3.220 \(\int \frac {x^2 \log (c (a+b x)^p)}{d+e x} \, dx\)

Optimal. Leaf size=159 \[ -\frac {a^2 p \log (a+b x)}{2 b^2 e}+\frac {d^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^3}-\frac {d (a+b x) \log \left (c (a+b x)^p\right )}{b e^2}+\frac {x^2 \log \left (c (a+b x)^p\right )}{2 e}+\frac {d^2 p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e^3}+\frac {a p x}{2 b e}+\frac {d p x}{e^2}-\frac {p x^2}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {43, 2416, 2389, 2295, 2395, 2394, 2393, 2391} \[ \frac {d^2 p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^3}-\frac {a^2 p \log (a+b x)}{2 b^2 e}+\frac {d^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^3}-\frac {d (a+b x) \log \left (c (a+b x)^p\right )}{b e^2}+\frac {x^2 \log \left (c (a+b x)^p\right )}{2 e}+\frac {a p x}{2 b e}+\frac {d p x}{e^2}-\frac {p x^2}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 2295

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

Rule 2389

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 2391

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 2393

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 2394

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 2395

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 2416

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

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Mathematica [A]  time = 0.10, size = 127, normalized size = 0.80 \[ \frac {-2 a^2 e^2 p \log (a+b x)+4 b^2 d^2 p \text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )+b \log \left (c (a+b x)^p\right ) \left (4 b d^2 \log \left (\frac {b (d+e x)}{b d-a e}\right )-4 a d e+2 b e x (e x-2 d)\right )+b e p x (2 a e+4 b d-b e x)}{4 b^2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(c*(b*x+a)^p)/(e*x+d),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(c*(b*x+a)^p)/(e*x+d),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.32, size = 666, normalized size = 4.19 \[ -\frac {p \,x^{2}}{4 e}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{4 e}+\frac {x^{2} \ln \relax (c )}{2 e}-\frac {a d p \ln \left (a e -b d +\left (e x +d \right ) b \right )}{b \,e^{2}}-\frac {a^{2} p \ln \left (a e -b d +\left (e x +d \right ) b \right )}{2 b^{2} e}+\frac {a d p}{2 b \,e^{2}}-\frac {d^{2} p \ln \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right ) \ln \left (e x +d \right )}{e^{3}}+\frac {5 d^{2} p}{4 e^{3}}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4 e}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4 e}+\frac {d p x}{e^{2}}-\frac {d x \ln \relax (c )}{e^{2}}+\frac {d^{2} \ln \relax (c ) \ln \left (e x +d \right )}{e^{3}}+\frac {x^{2} \ln \left (\left (b x +a \right )^{p}\right )}{2 e}-\frac {d^{2} p \dilog \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right )}{e^{3}}+\frac {d^{2} \ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e^{3}}-\frac {d x \ln \left (\left (b x +a \right )^{p}\right )}{e^{2}}-\frac {i \pi \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{2 e^{2}}+\frac {i \pi d x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2 e^{2}}-\frac {i \pi \,d^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 e^{3}}+\frac {a p x}{2 b e}+\frac {i \pi \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi \,d^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e^{3}}-\frac {i \pi d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2 e^{2}}-\frac {i \pi d x \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2 e^{2}}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{4 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*ln(c*(b*x+a)^p)/(e*x+d),x)

[Out]

-1/4*I*Pi*csgn(I*c*(b*x+a)^p)^3/e*x^2-1/2*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)*csgn(I*c)*d^2/e^3*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^2*x*d-1/4*p*x^2/e+1/2*ln(c)/e*x^2+1/4*I*Pi*csgn(
I*c*(b*x+a)^p)^2*csgn(I*c)/e*x^2-1/2*I*Pi*csgn(I*c*(b*x+a)^p)^3*d^2/e^3*ln(e*x+d)+1/2*I*Pi*csgn(I*c*(b*x+a)^p)
^3/e^2*x*d+1/4*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2/e*x^2-1/2*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*c)/e^2
*x*d-1/b*p/e^2*a*ln(a*e-b*d+(e*x+d)*b)*d-1/2/b^2*p/e*a^2*ln(a*e-b*d+(e*x+d)*b)+1/2/b*p/e^2*a*d-p/e^3*d^2*ln(e*
x+d)*ln((a*e-b*d+(e*x+d)*b)/(a*e-b*d))+5/4*p/e^3*d^2+d*p*x/e^2-ln(c)/e^2*x*d+ln(c)*d^2/e^3*ln(e*x+d)-1/2*I*Pi*
csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2/e^2*x*d+1/2*I*Pi*csgn(I*c*(b*x+a)^p)^2*csgn(I*c)*d^2/e^3*ln(e*x+d)+1/2
*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2*d^2/e^3*ln(e*x+d)+1/2*ln((b*x+a)^p)/e*x^2+ln((b*x+a)^p)*d^2/e^3*
ln(e*x+d)-ln((b*x+a)^p)/e^2*x*d-1/4*I*Pi*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)*csgn(I*c)/e*x^2-p/e^3*d^2*dilog
((a*e-b*d+(e*x+d)*b)/(a*e-b*d))+1/2*a*p*x/b/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*ln(c*(b*x+a)**p)/(e*x+d),x)

[Out]

Integral(x**2*log(c*(a + b*x)**p)/(d + e*x), x)

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