3.4.63 \(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x^2} \, dx\) [363]

Optimal. Leaf size=102 \[ \frac {b e m n \log (x)}{d}-\frac {b e n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{d}-\frac {b e m n \log (d+e x)}{d}-\left (\frac {m}{x}+\frac {\log \left (f x^m\right )}{x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e m n \text {Li}_2\left (-\frac {d}{e x}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2473, 2379, 2438, 36, 29, 31} \begin {gather*} \frac {b e m n \text {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\left (\frac {\log \left (f x^m\right )}{x}+\frac {m}{x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e n \log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}+\frac {b e m n \log (x)}{d}-\frac {b e m n \log (d+e x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2473

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

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 111, normalized size = 1.09 \begin {gather*} -\frac {b e m n x \log ^2(x)+2 \left (m+\log \left (f x^m\right )\right ) \left (a d+b e n x \log (d+e x)+b d \log \left (c (d+e x)^n\right )\right )-2 b e n x \log (x) \left (m+\log \left (f x^m\right )+m \log (d+e x)-m \log \left (1+\frac {e x}{d}\right )\right )+2 b e m n x \text {Li}_2\left (-\frac {e x}{d}\right )}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.60, size = 1859, normalized size = 18.23

method	result	size
risch	\(\text {Expression too large to display}\)	\(1859\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.33, size = 175, normalized size = 1.72 \begin {gather*} -\frac {1}{2} \, {\left (\frac {2 \, {\left (\log \left (x\right ) \log \left (\frac {x e}{d} + 1\right ) + {\rm Li}_2\left (-\frac {x e}{d}\right )\right )} b n e}{d} + \frac {2 \, b n e \log \left (x e + d\right )}{d} - \frac {2 \, b n x e \log \left (x e + d\right ) \log \left (x\right ) - b n x e \log \left (x\right )^{2} + 2 \, b n x e \log \left (x\right ) - 2 \, b d \log \left ({\left (x e + d\right )}^{n}\right ) - 2 \, b d \log \left (c\right ) - 2 \, a d}{d x}\right )} m - {\left (b n {\left (\frac {\log \left (x e + d\right )}{d} - \frac {\log \left (x\right )}{d}\right )} e + \frac {b \log \left ({\left (x e + d\right )}^{n} c\right )}{x} + \frac {a}{x}\right )} \log \left (f x^{m}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(log(x)*log(x*e/d + 1) + dilog(-x*e/d))*b*n*e/d + 2*b*n*e*log(x*e + d)/d - (2*b*n*x*e*log(x*e + d)*log
(x) - b*n*x*e*log(x)^2 + 2*b*n*x*e*log(x) - 2*b*d*log((x*e + d)^n) - 2*b*d*log(c) - 2*a*d)/(d*x))*m - (b*n*(lo
g(x*e + d)/d - log(x)/d)*e + b*log((x*e + d)^n*c)/x + a/x)*log(f*x^m)

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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