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

Optimal. Leaf size=202 \[ -\frac {b n (d g+e f) \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e^2}+\frac {f^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 d^2 (e f-d g)}-\frac {b n x (e f-d g) \left (a+b \log \left (c x^n\right )\right )}{d^2 e (d+e x)}-\frac {(f+g x)^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 (d+e x)^2 (e f-d g)}-\frac {b^2 n^2 (d g+e f) \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2 e^2}+\frac {b^2 n^2 (e f-d g) \log (d+e x)}{d^2 e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 278, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2357, 2319, 2347, 2344, 2301, 2317, 2391, 2314, 31, 2318} \[ -\frac {b^2 n^2 (e f-d g) \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2 e^2}-\frac {2 b^2 g n^2 \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d e^2}-\frac {b n (e f-d g) \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 e^2}+\frac {(e f-d g) \left (a+b \log \left (c x^n\right )\right )^2}{2 d^2 e^2}-\frac {b n x (e f-d g) \left (a+b \log \left (c x^n\right )\right )}{d^2 e (d+e x)}-\frac {(e f-d g) \left (a+b \log \left (c x^n\right )\right )^2}{2 e^2 (d+e x)^2}-\frac {2 b g n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d e^2}+\frac {g x \left (a+b \log \left (c x^n\right )\right )^2}{d e (d+e x)}+\frac {b^2 n^2 (e f-d g) \log (d+e x)}{d^2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2314

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

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2318

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])
^p)/(d*(d + e*x)), x] - Dist[(b*n*p)/d, Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 244, normalized size = 1.21 \[ \frac {\frac {(e f-d g) \left (-2 b n (d+e x) \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+(d+e x) \left (a+b \log \left (c x^n\right )\right )^2+2 b d n \left (a+b \log \left (c x^n\right )\right )-2 b^2 n^2 (d+e x) \text {Li}_2\left (-\frac {e x}{d}\right )-2 b^2 n^2 (d+e x) (\log (x)-\log (d+e x))\right )}{d^2 (d+e x)}+\frac {2 g \left (\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-2 b n \log \left (\frac {e x}{d}+1\right )\right )-2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )\right )}{d}-\frac {(e f-d g) \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}-\frac {2 g \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} g x + a^{2} f + {\left (b^{2} g x + b^{2} f\right )} \log \left (c x^{n}\right )^{2} + 2 \, {\left (a b g x + a b f\right )} \log \left (c x^{n}\right )}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.39, size = 2163, normalized size = 10.71 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a b f n {\left (\frac {1}{d e^{2} x + d^{2} e} - \frac {\log \left (e x + d\right )}{d^{2} e} + \frac {\log \relax (x)}{d^{2} e}\right )} - a b g n {\left (\frac {1}{e^{3} x + d e^{2}} + \frac {\log \left (e x + d\right )}{d e^{2}} - \frac {\log \relax (x)}{d e^{2}}\right )} - \frac {{\left (2 \, e x + d\right )} a b g \log \left (c x^{n}\right )}{e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}} - \frac {{\left (2 \, e x + d\right )} a^{2} g}{2 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} - \frac {a b f \log \left (c x^{n}\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e} - \frac {a^{2} f}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {{\left (2 \, b^{2} e g x + {\left (e f + d g\right )} b^{2}\right )} \log \left (x^{n}\right )^{2}}{2 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \int \frac {b^{2} e^{2} g x^{2} \log \relax (c)^{2} + b^{2} e^{2} f x \log \relax (c)^{2} + {\left (2 \, {\left (e^{2} g n + e^{2} g \log \relax (c)\right )} b^{2} x^{2} + {\left (e^{2} f n + 3 \, d e g n + 2 \, e^{2} f \log \relax (c)\right )} b^{2} x + {\left (d e f n + d^{2} g n\right )} b^{2}\right )} \log \left (x^{n}\right )}{e^{5} x^{4} + 3 \, d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + d^{3} e^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________