3.1.2 \(\int x^3 (a+b \log (c x^n)) \log (1+e x) \, dx\) [2]

Optimal. Leaf size=210 \[ -\frac {5 b n x}{16 e^3}+\frac {3 b n x^2}{32 e^2}-\frac {7 b n x^3}{144 e}+\frac {1}{32} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1+e x)}{16 e^4}-\frac {1}{16} b n x^4 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{4 e^4} \]

[Out]

-5/16*b*n*x/e^3+3/32*b*n*x^2/e^2-7/144*b*n*x^3/e+1/32*b*n*x^4+1/4*x*(a+b*ln(c*x^n))/e^3-1/8*x^2*(a+b*ln(c*x^n)
)/e^2+1/12*x^3*(a+b*ln(c*x^n))/e-1/16*x^4*(a+b*ln(c*x^n))+1/16*b*n*ln(e*x+1)/e^4-1/16*b*n*x^4*ln(e*x+1)-1/4*(a
+b*ln(c*x^n))*ln(e*x+1)/e^4+1/4*x^4*(a+b*ln(c*x^n))*ln(e*x+1)-1/4*b*n*polylog(2,-e*x)/e^4

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Rubi [A]
time = 0.09, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2442, 45, 2423, 2438} \begin {gather*} -\frac {b n \text {PolyLog}(2,-e x)}{4 e^4}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{4 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {1}{4} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (e x+1)}{16 e^4}-\frac {5 b n x}{16 e^3}+\frac {3 b n x^2}{32 e^2}-\frac {1}{16} b n x^4 \log (e x+1)-\frac {7 b n x^3}{144 e}+\frac {1}{32} b n x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Log[c*x^n])*Log[1 + e*x],x]

[Out]

(-5*b*n*x)/(16*e^3) + (3*b*n*x^2)/(32*e^2) - (7*b*n*x^3)/(144*e) + (b*n*x^4)/32 + (x*(a + b*Log[c*x^n]))/(4*e^
3) - (x^2*(a + b*Log[c*x^n]))/(8*e^2) + (x^3*(a + b*Log[c*x^n]))/(12*e) - (x^4*(a + b*Log[c*x^n]))/16 + (b*n*L
og[1 + e*x])/(16*e^4) - (b*n*x^4*Log[1 + e*x])/16 - ((a + b*Log[c*x^n])*Log[1 + e*x])/(4*e^4) + (x^4*(a + b*Lo
g[c*x^n])*Log[1 + e*x])/4 - (b*n*PolyLog[2, -(e*x)])/(4*e^4)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx &=\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-(b n) \int \left (\frac {1}{4 e^3}-\frac {x}{8 e^2}+\frac {x^2}{12 e}-\frac {x^3}{16}-\frac {\log (1+e x)}{4 e^4 x}+\frac {1}{4} x^3 \log (1+e x)\right ) \, dx\\ &=-\frac {b n x}{4 e^3}+\frac {b n x^2}{16 e^2}-\frac {b n x^3}{36 e}+\frac {1}{64} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {1}{4} (b n) \int x^3 \log (1+e x) \, dx+\frac {(b n) \int \frac {\log (1+e x)}{x} \, dx}{4 e^4}\\ &=-\frac {b n x}{4 e^3}+\frac {b n x^2}{16 e^2}-\frac {b n x^3}{36 e}+\frac {1}{64} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b n x^4 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{4 e^4}+\frac {1}{16} (b e n) \int \frac {x^4}{1+e x} \, dx\\ &=-\frac {b n x}{4 e^3}+\frac {b n x^2}{16 e^2}-\frac {b n x^3}{36 e}+\frac {1}{64} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b n x^4 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{4 e^4}+\frac {1}{16} (b e n) \int \left (-\frac {1}{e^4}+\frac {x}{e^3}-\frac {x^2}{e^2}+\frac {x^3}{e}+\frac {1}{e^4 (1+e x)}\right ) \, dx\\ &=-\frac {5 b n x}{16 e^3}+\frac {3 b n x^2}{32 e^2}-\frac {7 b n x^3}{144 e}+\frac {1}{32} b n x^4+\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1+e x)}{16 e^4}-\frac {1}{16} b n x^4 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^4}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{4 e^4}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 188, normalized size = 0.90 \begin {gather*} \frac {72 a e x-90 b e n x-36 a e^2 x^2+27 b e^2 n x^2+24 a e^3 x^3-14 b e^3 n x^3-18 a e^4 x^4+9 b e^4 n x^4-72 a \log (1+e x)+18 b n \log (1+e x)+72 a e^4 x^4 \log (1+e x)-18 b e^4 n x^4 \log (1+e x)+6 b \log \left (c x^n\right ) \left (e x \left (12-6 e x+4 e^2 x^2-3 e^3 x^3\right )+12 \left (-1+e^4 x^4\right ) \log (1+e x)\right )-72 b n \text {Li}_2(-e x)}{288 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Log[c*x^n])*Log[1 + e*x],x]

[Out]

(72*a*e*x - 90*b*e*n*x - 36*a*e^2*x^2 + 27*b*e^2*n*x^2 + 24*a*e^3*x^3 - 14*b*e^3*n*x^3 - 18*a*e^4*x^4 + 9*b*e^
4*n*x^4 - 72*a*Log[1 + e*x] + 18*b*n*Log[1 + e*x] + 72*a*e^4*x^4*Log[1 + e*x] - 18*b*e^4*n*x^4*Log[1 + e*x] +
6*b*Log[c*x^n]*(e*x*(12 - 6*e*x + 4*e^2*x^2 - 3*e^3*x^3) + 12*(-1 + e^4*x^4)*Log[1 + e*x]) - 72*b*n*PolyLog[2,
-(e*x)])/(288*e^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 1014, normalized size = 4.83

method	result	size
risch	\(\text {Expression too large to display}\)	\(1014\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))*ln(e*x+1),x,method=_RETURNVERBOSE)

[Out]

25/48*a/e^4-1/8*I*Pi*b*csgn(I*c*x^n)^3*ln(e*x+1)*x^4+1/24*I/e*x^3*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2+1/12*a/e*x^3+
1/4*a*ln(e*x+1)*x^4-1/4*a/e^4*ln(e*x+1)+1/32*I*Pi*b*x^4*csgn(I*c*x^n)^3-25/96*I/e^4*Pi*b*csgn(I*c*x^n)^3+1/4*a
/e^3*x-1/16*ln(c)*b*x^4+1/12*b*ln(c)/e*x^3-1/8*a/e^2*x^2-1/8*b*ln(c)/e^2*x^2+1/32*b*n*x^4-1/16*a*x^4+(1/4*x^4*
b*ln(e*x+1)-1/48*b*(3*e^4*x^4-4*e^3*x^3+6*e^2*x^2-12*e*x+12*ln(e*x+1))/e^4)*ln(x^n)-1/4*b*n/e^4*dilog(e*x+1)+1
/4*b*ln(c)*ln(e*x+1)*x^4-1/4/e^4*ln(e*x+1)*b*ln(c)+1/8*I/e^4*ln(e*x+1)*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n
)+1/4*b*ln(c)/e^3*x+1/16*I/e^2*x^2*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/24*I/e*x^3*Pi*b*csgn(I*c)*csgn(I
*x^n)*csgn(I*c*x^n)-1/8*I/e^3*x*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)*ln(e*x+1)*x^4+25/48/e^4*b*ln(c)-35/72*b*n/e^4-7/144*b*n*x^3/e+3/32*b*n*x^2/e^2+1/8*I*Pi*b*csgn(I*c)*
csgn(I*c*x^n)^2*ln(e*x+1)*x^4+1/8*I/e^4*ln(e*x+1)*Pi*b*csgn(I*c*x^n)^3-1/32*I*Pi*b*x^4*csgn(I*x^n)*csgn(I*c*x^
n)^2+25/96*I/e^4*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*I/e^4*ln(e*x+1)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/16*b*
n*ln(e*x+1)/e^4-1/16*b*n*x^4*ln(e*x+1)+1/24*I/e*x^3*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*I/e^4*ln(e*x+1)*Pi*b*
csgn(I*c)*csgn(I*c*x^n)^2+1/8*I/e^3*x*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2-25/96*I/e^4*Pi*b*csgn(I*c)*csgn(I*x^n)*cs
gn(I*c*x^n)-5/16*b*n*x/e^3+1/8*I/e^3*x*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/16*I/e^2*x^2*Pi*b*csgn(I*c)*csgn(I*c
*x^n)^2-1/24*I/e*x^3*Pi*b*csgn(I*c*x^n)^3-1/32*I*Pi*b*x^4*csgn(I*c)*csgn(I*c*x^n)^2+1/16*I/e^2*x^2*Pi*b*csgn(I
*c*x^n)^3+25/96*I/e^4*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2-1/8*I/e^3*x*Pi*b*csgn(I*c*x^n)^3+1/32*I*Pi*b*x^4*csgn(I*c
)*csgn(I*x^n)*csgn(I*c*x^n)-1/16*I/e^2*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/8*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^
n)^2*ln(e*x+1)*x^4

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Maxima [A]
time = 0.33, size = 223, normalized size = 1.06 \begin {gather*} -\frac {1}{4} \, {\left (\log \left (x e + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-x e\right )\right )} b n e^{\left (-4\right )} + \frac {1}{16} \, {\left (b {\left (n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} e^{\left (-4\right )} \log \left (x e + 1\right ) + \frac {1}{288} \, {\left (9 \, {\left (b {\left (n - 2 \, \log \left (c\right )\right )} - 2 \, a\right )} x^{4} e^{4} - 2 \, {\left (b {\left (7 \, n - 12 \, \log \left (c\right )\right )} - 12 \, a\right )} x^{3} e^{3} + 9 \, {\left (b {\left (3 \, n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} x^{2} e^{2} - 18 \, {\left (b {\left (5 \, n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} x e - 18 \, {\left ({\left (b {\left (n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} x^{4} e^{4} - 4 \, b n \log \left (x\right )\right )} \log \left (x e + 1\right ) - 6 \, {\left (3 \, b x^{4} e^{4} - 4 \, b x^{3} e^{3} + 6 \, b x^{2} e^{2} - 12 \, b x e - 12 \, {\left (b x^{4} e^{4} - b\right )} \log \left (x e + 1\right )\right )} \log \left (x^{n}\right )\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="maxima")

[Out]

-1/4*(log(x*e + 1)*log(x) + dilog(-x*e))*b*n*e^(-4) + 1/16*(b*(n - 4*log(c)) - 4*a)*e^(-4)*log(x*e + 1) + 1/28
8*(9*(b*(n - 2*log(c)) - 2*a)*x^4*e^4 - 2*(b*(7*n - 12*log(c)) - 12*a)*x^3*e^3 + 9*(b*(3*n - 4*log(c)) - 4*a)*
x^2*e^2 - 18*(b*(5*n - 4*log(c)) - 4*a)*x*e - 18*((b*(n - 4*log(c)) - 4*a)*x^4*e^4 - 4*b*n*log(x))*log(x*e + 1
) - 6*(3*b*x^4*e^4 - 4*b*x^3*e^3 + 6*b*x^2*e^2 - 12*b*x*e - 12*(b*x^4*e^4 - b)*log(x*e + 1))*log(x^n))*e^(-4)

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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(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="fricas")

[Out]

integral(b*x^3*log(c*x^n)*log(x*e + 1) + a*x^3*log(x*e + 1), 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(x**3*(a+b*ln(c*x**n))*ln(e*x+1),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(x^3*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^3*log(x*e + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(e*x + 1)*(a + b*log(c*x^n)),x)

[Out]

int(x^3*log(e*x + 1)*(a + b*log(c*x^n)), x)

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