^3}{d+e x} \, dx}{3 f}\\ &=\frac {(g+f x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 f}-\frac {(b e n) \int \left (\frac {f (-d f+e g)^2}{e^3}+\frac {(-d f+e g)^3}{e^3 (d+e x)}+\frac {f (-d f+e g) (g+f x)}{e^2}+\frac {f (g+f x)^2}{e}\right ) \, dx}{3 f}\\ &=-\frac {b (d f-e g)^2 n x}{3 e^2}+\frac {b (d f-e g) n (g+f x)^2}{6 e f}-\frac {b n (g+f x)^3}{9 f}+\frac {b (d f-e g)^3 n \log (d+e x)}{3 e^3 f}+\frac {(g+f x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 f}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 150, normalized size = 1.25 \[ \frac {e \left (x \left (6 a e^2 \left (f^2 x^2+3 f g x+3 g^2\right )-b n \left (6 d^2 f^2-3 d e f (f x+6 g)+e^2 \left (2 f^2 x^2+9 f g x+18 g^2\right )\right )\right )+6 b e \left (3 d g^2+e x \left (f^2 x^2+3 f g x+3 g^2\right )\right ) \log \left (c (d+e x)^n\right )\right )+6 b d^2 f n (d f-3 e g) \log (d+e x)}{18 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*d^2*f*(d*f - 3*e*g)*n*Log[d + e*x] + e*(x*(6*a*e^2*(3*g^2 + 3*f*g*x + f^2*x^2) - b*n*(6*d^2*f^2 - 3*d*e*f
*(6*g + f*x) + e^2*(18*g^2 + 9*f*g*x + 2*f^2*x^2))) + 6*b*e*(3*d*g^2 + e*x*(3*g^2 + 3*f*g*x + f^2*x^2))*Log[c*
(d + e*x)^n]))/(18*e^3)

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fricas [A]  time = 0.43, size = 219, normalized size = 1.82 \[ -\frac {2 \, {\left (b e^{3} f^{2} n - 3 \, a e^{3} f^{2}\right )} x^{3} - 3 \, {\left (6 \, a e^{3} f g + {\left (b d e^{2} f^{2} - 3 \, b e^{3} f g\right )} n\right )} x^{2} - 6 \, {\left (3 \, a e^{3} g^{2} - {\left (b d^{2} e f^{2} - 3 \, b d e^{2} f g + 3 \, b e^{3} g^{2}\right )} n\right )} x - 6 \, {\left (b e^{3} f^{2} n x^{3} + 3 \, b e^{3} f g n x^{2} + 3 \, b e^{3} g^{2} n x + {\left (b d^{3} f^{2} - 3 \, b d^{2} e f g + 3 \, b d e^{2} g^{2}\right )} n\right )} \log \left (e x + d\right ) - 6 \, {\left (b e^{3} f^{2} x^{3} + 3 \, b e^{3} f g x^{2} + 3 \, b e^{3} g^{2} x\right )} \log \relax (c)}{18 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*(b*e^3*f^2*n - 3*a*e^3*f^2)*x^3 - 3*(6*a*e^3*f*g + (b*d*e^2*f^2 - 3*b*e^3*f*g)*n)*x^2 - 6*(3*a*e^3*g^
2 - (b*d^2*e*f^2 - 3*b*d*e^2*f*g + 3*b*e^3*g^2)*n)*x - 6*(b*e^3*f^2*n*x^3 + 3*b*e^3*f*g*n*x^2 + 3*b*e^3*g^2*n*
x + (b*d^3*f^2 - 3*b*d^2*e*f*g + 3*b*d*e^2*g^2)*n)*log(e*x + d) - 6*(b*e^3*f^2*x^3 + 3*b*e^3*f*g*x^2 + 3*b*e^3
*g^2*x)*log(c))/e^3

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giac [B]  time = 0.20, size = 430, normalized size = 3.58 \[ \frac {1}{3} \, {\left (x e + d\right )}^{3} b f^{2} n e^{\left (-3\right )} \log \left (x e + d\right ) - {\left (x e + d\right )}^{2} b d f^{2} n e^{\left (-3\right )} \log \left (x e + d\right ) + {\left (x e + d\right )} b d^{2} f^{2} n e^{\left (-3\right )} \log \left (x e + d\right ) - \frac {1}{9} \, {\left (x e + d\right )}^{3} b f^{2} n e^{\left (-3\right )} + \frac {1}{2} \, {\left (x e + d\right )}^{2} b d f^{2} n e^{\left (-3\right )} - {\left (x e + d\right )} b d^{2} f^{2} n e^{\left (-3\right )} + {\left (x e + d\right )}^{2} b f g n e^{\left (-2\right )} \log \left (x e + d\right ) - 2 \, {\left (x e + d\right )} b d f g n e^{\left (-2\right )} \log \left (x e + d\right ) + \frac {1}{3} \, {\left (x e + d\right )}^{3} b f^{2} e^{\left (-3\right )} \log \relax (c) - {\left (x e + d\right )}^{2} b d f^{2} e^{\left (-3\right )} \log \relax (c) + {\left (x e + d\right )} b d^{2} f^{2} e^{\left (-3\right )} \log \relax (c) - \frac {1}{2} \, {\left (x e + d\right )}^{2} b f g n e^{\left (-2\right )} + 2 \, {\left (x e + d\right )} b d f g n e^{\left (-2\right )} + \frac {1}{3} \, {\left (x e + d\right )}^{3} a f^{2} e^{\left (-3\right )} - {\left (x e + d\right )}^{2} a d f^{2} e^{\left (-3\right )} + {\left (x e + d\right )} a d^{2} f^{2} e^{\left (-3\right )} + {\left (x e + d\right )} b g^{2} n e^{\left (-1\right )} \log \left (x e + d\right ) + {\left (x e + d\right )}^{2} b f g e^{\left (-2\right )} \log \relax (c) - 2 \, {\left (x e + d\right )} b d f g e^{\left (-2\right )} \log \relax (c) - {\left (x e + d\right )} b g^{2} n e^{\left (-1\right )} + {\left (x e + d\right )}^{2} a f g e^{\left (-2\right )} - 2 \, {\left (x e + d\right )} a d f g e^{\left (-2\right )} + {\left (x e + d\right )} b g^{2} e^{\left (-1\right )} \log \relax (c) + {\left (x e + d\right )} a g^{2} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.34, size = 585, normalized size = 4.88 \[ \frac {a \,f^{2} x^{3}}{3}+b \,g^{2} x \ln \relax (c )+\frac {b \,f^{2} x^{3} \ln \relax (c )}{3}+\frac {\left (f x +g \right )^{3} b \ln \left (\left (e x +d \right )^{n}\right )}{3 f}+\frac {b d \,g^{2} n \ln \left (e x +d \right )}{e}+\frac {b \,d^{3} f^{2} n \ln \left (e x +d \right )}{3 e^{3}}+\frac {b d \,f^{2} n \,x^{2}}{6 e}-\frac {b f g n \,x^{2}}{2}-\frac {b \,d^{2} f^{2} n x}{3 e^{2}}-b \,g^{2} n x +a f g \,x^{2}+a \,g^{2} x +b f g \,x^{2} \ln \relax (c )-\frac {b \,g^{3} n \ln \left (e x +d \right )}{3 f}-\frac {b \,f^{2} n \,x^{3}}{9}+\frac {b d f g n x}{e}-\frac {b \,d^{2} f g n \ln \left (e x +d \right )}{e^{2}}+\frac {i \pi b \,f^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{6}+\frac {i \pi b \,f^{2} x^{3} \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{6}-\frac {i \pi b f g \,x^{2} \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i \pi b \,g^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i \pi b \,g^{2} x \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i \pi b f g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2}-\frac {i \pi b \,f^{2} x^{3} \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{6}-\frac {i \pi b \,g^{2} x \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}-\frac {i \pi b \,f^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{6}+\frac {i \pi b f g \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i \pi b f g \,x^{2} \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i \pi b \,g^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*f^2*a*x^3+ln(c)*b*g^2*x+1/3*f^2*ln(c)*b*x^3+1/3*(f*x+g)^3*b/f*ln((e*x+d)^n)-1/6*I*f^2*Pi*b*x^3*csgn(I*c)*c
sgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/2*I*Pi*b*g^2*x*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*f*
Pi*b*g*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*f*Pi*b*g*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/e*ln(e
*x+d)*b*d*g^2*n+1/3/e^3*f^2*ln(e*x+d)*b*d^3*n-1/2*I*Pi*b*g^2*x*csgn(I*c*(e*x+d)^n)^3-1/6*I*f^2*Pi*b*x^3*csgn(I
*c*(e*x+d)^n)^3+1/2*I*Pi*b*g^2*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*Pi*b*g^2*x*csgn(I*(e*x+d)^n)*csgn(I*c*(
e*x+d)^n)^2+1/6*I*f^2*Pi*b*x^3*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/6*I*f^2*Pi*b*x^3*csgn(I*(e*x+d)^n)*csgn(I*c*(
e*x+d)^n)^2-1/2*I*f*Pi*b*g*x^2*csgn(I*c*(e*x+d)^n)^3-1/2*I*f*Pi*b*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(
e*x+d)^n)+1/6/e*f^2*b*d*n*x^2-1/2*f*b*g*n*x^2-1/3/e^2*f^2*b*d^2*n*x-b*g^2*n*x+f*a*g*x^2+a*g^2*x+f*ln(c)*b*g*x^
2-1/3/f*ln(e*x+d)*b*g^3*n-1/9*f^2*b*n*x^3+1/e*f*b*d*g*n*x-1/e^2*f*ln(e*x+d)*b*d^2*g*n

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maxima [A]  time = 0.49, size = 187, normalized size = 1.56 \[ \frac {1}{3} \, b f^{2} x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {1}{3} \, a f^{2} x^{3} - b e g^{2} n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + \frac {1}{18} \, b e f^{2} n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} - \frac {1}{2} \, b e f g n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} + b f g x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + a f g x^{2} + b g^{2} x \log \left ({\left (e x + d\right )}^{n} c\right ) + a g^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*f^2*x^3*log((e*x + d)^n*c) + 1/3*a*f^2*x^3 - b*e*g^2*n*(x/e - d*log(e*x + d)/e^2) + 1/18*b*e*f^2*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) - 1/2*b*e*f*g*n*(2*d^2*log(e*x + d)/e^3 + (e*x^2 -
2*d*x)/e^2) + b*f*g*x^2*log((e*x + d)^n*c) + a*f*g*x^2 + b*g^2*x*log((e*x + d)^n*c) + a*g^2*x

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mupad [B]  time = 0.30, size = 212, normalized size = 1.77 \[ x^2\,\left (\frac {f\,\left (a\,d\,f+2\,a\,e\,g-b\,e\,g\,n\right )}{2\,e}-\frac {d\,f^2\,\left (3\,a-b\,n\right )}{6\,e}\right )+x\,\left (\frac {3\,a\,e\,g^2-3\,b\,e\,g^2\,n+6\,a\,d\,f\,g}{3\,e}-\frac {d\,\left (\frac {f\,\left (a\,d\,f+2\,a\,e\,g-b\,e\,g\,n\right )}{e}-\frac {d\,f^2\,\left (3\,a-b\,n\right )}{3\,e}\right )}{e}\right )+\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (\frac {b\,f^2\,x^3}{3}+b\,f\,g\,x^2+b\,g^2\,x\right )+\frac {f^2\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {\ln \left (d+e\,x\right )\,\left (b\,n\,d^3\,f^2-3\,b\,n\,d^2\,e\,f\,g+3\,b\,n\,d\,e^2\,g^2\right )}{3\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 16.43, size = 277, normalized size = 2.31 \[ \begin {cases} \frac {a f^{2} x^{3}}{3} + a f g x^{2} + a g^{2} x + \frac {b d^{3} f^{2} n \log {\left (d + e x \right )}}{3 e^{3}} - \frac {b d^{2} f^{2} n x}{3 e^{2}} - \frac {b d^{2} f g n \log {\left (d + e x \right )}}{e^{2}} + \frac {b d f^{2} n x^{2}}{6 e} + \frac {b d f g n x}{e} + \frac {b d g^{2} n \log {\left (d + e x \right )}}{e} + \frac {b f^{2} n x^{3} \log {\left (d + e x \right )}}{3} - \frac {b f^{2} n x^{3}}{9} + \frac {b f^{2} x^{3} \log {\relax (c )}}{3} + b f g n x^{2} \log {\left (d + e x \right )} - \frac {b f g n x^{2}}{2} + b f g x^{2} \log {\relax (c )} + b g^{2} n x \log {\left (d + e x \right )} - b g^{2} n x + b g^{2} x \log {\relax (c )} & \text {for}\: e \neq 0 \\\left (a + b \log {\left (c d^{n} \right )}\right ) \left (\frac {f^{2} x^{3}}{3} + f g x^{2} + g^{2} x\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g/x)**2*x**2*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Piecewise((a*f**2*x**3/3 + a*f*g*x**2 + a*g**2*x + b*d**3*f**2*n*log(d + e*x)/(3*e**3) - b*d**2*f**2*n*x/(3*e*
*2) - b*d**2*f*g*n*log(d + e*x)/e**2 + b*d*f**2*n*x**2/(6*e) + b*d*f*g*n*x/e + b*d*g**2*n*log(d + e*x)/e + b*f
**2*n*x**3*log(d + e*x)/3 - b*f**2*n*x**3/9 + b*f**2*x**3*log(c)/3 + b*f*g*n*x**2*log(d + e*x) - b*f*g*n*x**2/
2 + b*f*g*x**2*log(c) + b*g**2*n*x*log(d + e*x) - b*g**2*n*x + b*g**2*x*log(c), Ne(e, 0)), ((a + b*log(c*d**n)
)*(f**2*x**3/3 + f*g*x**2 + g**2*x), True))

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