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

Optimal. Leaf size=227 \[ -\frac{e^2 p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d^3}+\frac{e^2 p \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d^3}-\frac{b^2 p \log (x)}{2 a^2 d}+\frac{b^2 p \log (a+b x)}{2 a^2 d}+\frac{e^2 \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^3}-\frac{e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^3}+\frac{e \log \left (c (a+b x)^p\right )}{d^2 x}-\frac{\log \left (c (a+b x)^p\right )}{2 d x^2}-\frac{b e p \log (x)}{a d^2}+\frac{b e p \log (a+b x)}{a d^2}-\frac{b p}{2 a d x} \]

[Out]

-(b*p)/(2*a*d*x) - (b^2*p*Log[x])/(2*a^2*d) - (b*e*p*Log[x])/(a*d^2) + (b^2*p*Log[a + b*x])/(2*a^2*d) + (b*e*p
*Log[a + b*x])/(a*d^2) - Log[c*(a + b*x)^p]/(2*d*x^2) + (e*Log[c*(a + b*x)^p])/(d^2*x) + (e^2*Log[-((b*x)/a)]*
Log[c*(a + b*x)^p])/d^3 - (e^2*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/d^3 - (e^2*p*PolyLog[2, -((e
*(a + b*x))/(b*d - a*e))])/d^3 + (e^2*p*PolyLog[2, 1 + (b*x)/a])/d^3

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Rubi [A]  time = 0.222614, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -\frac{e^2 p \text{PolyLog}\left (2,-\frac{e (a+b x)}{b d-a e}\right )}{d^3}+\frac{e^2 p \text{PolyLog}\left (2,\frac{b x}{a}+1\right )}{d^3}-\frac{b^2 p \log (x)}{2 a^2 d}+\frac{b^2 p \log (a+b x)}{2 a^2 d}+\frac{e^2 \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^3}-\frac{e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )}{d^3}+\frac{e \log \left (c (a+b x)^p\right )}{d^2 x}-\frac{\log \left (c (a+b x)^p\right )}{2 d x^2}-\frac{b e p \log (x)}{a d^2}+\frac{b e p \log (a+b x)}{a d^2}-\frac{b p}{2 a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*p)/(2*a*d*x) - (b^2*p*Log[x])/(2*a^2*d) - (b*e*p*Log[x])/(a*d^2) + (b^2*p*Log[a + b*x])/(2*a^2*d) + (b*e*p
*Log[a + b*x])/(a*d^2) - Log[c*(a + b*x)^p]/(2*d*x^2) + (e*Log[c*(a + b*x)^p])/(d^2*x) + (e^2*Log[-((b*x)/a)]*
Log[c*(a + b*x)^p])/d^3 - (e^2*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/d^3 - (e^2*p*PolyLog[2, -((e
*(a + b*x))/(b*d - a*e))])/d^3 + (e^2*p*PolyLog[2, 1 + (b*x)/a])/d^3

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

Mathematica [A]  time = 0.168071, size = 188, normalized size = 0.83 \[ -\frac{2 e^2 p \text{PolyLog}\left (2,\frac{e (a+b x)}{a e-b d}\right )-2 e^2 p \text{PolyLog}\left (2,\frac{b x}{a}+1\right )+\frac{b d^2 p (-b x \log (a+b x)+a+b x \log (x))}{a^2 x}+\frac{d^2 \log \left (c (a+b x)^p\right )}{x^2}+2 e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac{b (d+e x)}{b d-a e}\right )-\frac{2 d e \log \left (c (a+b x)^p\right )}{x}-2 e^2 \log \left (-\frac{b x}{a}\right ) \log \left (c (a+b x)^p\right )+\frac{2 b d e p (\log (x)-\log (a+b x))}{a}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.608, size = 850, normalized size = 3.7 \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^3/(e*x+d),x)

[Out]

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

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Maxima [A]  time = 1.23295, size = 292, normalized size = 1.29 \begin{align*} \frac{1}{2} \,{\left (2 \, e{\left (\frac{\log \left (b x + a\right )}{a d^{2}} - \frac{\log \left (x\right )}{a d^{2}}\right )} - \frac{2 \,{\left (\log \left (\frac{b x}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a}\right )\right )} e^{2}}{b d^{3}} + \frac{2 \,{\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^{2}}{b d^{3}} + \frac{b \log \left (b x + a\right )}{a^{2} d} - \frac{b \log \left (x\right )}{a^{2} d} - \frac{1}{a d x}\right )} b p - \frac{1}{2} \,{\left (\frac{2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac{2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac{2 \, e x - d}{d^{2} x^{2}}\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^3/(e*x+d),x, algorithm="maxima")

[Out]

1/2*(2*e*(log(b*x + a)/(a*d^2) - log(x)/(a*d^2)) - 2*(log(b*x/a + 1)*log(x) + dilog(-b*x/a))*e^2/(b*d^3) + 2*(
log(e*x + d)*log(-(b*e*x + b*d)/(b*d - a*e) + 1) + dilog((b*e*x + b*d)/(b*d - a*e)))*e^2/(b*d^3) + b*log(b*x +
a)/(a^2*d) - b*log(x)/(a^2*d) - 1/(a*d*x))*b*p - 1/2*(2*e^2*log(e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)
/(d^2*x^2))*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^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x + a)^p*c)/(e*x^4 + d*x^3), 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**3/(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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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