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

Optimal. Leaf size=58 \[ \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2394, 2393, 2391} \[ \frac {p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e}+\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 57, normalized size = 0.98 \[ \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.10, size = 242, normalized size = 4.17 \[ -\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{2 e}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e}-\frac {i \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 e}-\frac {p \ln \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right ) \ln \left (e x +d \right )}{e}-\frac {p \dilog \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right )}{e}+\frac {\ln \relax (c ) \ln \left (e x +d \right )}{e}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 118, normalized size = 2.03 \[ \frac {b p {\left (\frac {\log \left (b x + a\right ) \log \left (e x + d\right )}{b} - \frac {\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )}{b}\right )}}{e} - \frac {p \log \left (b x + a\right ) \log \left (e x + d\right )}{e} + \frac {\log \left ({\left (b x + a\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x)^p)/(d + e*x),x)

[Out]

int(log(c*(a + b*x)^p)/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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