3.3.24 \(\int \frac {\log (c (a+b x)^p)}{x^2 (d+e x)} \, dx\) [224]

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

[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 = 0.11, 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 = {46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \begin {gather*} \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} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2438

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 2440

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 2441

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 2442

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 2463

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) \text {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 = 0.04, size = 139, normalized size = 0.95 \begin {gather*} \frac {b d p x \log (x)-b d p x \log (a+b x)-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 x \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )+a e p x \text {Li}_2\left (\frac {e (a+b x)}{-b d+a e}\right )-a e p x \text {Li}_2\left (1+\frac {b x}{a}\right )}{a d^2 x} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.48, size = 615, normalized size = 4.21

method	result	size
risch	\(\frac {\ln \left (\left (b x +a \right )^{p}\right ) e \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{d x}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) e \ln \left (x \right )}{d^{2}}-\frac {p e \dilog \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{2}}-\frac {p e \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{2}}+\frac {b p \ln \left (x \right )}{a d}-\frac {b p \ln \left (b x +a \right )}{a d}+\frac {p e \dilog \left (\frac {b x +a}{a}\right )}{d^{2}}+\frac {p e \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{2}}+\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 ) \mathrm {csgn}\left (i c \right )}{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 \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) e \ln \left (x \right )}{2 d^{2}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) e \ln \left (x \right )}{2 d^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{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} e \ln \left (x \right )}{2 d^{2}}+\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} e \ln \left (e x +d \right )}{2 d^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) e \ln \left (e x +d \right )}{2 d^{2}}+\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} e \ln \left (x \right )}{2 d^{2}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} e \ln \left (e x +d \right )}{2 d^{2}}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{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 ) \mathrm {csgn}\left (i c \right ) e \ln \left (e x +d \right )}{2 d^{2}}+\frac {\ln \left (c \right ) e \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (c \right )}{d x}-\frac {\ln \left (c \right ) e \ln \left (x \right )}{d^{2}}\)	\(615\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(((e*x+d)*b+a*e-b*d)/(a
*e-b*d))-p*e/d^2*ln(e*x+d)*ln(((e*x+d)*b+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*(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/d/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(
x)-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*c*(b*x+a)^p)^3/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*(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)^3*e/d^2*ln(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)^2*csgn(I*c)/d/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)+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 = 0.36, size = 166, normalized size = 1.14 \begin {gather*} 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 (x e + d\right ) \log \left (-\frac {b x e + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b x e + 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 (x e + 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 {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x^3*e + d*x^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/((x*e + d)*x^2), x)

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

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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