3.3.55 \(\int \frac {a+b \log (c (d+e x)^n)}{x^3 (f+g x)^2} \, dx\) [255]

Optimal. Leaf size=335 \[ -\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^4} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \begin {gather*} -\frac {3 b g^2 n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^4}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {2 b e g n \log (x)}{d f^3}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e n}{2 d f^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*e*n)/(d*f^2*x) - (b*e^2*n*Log[x])/(2*d^2*f^2) - (2*b*e*g*n*Log[x])/(d*f^3) + (b*e^2*n*Log[d + e*x])/(2
*d^2*f^2) + (2*b*e*g*n*Log[d + e*x])/(d*f^3) - (b*e*g^2*n*Log[d + e*x])/(f^3*(e*f - d*g)) - (a + b*Log[c*(d +
e*x)^n])/(2*f^2*x^2) + (2*g*(a + b*Log[c*(d + e*x)^n]))/(f^3*x) + (g^2*(a + b*Log[c*(d + e*x)^n]))/(f^3*(f + g
*x)) + (3*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^4 + (b*e*g^2*n*Log[f + g*x])/(f^3*(e*f - d*g)) - (
3*g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^4 - (3*b*g^2*n*PolyLog[2, -((g*(d + e*x))/(
e*f - d*g))])/f^4 + (3*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

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

Rule 2441

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

Rule 2442

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 2463

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

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx &=\int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}-\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x^2}+\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 x}-\frac {g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)^2}-\frac {3 g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 (f+g x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3} \, dx}{f^2}-\frac {(2 g) \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2} \, dx}{f^3}+\frac {\left (3 g^2\right ) \int \frac {a+b \log \left (c (d+e x)^n\right )}{x} \, dx}{f^4}-\frac {\left (3 g^3\right ) \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{f^4}-\frac {g^3 \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{f^3}\\ &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}+\frac {(b e n) \int \frac {1}{x^2 (d+e x)} \, dx}{2 f^2}-\frac {(2 b e g n) \int \frac {1}{x (d+e x)} \, dx}{f^3}-\frac {\left (3 b e g^2 n\right ) \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx}{f^4}+\frac {\left (3 b e g^2 n\right ) \int \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{f^4}-\frac {\left (b e g^2 n\right ) \int \frac {1}{(d+e x) (f+g x)} \, dx}{f^3}\\ &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^4}+\frac {(b e n) \int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx}{2 f^2}-\frac {(2 b e g n) \int \frac {1}{x} \, dx}{d f^3}+\frac {\left (2 b e^2 g n\right ) \int \frac {1}{d+e x} \, dx}{d f^3}+\frac {\left (3 b g^2 n\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{f^4}-\frac {\left (b e^2 g^2 n\right ) \int \frac {1}{d+e x} \, dx}{f^3 (e f-d g)}+\frac {\left (b e g^3 n\right ) \int \frac {1}{f+g x} \, dx}{f^3 (e f-d g)}\\ &=-\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^4}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 269, normalized size = 0.80 \begin {gather*} -\frac {\frac {4 b e f g n (\log (x)-\log (d+e x))}{d}+\frac {b e f^2 n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac {4 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-\frac {2 f g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-6 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {2 b e f g^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}+6 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b g^2 n \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )-6 b g^2 n \text {Li}_2\left (1+\frac {e x}{d}\right )}{2 f^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((4*b*e*f*g*n*(Log[x] - Log[d + e*x]))/d + (b*e*f^2*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x) + (f^2
*(a + b*Log[c*(d + e*x)^n]))/x^2 - (4*f*g*(a + b*Log[c*(d + e*x)^n]))/x - (2*f*g^2*(a + b*Log[c*(d + e*x)^n]))
/(f + g*x) - 6*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + (2*b*e*f*g^2*n*(Log[d + e*x] - Log[f + g*x]))/
(e*f - d*g) + 6*g^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] + 6*b*g^2*n*PolyLog[2, (g*(d + e
*x))/(-(e*f) + d*g)] - 6*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.44, size = 1224, normalized size = 3.65

method	result	size
risch	\(\text {Expression too large to display}\)	\(1224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/x^3/(g*x+f)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x^{3} \left (f + g x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/x**3/(g*x+f)**2,x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))/(x**3*(f + g*x)**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^3/(g*x+f)^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,{\left (f+g\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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