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

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

[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

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Rubi [A]  time = 0.107653, antiderivative size = 91, normalized size of antiderivative = 1., 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 = {43, 2416, 2389, 2295, 2394, 2393, 2391} \[ -\frac{d p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{e^2}-\frac{d \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{e^2}+\frac{(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac{p x}{e} \]

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 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 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]

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 2295

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

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 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 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]

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{\operatorname{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) \operatorname{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 = 0.0327032, size = 79, normalized size = 0.87 \[ \frac{-b d p \text{PolyLog}\left (2,\frac{e (a+b x)}{a e-b d}\right )+\log \left (c (a+b x)^p\right ) \left (-b d \log \left (\frac{b (d+e x)}{b d-a e}\right )+a e+b e x\right )-b e p x}{b e^2} \]

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)

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Maple [C]  time = 0.622, size = 427, normalized size = 4.7 \begin{align*}{\frac{\ln \left ( \left ( bx+a \right ) ^{p} \right ) x}{e}}-{\frac{\ln \left ( \left ( bx+a \right ) ^{p} \right ) d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{px}{e}}-{\frac{dp}{{e}^{2}}}+{\frac{ap\ln \left ( b \left ( ex+d \right ) +ae-bd \right ) }{be}}+{\frac{dp}{{e}^{2}}{\it dilog} \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }+{\frac{dp\ln \left ( ex+d \right ) }{{e}^{2}}\ln \left ({\frac{b \left ( ex+d \right ) +ae-bd}{ae-bd}} \right ) }+{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{3}x}{e}}-{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) x}{e}}+{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}x}{e}}+{\frac{{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{3}d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}d\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( bx+a \right ) ^{p} \right ) \right ) ^{2}x}{e}}+{\frac{\ln \left ( c \right ) x}{e}}-{\frac{\ln \left ( c \right ) d\ln \left ( ex+d \right ) }{{e}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \end{align*}

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)/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}, x\right ) \end{align*}

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)/(e*x + d), x)

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

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \end{align*}

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)/(e*x + d), x)