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

Optimal. Leaf size=97 \[ -\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d}-\frac {p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {p \text {Li}_2\left (\frac {b x}{a}+1\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {36, 29, 31, 2416, 2394, 2315, 2393, 2391} \[ -\frac {p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {p \text {PolyLog}\left (2,\frac {b x}{a}+1\right )}{d}-\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 98, normalized size = 1.01 \[ -\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d}-\frac {p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {p \text {Li}_2\left (\frac {a+b x}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 420, normalized size = 4.33 \[ -\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \ln \relax (x )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \relax (x )}{2 d}-\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 d}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \relax (x )}{2 d}-\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 d}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \relax (x )}{2 d}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 d}-\frac {p \ln \relax (x ) \ln \left (\frac {b x +a}{a}\right )}{d}+\frac {p \ln \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right ) \ln \left (e x +d \right )}{d}-\frac {p \dilog \left (\frac {b x +a}{a}\right )}{d}+\frac {p \dilog \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right )}{d}+\frac {\ln \relax (c ) \ln \relax (x )}{d}-\frac {\ln \relax (c ) \ln \left (e x +d \right )}{d}+\frac {\ln \relax (x ) \ln \left (\left (b x +a \right )^{p}\right )}{d}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 123, normalized size = 1.27 \[ -b p {\left (\frac {\log \left (\frac {b x}{a} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a}\right )}{b d} - \frac {\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )}{b d}\right )} - {\left (\frac {\log \left (e x + d\right )}{d} - \frac {\log \relax (x)}{d}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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