3.3.19 \(\int \frac {x^3 \log (c (a+b x)^p)}{d+e x} \, dx\) [219]

Optimal. Leaf size=250 \[ -\frac {d^2 p x}{e^3}-\frac {a d p x}{2 b e^2}-\frac {a^2 p x}{3 b^2 e}+\frac {d p x^2}{4 e^2}+\frac {a p x^2}{6 b e}-\frac {p x^3}{9 e}+\frac {a^2 d p \log (a+b x)}{2 b^2 e^2}+\frac {a^3 p \log (a+b x)}{3 b^3 e}-\frac {d x^2 \log \left (c (a+b x)^p\right )}{2 e^2}+\frac {x^3 \log \left (c (a+b x)^p\right )}{3 e}+\frac {d^2 (a+b x) \log \left (c (a+b x)^p\right )}{b e^3}-\frac {d^3 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^4}-\frac {d^3 p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {45, 2463, 2436, 2332, 2442, 2441, 2440, 2438} \begin {gather*} -\frac {d^3 p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^4}+\frac {a^3 p \log (a+b x)}{3 b^3 e}+\frac {a^2 d p \log (a+b x)}{2 b^2 e^2}-\frac {a^2 p x}{3 b^2 e}-\frac {d^3 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^4}+\frac {d^2 (a+b x) \log \left (c (a+b x)^p\right )}{b e^3}-\frac {d x^2 \log \left (c (a+b x)^p\right )}{2 e^2}+\frac {x^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {a d p x}{2 b e^2}+\frac {a p x^2}{6 b e}-\frac {d^2 p x}{e^3}+\frac {d p x^2}{4 e^2}-\frac {p x^3}{9 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2*p*x)/e^3) - (a*d*p*x)/(2*b*e^2) - (a^2*p*x)/(3*b^2*e) + (d*p*x^2)/(4*e^2) + (a*p*x^2)/(6*b*e) - (p*x^3)
/(9*e) + (a^2*d*p*Log[a + b*x])/(2*b^2*e^2) + (a^3*p*Log[a + b*x])/(3*b^3*e) - (d*x^2*Log[c*(a + b*x)^p])/(2*e
^2) + (x^3*Log[c*(a + b*x)^p])/(3*e) + (d^2*(a + b*x)*Log[c*(a + b*x)^p])/(b*e^3) - (d^3*Log[c*(a + b*x)^p]*Lo
g[(b*(d + e*x))/(b*d - a*e)])/e^4 - (d^3*p*PolyLog[2, -((e*(a + b*x))/(b*d - a*e))])/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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

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

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 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

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

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]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 183, normalized size = 0.73 \begin {gather*} \frac {6 a^2 e^2 (3 b d+2 a e) p \log (a+b x)+b \left (-e p x \left (12 a^2 e^2-6 a b e (-3 d+e x)+b^2 \left (36 d^2-9 d e x+4 e^2 x^2\right )\right )+6 b \log \left (c (a+b x)^p\right ) \left (6 a d^2 e+b e x \left (6 d^2-3 d e x+2 e^2 x^2\right )-6 b d^3 \log \left (\frac {b (d+e x)}{b d-a e}\right )\right )\right )-36 b^3 d^3 p \text {Li}_2\left (\frac {e (a+b x)}{-b d+a e}\right )}{36 b^3 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*e^2*(3*b*d + 2*a*e)*p*Log[a + b*x] + b*(-(e*p*x*(12*a^2*e^2 - 6*a*b*e*(-3*d + e*x) + b^2*(36*d^2 - 9*d*
e*x + 4*e^2*x^2))) + 6*b*Log[c*(a + b*x)^p]*(6*a*d^2*e + b*e*x*(6*d^2 - 3*d*e*x + 2*e^2*x^2) - 6*b*d^3*Log[(b*
(d + e*x))/(b*d - a*e)])) - 36*b^3*d^3*p*PolyLog[2, (e*(a + b*x))/(-(b*d) + a*e)])/(36*b^3*e^4)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.48, size = 919, normalized size = 3.68

method	result	size
risch	\(\frac {p \,a^{3} \ln \left (\left (e x +d \right ) b +a e -b d \right )}{3 b^{3} e}+\frac {p \,d^{3} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e^{4}}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) d \,x^{2}}{4 e^{2}}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) d^{3} \ln \left (e x +d \right )}{2 e^{4}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) x \,d^{2}}{2 e^{3}}-\frac {d^{2} p x}{e^{3}}+\frac {p \,a^{2} \ln \left (\left (e x +d \right ) b +a e -b d \right ) d}{2 b^{2} e^{2}}+\frac {p a \ln \left (\left (e x +d \right ) b +a e -b d \right ) d^{2}}{b \,e^{3}}+\frac {p \,d^{3} \dilog \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e^{4}}+\frac {\ln \left (c \right ) x^{3}}{3 e}-\frac {2 p a \,d^{2}}{3 b \,e^{3}}-\frac {p \,a^{2} d}{3 b^{2} e^{2}}-\frac {49 p \,d^{3}}{36 e^{4}}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d^{3} \ln \left (e x +d \right )}{e^{4}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d^{3} \ln \left (e x +d \right )}{2 e^{4}}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} x \,d^{2}}{2 e^{3}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) x^{3}}{6 e}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) x \,d^{2}}{e^{3}}-\frac {\ln \left (c \right ) d \,x^{2}}{2 e^{2}}+\frac {\ln \left (c \right ) x \,d^{2}}{e^{3}}-\frac {\ln \left (c \right ) d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} x^{3}}{6 e}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d \,x^{2}}{2 e^{2}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} x \,d^{2}}{2 e^{3}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} d^{3} \ln \left (e x +d \right )}{2 e^{4}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} x^{3}}{6 e}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} d \,x^{2}}{4 e^{2}}-\frac {a^{2} p x}{3 e \,b^{2}}+\frac {a p \,x^{2}}{6 b e}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x^{3}}{6 e}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x \,d^{2}}{2 e^{3}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d \,x^{2}}{4 e^{2}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} d^{3} \ln \left (e x +d \right )}{2 e^{4}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} d \,x^{2}}{4 e^{2}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) x^{3}}{3 e}-\frac {p \,x^{3}}{9 e}+\frac {d p \,x^{2}}{4 e^{2}}-\frac {a d p x}{2 b \,e^{2}}\)	\(919\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*p/b^3/e*a^3*ln((e*x+d)*b+a*e-b*d)+p/e^4*d^3*ln(e*x+d)*ln(((e*x+d)*b+a*e-b*d)/(a*e-b*d))-d^2*p*x/e^3+1/2*p/
b^2/e^2*a^2*ln((e*x+d)*b+a*e-b*d)*d+p/b/e^3*a*ln((e*x+d)*b+a*e-b*d)*d^2+1/6*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b
*x+a)^p)^2/e*x^3+1/4*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)*csgn(I*c)/e^2*d*x^2+1/2*I*Pi*csgn(I*(b*x+a)^p)
*csgn(I*c*(b*x+a)^p)*csgn(I*c)*d^3/e^4*ln(e*x+d)-1/2*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)*csgn(I*c)/e^3*
x*d^2+1/6*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*c)/e*x^3+1/3*ln(c)/e*x^3-2/3*p/b/e^3*a*d^2-1/3*p/b^2/e^2*a^2*d-1/2
*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*c)*d^3/e^4*ln(e*x+d)-1/6*I*Pi*csgn(I*c*(b*x+a)^p)^3/e*x^3-49/36*p/e^4*d^3+1
/2*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2/e^3*x*d^2-ln((b*x+a)^p)*d^3/e^4*ln(e*x+d)+1/4*I*Pi*csgn(I*c*(b
*x+a)^p)^3/e^2*d*x^2+p/e^4*d^3*dilog(((e*x+d)*b+a*e-b*d)/(a*e-b*d))+1/2*I*Pi*csgn(I*c*(b*x+a)^p)^3*d^3/e^4*ln(
e*x+d)+ln((b*x+a)^p)/e^3*x*d^2-1/2*ln(c)/e^2*d*x^2+ln(c)/e^3*x*d^2-ln(c)*d^3/e^4*ln(e*x+d)-1/2*ln((b*x+a)^p)/e
^2*d*x^2-1/6*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)*csgn(I*c)/e*x^3+1/2*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*
c)/e^3*x*d^2-1/2*I*Pi*csgn(I*c*(b*x+a)^p)^3/e^3*x*d^2-1/3*a^2*p*x/e/b^2+1/6*a*p*x^2/b/e+1/3*ln((b*x+a)^p)/e*x^
3-1/9*p*x^3/e+1/4*d*p*x^2/e^2-1/4*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*c)/e^2*d*x^2-1/2*I*Pi*csgn(I*(b*x+a)^p)*cs
gn(I*c*(b*x+a)^p)^2*d^3/e^4*ln(e*x+d)-1/2*a*d*p*x/b/e^2-1/4*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2/e^2*d
*x^2

________________________________________________________________________________________

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(x^3*log(c*(b*x+a)^p)/(e*x+d),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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*log(c*(b*x+a)^p)/(e*x+d),x, algorithm="fricas")

[Out]

integral(x^3*log((b*x + a)^p*c)/(x*e + d), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*ln(c*(b*x+a)**p)/(e*x+d),x)

[Out]

Integral(x**3*log(c*(a + b*x)**p)/(d + e*x), x)

________________________________________________________________________________________

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*log(c*(b*x+a)^p)/(e*x+d),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________