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

Optimal. Leaf size=146 \[ -\frac {e \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac {e \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^2}-\frac {\log \left (c (a+b x)^p\right )}{d x}+\frac {e p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d^2}-\frac {e p \text {Li}_2\left (\frac {b x}{a}+1\right )}{d^2}+\frac {b p \log (x)}{a d}-\frac {b p \log (a+b x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 146, normalized size of antiderivative = 1.00, 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 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 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 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 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 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^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*}

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Mathematica [A]  time = 0.05, size = 139, normalized size = 0.95 \[ \frac {a e x \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )-a d \log \left (c (a+b x)^p\right )-a e x \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )+a e p x \text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )-b d p x \log (a+b x)-a e p x \text {Li}_2\left (\frac {b x}{a}+1\right )+b d p x \log (x)}{a d^2 x} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{e x^{3} + d x^{2}}, x\right ) \]

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

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)

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maple [C]  time = 0.25, size = 615, normalized size = 4.21 \[ \frac {i \pi e \,\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^{2}}-\frac {i \pi e \,\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^{2}}-\frac {i \pi e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \relax (x )}{2 d^{2}}+\frac {i \pi e \,\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^{2}}-\frac {i \pi e \,\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^{2}}+\frac {i \pi e \,\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^{2}}+\frac {i \pi e \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \relax (x )}{2 d^{2}}-\frac {i \pi e \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 d^{2}}+\frac {e p \ln \relax (x ) \ln \left (\frac {b x +a}{a}\right )}{d^{2}}-\frac {e 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^{2}}+\frac {b p \ln \relax (x )}{a d}-\frac {b p \ln \left (b x +a \right )}{a 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 )}{2 d x}-\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2 d x}-\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}}{2 d x}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2 d x}+\frac {e p \dilog \left (\frac {b x +a}{a}\right )}{d^{2}}-\frac {e p \dilog \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right )}{d^{2}}-\frac {e \ln \relax (c ) \ln \relax (x )}{d^{2}}+\frac {e \ln \relax (c ) \ln \left (e x +d \right )}{d^{2}}-\frac {e \ln \relax (x ) \ln \left (\left (b x +a \right )^{p}\right )}{d^{2}}+\frac {e \ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \relax (c )}{d x}-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{d x} \]

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

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maxima [A]  time = 0.65, size = 156, normalized size = 1.07 \[ b p {\left (\frac {{\left (\log \left (\frac {b x}{a} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} e}{b d^{2}} - \frac {{\left (\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )\right )} e}{b d^{2}} - \frac {\log \left (b x + a\right )}{a d} + \frac {\log \relax (x)}{a d}\right )} + {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \relax (x)}{d^{2}} - \frac {1}{d x}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/x^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{x^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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