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

Optimal. Leaf size=146 \[ \frac{e p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d^2}-\frac{e p \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d^2}-\frac{e \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac{e \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^2}-\frac{\log \left (c (a+b x)^p\right )}{d x}+\frac{b p \log (x)}{a d}-\frac{b p \log (a+b x)}{a d} \]

[Out]

(b*p*Log[x])/(a*d) - (b*p*Log[a + b*x])/(a*d) - Log[c*(a + b*x)^p]/(d*x) - (e*Log[-((b*x)/a)]*Log[c*(a + b*x)^
p])/d^2 + (e*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/d^2 + (e*p*PolyLog[2, -((e*(a + b*x))/(b*d - a
*e))])/d^2 - (e*p*PolyLog[2, 1 + (b*x)/a])/d^2

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Rubi [A]  time = 0.166364, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ \frac{e p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d^2}-\frac{e p \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d^2}-\frac{e \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac{e \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^2}-\frac{\log \left (c (a+b x)^p\right )}{d x}+\frac{b p \log (x)}{a d}-\frac{b p \log (a+b x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*p*Log[x])/(a*d) - (b*p*Log[a + b*x])/(a*d) - Log[c*(a + b*x)^p]/(d*x) - (e*Log[-((b*x)/a)]*Log[c*(a + b*x)^
p])/d^2 + (e*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/d^2 + (e*p*PolyLog[2, -((e*(a + b*x))/(b*d - a
*e))])/d^2 - (e*p*PolyLog[2, 1 + (b*x)/a])/d^2

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 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{\log \left (c (a+b x)^p\right )}{x^2 (d+e x)} \, dx &=\int \left (\frac{\log \left (c (a+b x)^p\right )}{d x^2}-\frac{e \log \left (c (a+b x)^p\right )}{d^2 x}+\frac{e^2 \log \left (c (a+b x)^p\right )}{d^2 (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{\log \left (c (a+b x)^p\right )}{x^2} \, dx}{d}-\frac{e \int \frac{\log \left (c (a+b x)^p\right )}{x} \, dx}{d^2}+\frac{e^2 \int \frac{\log \left (c (a+b x)^p\right )}{d+e x} \, dx}{d^2}\\ &=-\frac{\log \left (c (a+b x)^p\right )}{d x}-\frac{e \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac{e \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^2}+\frac{(b p) \int \frac{1}{x (a+b x)} \, dx}{d}+\frac{(b e p) \int \frac{\log \left (-\frac{b x}{a}\right )}{a+b x} \, dx}{d^2}-\frac{(b e p) \int \frac{\log \left (\frac{b (d+e x)}{b d-a e}\right )}{a+b x} \, dx}{d^2}\\ &=-\frac{\log \left (c (a+b x)^p\right )}{d x}-\frac{e \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac{e \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^2}-\frac{e p \text{Li}_2\left (1+\frac{b x}{a}\right )}{d^2}+\frac{(b p) \int \frac{1}{x} \, dx}{a d}-\frac{\left (b^2 p\right ) \int \frac{1}{a+b x} \, dx}{a d}-\frac{(e p) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e x}{b d-a e}\right )}{x} \, dx,x,a+b x\right )}{d^2}\\ &=\frac{b p \log (x)}{a d}-\frac{b p \log (a+b x)}{a d}-\frac{\log \left (c (a+b x)^p\right )}{d x}-\frac{e \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac{e \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^2}+\frac{e p \text{Li}_2\left (-\frac{e (a+b x)}{b d-a e}\right )}{d^2}-\frac{e p \text{Li}_2\left (1+\frac{b x}{a}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.0482952, size = 139, normalized size = 0.95 \[ \frac{a e p x \text{PolyLog}\left (2,\frac{e (a+b x)}{a e-b d}\right )-a e p x \text{PolyLog}\left (2,\frac{b x}{a}+1\right )+a e x \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )-a d \log \left (c (a+b x)^p\right )-a e x \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )-b d p x \log (a+b x)+b d p x \log (x)}{a d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*p*x*Log[x] - b*d*p*x*Log[a + b*x] - a*d*Log[c*(a + b*x)^p] - a*e*x*Log[-((b*x)/a)]*Log[c*(a + b*x)^p] + a
*e*x*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)] + a*e*p*x*PolyLog[2, (e*(a + b*x))/(-(b*d) + a*e)] - a*
e*p*x*PolyLog[2, 1 + (b*x)/a])/(a*d^2*x)

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Maple [C]  time = 0.595, size = 615, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.23382, size = 211, normalized size = 1.45 \begin{align*} b p{\left (\frac{{\left (\log \left (\frac{b x}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a}\right )\right )} e}{b d^{2}} - \frac{{\left (\log \left (e x + d\right ) \log \left (-\frac{b e x + b d}{b d - a e} + 1\right ) +{\rm Li}_2\left (\frac{b e x + b d}{b d - a e}\right )\right )} e}{b d^{2}} - \frac{\log \left (b x + a\right )}{a d} + \frac{\log \left (x\right )}{a d}\right )} +{\left (\frac{e \log \left (e x + d\right )}{d^{2}} - \frac{e \log \left (x\right )}{d^{2}} - \frac{1}{d x}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*p*((log(b*x/a + 1)*log(x) + dilog(-b*x/a))*e/(b*d^2) - (log(e*x + d)*log(-(b*e*x + b*d)/(b*d - a*e) + 1) + d
ilog((b*e*x + b*d)/(b*d - a*e)))*e/(b*d^2) - log(b*x + a)/(a*d) + log(x)/(a*d)) + (e*log(e*x + d)/d^2 - e*log(
x)/d^2 - 1/(d*x))*log((b*x + a)^p*c)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (b x + a\right )}^{p} c\right )}{e x^{3} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x + a)^p*c)/(e*x^3 + d*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (b x + a\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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