3.4.84 \(\int \frac {(a+b \log (c (d+e x)^n)) (f+g \log (c (d+e x)^n))}{x^3} \, dx\) [384]

Optimal. Leaf size=156 \[ \frac {b e^2 g n^2 \log (x)}{d^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )}{2 x^2}-\frac {e n (d+e x) \left (b f+a g+2 b g \log \left (c (d+e x)^n\right )\right )}{2 d^2 x}-\frac {e^2 n \left (b f+a g+2 b g \log \left (c (d+e x)^n\right )\right ) \log \left (1-\frac {d}{d+e x}\right )}{2 d^2}+\frac {b e^2 g n^2 \text {Li}_2\left (\frac {d}{d+e x}\right )}{d^2} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2483, 2458, 2389, 2379, 2438, 2351, 31} \begin {gather*} \frac {b e^2 g n^2 \text {PolyLog}\left (2,\frac {d}{d+e x}\right )}{d^2}-\frac {e^2 n \log \left (1-\frac {d}{d+e x}\right ) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 d^2}-\frac {e n (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 d^2 x}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )}{2 x^2}+\frac {b e^2 g n^2 \log (x)}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 2351

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 2379

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

Rule 2389

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 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 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2483

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(g_.))*
(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(a + b*Log[c*(d + e*x)^n])*((f + g*Log[c*(d + e*x)^n])/(m + 1)), x] -
Dist[e*(n/(m + 1)), Int[(x^(m + 1)*(b*f + a*g + 2*b*g*Log[c*(d + e*x)^n]))/(d + e*x), x], x] /; FreeQ[{a, b, c
, d, e, f, g, n, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 254, normalized size = 1.63 \begin {gather*} -\frac {a f}{2 x^2}+\frac {1}{2} b e f n \left (-\frac {1}{d x}-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}\right )+\frac {1}{2} a e g n \left (-\frac {1}{d x}-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}\right )-\frac {b f \log \left (c (d+e x)^n\right )}{2 x^2}-\frac {a g \log \left (c (d+e x)^n\right )}{2 x^2}-\frac {b g \log ^2\left (c (d+e x)^n\right )}{2 x^2}+b e g n \left (\frac {e n \left (\frac {\log (x)}{d}-\frac {\log (d+e x)}{d}\right )}{d}-\frac {\log \left (c (d+e x)^n\right )}{d x}-\frac {e \log \left (-\frac {e x}{d}\right ) \log \left (c (d+e x)^n\right )}{d^2}+\frac {e \log ^2\left (c (d+e x)^n\right )}{2 d^2 n}-\frac {e n \text {Li}_2\left (\frac {d+e x}{d}\right )}{d^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.26, size = 1201, normalized size = 7.70

method	result	size
risch	\(\text {Expression too large to display}\)	\(1201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )\,\left (f+g\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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