\(\int \frac {\log (c (a+b x)^p)}{x^3 (d+e x)} \, dx\) [225]

Optimal result

Integrand size = 21, antiderivative size = 227 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=-\frac {b p}{2 a d x}-\frac {b^2 p \log (x)}{2 a^2 d}-\frac {b e p \log (x)}{a d^2}+\frac {b^2 p \log (a+b x)}{2 a^2 d}+\frac {b e p \log (a+b x)}{a d^2}-\frac {\log \left (c (a+b x)^p\right )}{2 d x^2}+\frac {e \log \left (c (a+b x)^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^3}-\frac {e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{d^3} \]

[Out]

-1/2*b*p/a/d/x-1/2*b^2*p*ln(x)/a^2/d-b*e*p*ln(x)/a/d^2+1/2*b^2*p*ln(b*x+a)/a^2/d+b*e*p*ln(b*x+a)/a/d^2-1/2*ln(
c*(b*x+a)^p)/d/x^2+e*ln(c*(b*x+a)^p)/d^2/x+e^2*ln(-b*x/a)*ln(c*(b*x+a)^p)/d^3-e^2*ln(c*(b*x+a)^p)*ln(b*(e*x+d)
/(-a*e+b*d))/d^3-e^2*p*polylog(2,-e*(b*x+a)/(-a*e+b*d))/d^3+e^2*p*polylog(2,1+b*x/a)/d^3

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=-\frac {b^2 p \log (x)}{2 a^2 d}+\frac {b^2 p \log (a+b x)}{2 a^2 d}+\frac {e^2 \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^3}-\frac {e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^3}+\frac {e \log \left (c (a+b x)^p\right )}{d^2 x}-\frac {\log \left (c (a+b x)^p\right )}{2 d x^2}-\frac {e^2 p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {b x}{a}+1\right )}{d^3}-\frac {b e p \log (x)}{a d^2}+\frac {b e p \log (a+b x)}{a d^2}-\frac {b p}{2 a d x} \]

[In]

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

[Out]

-1/2*(b*p)/(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 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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=-\frac {\frac {2 b d e p \log (x)}{a}-\frac {2 b d e p \log (a+b x)}{a}+\frac {b d^2 p (a+b x \log (x)-b x \log (a+b x))}{a^2 x}+\frac {d^2 \log \left (c (a+b x)^p\right )}{x^2}-\frac {2 d e \log \left (c (a+b x)^p\right )}{x}-2 e^2 \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )+2 e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )+2 e^2 p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )-2 e^2 p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{2 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.15

method	result	size
parts	\(-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (c \left (b x +a \right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {e \ln \left (c \left (b x +a \right )^{p}\right )}{d^{2} x}-\frac {p b \left (\frac {2 e^{2} \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{3} b}+\frac {2 e^{2} \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{3} b}-\frac {\frac {\left (2 a e +b d \right ) \ln \left (b x +a \right )}{a^{2}}-\frac {d}{a x}+\frac {\left (-2 a e -b d \right ) \ln \left (x \right )}{a^{2}}}{d^{2}}\right )}{2}\)	\(260\)
risch	\(-\frac {\ln \left (\left (b x +a \right )^{p}\right ) e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) e}{d^{2} x}+\frac {b e p \ln \left (b x +a \right )}{a \,d^{2}}+\frac {b^{2} p \ln \left (b x +a \right )}{2 a^{2} d}-\frac {b e p \ln \left (x \right )}{a \,d^{2}}-\frac {b^{2} p \ln \left (x \right )}{2 a^{2} d}-\frac {b p}{2 a d x}-\frac {p \,e^{2} \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{3}}-\frac {p \,e^{2} \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{3}}+\frac {p \,e^{2} \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{3}}+\frac {p \,e^{2} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{d^{3}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {1}{2 d \,x^{2}}+\frac {e^{2} \ln \left (x \right )}{d^{3}}+\frac {e}{d^{2} x}\right )\)	\(409\)

[In]

int(ln(c*(b*x+a)^p)/x^3/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\log {\left (c \left (a + b x\right )^{p} \right )}}{x^{3} \left (d + e x\right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\frac {1}{2} \, {\left (2 \, e {\left (\frac {\log \left (b x + a\right )}{a d^{2}} - \frac {\log \left (x\right )}{a d^{2}}\right )} - \frac {2 \, {\left (\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} e^{2}}{b d^{3}} + \frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )\right )} e^{2}}{b d^{3}} + \frac {b \log \left (b x + a\right )}{a^{2} d} - \frac {b \log \left (x\right )}{a^{2} d} - \frac {1}{a d x}\right )} b p - \frac {1}{2} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[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)

Giac [F]

\[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \]

[In]

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

[Out]

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