3.1.9 \(\int \frac {(a+b \log (c x^n)) \log (1+e x)}{x^4} \, dx\) [9]

Optimal. Leaf size=195 \[ -\frac {5 b e n}{36 x^2}+\frac {4 b e^2 n}{9 x}+\frac {1}{9} b e^3 n \log (x)-\frac {1}{6} b e^3 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b e^3 n \log (1+e x)-\frac {b n \log (1+e x)}{9 x^3}-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-\frac {1}{3} b e^3 n \text {Li}_2(-e x) \]

[Out]

-5/36*b*e*n/x^2+4/9*b*e^2*n/x+1/9*b*e^3*n*ln(x)-1/6*b*e^3*n*ln(x)^2-1/6*e*(a+b*ln(c*x^n))/x^2+1/3*e^2*(a+b*ln(
c*x^n))/x+1/3*e^3*ln(x)*(a+b*ln(c*x^n))-1/9*b*e^3*n*ln(e*x+1)-1/9*b*n*ln(e*x+1)/x^3-1/3*e^3*(a+b*ln(c*x^n))*ln
(e*x+1)-1/3*(a+b*ln(c*x^n))*ln(e*x+1)/x^3-1/3*b*e^3*n*polylog(2,-e*x)

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Rubi [A]
time = 0.08, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2442, 46, 2423, 2338, 2438} \begin {gather*} -\frac {1}{3} b e^3 n \text {PolyLog}(2,-e x)+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{3} e^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}-\frac {1}{6} b e^3 n \log ^2(x)+\frac {1}{9} b e^3 n \log (x)-\frac {1}{9} b e^3 n \log (e x+1)+\frac {4 b e^2 n}{9 x}-\frac {b n \log (e x+1)}{9 x^3}-\frac {5 b e n}{36 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[1 + e*x])/x^4,x]

[Out]

(-5*b*e*n)/(36*x^2) + (4*b*e^2*n)/(9*x) + (b*e^3*n*Log[x])/9 - (b*e^3*n*Log[x]^2)/6 - (e*(a + b*Log[c*x^n]))/(
6*x^2) + (e^2*(a + b*Log[c*x^n]))/(3*x) + (e^3*Log[x]*(a + b*Log[c*x^n]))/3 - (b*e^3*n*Log[1 + e*x])/9 - (b*n*
Log[1 + e*x])/(9*x^3) - (e^3*(a + b*Log[c*x^n])*Log[1 + e*x])/3 - ((a + b*Log[c*x^n])*Log[1 + e*x])/(3*x^3) -
(b*e^3*n*PolyLog[2, -(e*x)])/3

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 2338

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 2423

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx &=-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-(b n) \int \left (-\frac {e}{6 x^3}+\frac {e^2}{3 x^2}+\frac {e^3 \log (x)}{3 x}-\frac {\log (1+e x)}{3 x^4}-\frac {e^3 \log (1+e x)}{3 x}\right ) \, dx\\ &=-\frac {b e n}{12 x^2}+\frac {b e^2 n}{3 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}+\frac {1}{3} (b n) \int \frac {\log (1+e x)}{x^4} \, dx-\frac {1}{3} \left (b e^3 n\right ) \int \frac {\log (x)}{x} \, dx+\frac {1}{3} \left (b e^3 n\right ) \int \frac {\log (1+e x)}{x} \, dx\\ &=-\frac {b e n}{12 x^2}+\frac {b e^2 n}{3 x}-\frac {1}{6} b e^3 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \log (1+e x)}{9 x^3}-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-\frac {1}{3} b e^3 n \text {Li}_2(-e x)+\frac {1}{9} (b e n) \int \frac {1}{x^3 (1+e x)} \, dx\\ &=-\frac {b e n}{12 x^2}+\frac {b e^2 n}{3 x}-\frac {1}{6} b e^3 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \log (1+e x)}{9 x^3}-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-\frac {1}{3} b e^3 n \text {Li}_2(-e x)+\frac {1}{9} (b e n) \int \left (\frac {1}{x^3}-\frac {e}{x^2}+\frac {e^2}{x}-\frac {e^3}{1+e x}\right ) \, dx\\ &=-\frac {5 b e n}{36 x^2}+\frac {4 b e^2 n}{9 x}+\frac {1}{9} b e^3 n \log (x)-\frac {1}{6} b e^3 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b e^3 n \log (1+e x)-\frac {b n \log (1+e x)}{9 x^3}-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-\frac {1}{3} b e^3 n \text {Li}_2(-e x)\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 206, normalized size = 1.06 \begin {gather*} -\frac {6 a e x+5 b e n x-12 a e^2 x^2-16 b e^2 n x^2+6 b e^3 n x^3 \log ^2(x)+6 b e x \log \left (c x^n\right )-12 b e^2 x^2 \log \left (c x^n\right )-4 e^3 x^3 \log (x) \left (3 a+b n+3 b \log \left (c x^n\right )\right )+12 a \log (1+e x)+4 b n \log (1+e x)+12 a e^3 x^3 \log (1+e x)+4 b e^3 n x^3 \log (1+e x)+12 b \log \left (c x^n\right ) \log (1+e x)+12 b e^3 x^3 \log \left (c x^n\right ) \log (1+e x)+12 b e^3 n x^3 \text {Li}_2(-e x)}{36 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[1 + e*x])/x^4,x]

[Out]

-1/36*(6*a*e*x + 5*b*e*n*x - 12*a*e^2*x^2 - 16*b*e^2*n*x^2 + 6*b*e^3*n*x^3*Log[x]^2 + 6*b*e*x*Log[c*x^n] - 12*
b*e^2*x^2*Log[c*x^n] - 4*e^3*x^3*Log[x]*(3*a + b*n + 3*b*Log[c*x^n]) + 12*a*Log[1 + e*x] + 4*b*n*Log[1 + e*x]
+ 12*a*e^3*x^3*Log[1 + e*x] + 4*b*e^3*n*x^3*Log[1 + e*x] + 12*b*Log[c*x^n]*Log[1 + e*x] + 12*b*e^3*x^3*Log[c*x
^n]*Log[1 + e*x] + 12*b*e^3*n*x^3*PolyLog[2, -(e*x)])/x^3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 796, normalized size = 4.08

method	result	size
risch	\(-\frac {\ln \left (e x +1\right ) a}{3 x^{3}}+\left (-\frac {b \ln \left (e x +1\right )}{3 x^{3}}-\frac {b e \left (2 e^{2} \ln \left (e x +1\right ) x^{2}-2 e^{2} \ln \left (x \right ) x^{2}-2 e x +1\right )}{6 x^{2}}\right ) \ln \left (x^{n}\right )+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +1\right )}{6 x^{3}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{2}}{6 x}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{3} \ln \left (e x +1\right )}{6}+\frac {i e \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{12 x^{2}}-\frac {b \ln \left (c \right ) \ln \left (e x +1\right )}{3 x^{3}}+\frac {e^{3} b \ln \left (c \right ) \ln \left (e x \right )}{3}+\frac {n b \,e^{3} \ln \left (e x \right )}{9}-\frac {e a}{6 x^{2}}+\frac {e^{3} a \ln \left (e x \right )}{3}-\frac {e^{3} a \ln \left (e x +1\right )}{3}+\frac {e^{2} a}{3 x}-\frac {e^{3} b n \dilog \left (e x +1\right )}{3}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e^{3} \ln \left (e x +1\right )}{6}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) e^{2}}{6 x}-\frac {i e^{3} \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x \right )}{6}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +1\right )}{6 x^{3}}+\frac {i e \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{12 x^{2}}+\frac {b \ln \left (c \right ) e^{2}}{3 x}-\frac {e b \ln \left (c \right )}{6 x^{2}}-\frac {b \ln \left (c \right ) e^{3} \ln \left (e x +1\right )}{3}+\frac {4 b \,e^{2} n}{9 x}-\frac {5 b e n}{36 x^{2}}-\frac {b \,e^{3} n \ln \left (e x +1\right )}{9}-\frac {b n \ln \left (e x +1\right )}{9 x^{3}}-\frac {b \,e^{3} n \ln \left (x \right )^{2}}{6}+\frac {i e^{3} \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x \right )}{6}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{6 x^{3}}+\frac {i e^{3} \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x \right )}{6}-\frac {i e \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12 x^{2}}-\frac {i e \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12 x^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{3} \ln \left (e x +1\right )}{6}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2}}{6 x}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2}}{6 x}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{3} \ln \left (e x +1\right )}{6}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{6 x^{3}}-\frac {i e^{3} \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x \right )}{6}\)	\(796\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*ln(e*x+1)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*ln(e*x+1)/x^3*a+(-1/3*b/x^3*ln(e*x+1)-1/6*b*e*(2*e^2*ln(e*x+1)*x^2-2*e^2*ln(x)*x^2-2*e*x+1)/x^2)*ln(x^n)+
1/6*I*e^3*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*ln(e*x)-1/6*I*e^3*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*ln(e*x)+
1/6*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*ln(e*x+1)/x^3+1/12*I*e*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)
/x^2+1/6*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^3*ln(e*x+1)-1/3*b*ln(c)*ln(e*x+1)/x^3+1/3*e^3*b*ln(c)*ln
(e*x)+1/9*n*b*e^3*ln(e*x)-1/6*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^2/x-1/6*e*a/x^2+1/3*e^3*a*ln(e*x)-1
/3*e^3*a*ln(e*x+1)+1/3*e^2*a/x-1/6*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*ln(e*x+1)/x^3+1/6*I*e^3*Pi*b*csgn(I*c)*c
sgn(I*c*x^n)^2*ln(e*x)-1/6*I*e^3*Pi*b*csgn(I*c*x^n)^3*ln(e*x)-1/3*e^3*b*n*dilog(e*x+1)-1/12*I*e*Pi*b*csgn(I*x^
n)*csgn(I*c*x^n)^2/x^2-1/12*I*e*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2/x^2+1/3*b*ln(c)*e^2/x-1/6*e*b*ln(c)/x^2-1/3*b*l
n(c)*e^3*ln(e*x+1)+4/9*b*e^2*n/x-1/6*I*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2*e^3*ln(e*x+1)+1/6*I*Pi*b*csgn(I*x^n)*csg
n(I*c*x^n)^2*e^2/x-5/36*b*e*n/x^2-1/9*b*e^3*n*ln(e*x+1)-1/9*b*n*ln(e*x+1)/x^3+1/6*I*Pi*b*csgn(I*c)*csgn(I*c*x^
n)^2*e^2/x-1/6*b*e^3*n*ln(x)^2-1/6*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*e^3*ln(e*x+1)-1/6*I*Pi*b*csgn(I*c)*csgn(
I*c*x^n)^2*ln(e*x+1)/x^3+1/6*I*Pi*b*csgn(I*c*x^n)^3*ln(e*x+1)/x^3-1/6*I*Pi*b*csgn(I*c*x^n)^3*e^2/x+1/6*I*Pi*b*
csgn(I*c*x^n)^3*e^3*ln(e*x+1)+1/12*I*e*Pi*b*csgn(I*c*x^n)^3/x^2

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Maxima [A]
time = 0.34, size = 208, normalized size = 1.07 \begin {gather*} -\frac {1}{3} \, {\left (\log \left (x e + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-x e\right )\right )} b n e^{3} - \frac {1}{9} \, {\left (b {\left (n + 3 \, \log \left (c\right )\right )} + 3 \, a\right )} e^{3} \log \left (x e + 1\right ) - \frac {6 \, b n x^{3} e^{3} \log \left (x\right )^{2} - 4 \, {\left (b {\left (n + 3 \, \log \left (c\right )\right )} + 3 \, a\right )} x^{3} e^{3} \log \left (x\right ) - 4 \, {\left (b {\left (4 \, n + 3 \, \log \left (c\right )\right )} + 3 \, a\right )} x^{2} e^{2} + {\left (b {\left (5 \, n + 6 \, \log \left (c\right )\right )} + 6 \, a\right )} x e - 4 \, {\left (3 \, b n x^{3} e^{3} \log \left (x\right ) - b {\left (n + 3 \, \log \left (c\right )\right )} - 3 \, a\right )} \log \left (x e + 1\right ) - 6 \, {\left (2 \, b x^{3} e^{3} \log \left (x\right ) + 2 \, b x^{2} e^{2} - b x e - 2 \, {\left (b x^{3} e^{3} + b\right )} \log \left (x e + 1\right )\right )} \log \left (x^{n}\right )}{36 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^4,x, algorithm="maxima")

[Out]

-1/3*(log(x*e + 1)*log(x) + dilog(-x*e))*b*n*e^3 - 1/9*(b*(n + 3*log(c)) + 3*a)*e^3*log(x*e + 1) - 1/36*(6*b*n
*x^3*e^3*log(x)^2 - 4*(b*(n + 3*log(c)) + 3*a)*x^3*e^3*log(x) - 4*(b*(4*n + 3*log(c)) + 3*a)*x^2*e^2 + (b*(5*n
+ 6*log(c)) + 6*a)*x*e - 4*(3*b*n*x^3*e^3*log(x) - b*(n + 3*log(c)) - 3*a)*log(x*e + 1) - 6*(2*b*x^3*e^3*log(
x) + 2*b*x^2*e^2 - b*x*e - 2*(b*x^3*e^3 + b)*log(x*e + 1))*log(x^n))/x^3

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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*x^n))*log(e*x+1)/x^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n)*log(x*e + 1) + a*log(x*e + 1))/x^4, 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((a+b*ln(c*x**n))*ln(e*x+1)/x**4,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((a+b*log(c*x^n))*log(e*x+1)/x^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*log(x*e + 1)/x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(e*x + 1)*(a + b*log(c*x^n)))/x^4,x)

[Out]

int((log(e*x + 1)*(a + b*log(c*x^n)))/x^4, x)

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