3.3.21 \(\int \frac {x \log (c (a+b x)^p)}{d+e x} \, dx\) [221]

Optimal. Leaf size=91 \[ -\frac {p x}{e}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}-\frac {d p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {45, 2463, 2436, 2332, 2441, 2440, 2438} \begin {gather*} -\frac {d p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^2}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {p x}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 79, normalized size = 0.87 \begin {gather*} \frac {-b e p x+\log \left (c (a+b x)^p\right ) \left (a e+b e x-b d \log \left (\frac {b (d+e x)}{b d-a e}\right )\right )-b d p \text {Li}_2\left (\frac {e (a+b x)}{-b d+a e}\right )}{b e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.49, size = 427, normalized size = 4.69

method	result	size
risch	\(\frac {\ln \left (\left (b x +a \right )^{p}\right ) x}{e}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {p x}{e}-\frac {p d}{e^{2}}+\frac {p a \ln \left (\left (e x +d \right ) b +a e -b d \right )}{b e}+\frac {p d \dilog \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e^{2}}+\frac {p d \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} d \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} x}{2 e}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) d \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) x}{2 e}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} d \ln \left (e x +d \right )}{2 e^{2}}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} x}{2 e}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) d \ln \left (e x +d \right )}{2 e^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x}{2 e}+\frac {\ln \left (c \right ) x}{e}-\frac {\ln \left (c \right ) d \ln \left (e x +d \right )}{e^{2}}\)	\(427\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

[Out]

integral(x*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 \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*ln(c*(b*x+a)**p)/(e*x+d),x)

[Out]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\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*log(c*(a + b*x)^p))/(d + e*x),x)

[Out]

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

________________________________________________________________________________________