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

Optimal. Leaf size=228 \[ -\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )}{e}-\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}+\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e} \]

[Out]

-((n*Log[-((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e) - (n*Log
[-((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e + (Log[d + e*x]*L
og[d*(a + b*x + c*x^2)^n])/e - (n*PolyLog[2, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/e - (n*Poly
Log[2, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/e

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Rubi [A]  time = 0.411819, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2524, 2418, 2394, 2393, 2391} \[ -\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )}{e}-\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}+\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[d*(a + b*x + c*x^2)^n]/(d + e*x),x]

[Out]

-((n*Log[-((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e) - (n*Log
[-((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e + (Log[d + e*x]*L
og[d*(a + b*x + c*x^2)^n])/e - (n*PolyLog[2, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/e - (n*Poly
Log[2, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/e

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\log \left (d \left (a+b x+c x^2\right )^n\right )}{d+e x} \, dx &=\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}-\frac{n \int \frac{(b+2 c x) \log (d+e x)}{a+b x+c x^2} \, dx}{e}\\ &=\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}-\frac{n \int \left (\frac{2 c \log (d+e x)}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{2 c \log (d+e x)}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx}{e}\\ &=\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}-\frac{(2 c n) \int \frac{\log (d+e x)}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{e}-\frac{(2 c n) \int \frac{\log (d+e x)}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{e}\\ &=-\frac{n \log \left (-\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}-\frac{n \log \left (-\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}+\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}+n \int \frac{\log \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{d+e x} \, dx+n \int \frac{\log \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{d+e x} \, dx\\ &=-\frac{n \log \left (-\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}-\frac{n \log \left (-\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}+\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}+\frac{n \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 c x}{-2 c d+\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{x} \, dx,x,d+e x\right )}{e}+\frac{n \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 c x}{-2 c d+\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{x} \, dx,x,d+e x\right )}{e}\\ &=-\frac{n \log \left (-\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}-\frac{n \log \left (-\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \log (d+e x)}{e}+\frac{\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e}-\frac{n \text{Li}_2\left (\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{e}-\frac{n \text{Li}_2\left (\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.292592, size = 226, normalized size = 0.99 \[ -\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{e \sqrt{b^2-4 a c}-b e+2 c d}\right )}{e}-\frac{n \text{PolyLog}\left (2,\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )}{e}-\frac{n \log (d+e x) \log \left (-\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{e}+\frac{\log (d+e x) \log \left (d (a+x (b+c x))^n\right )}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[d*(a + b*x + c*x^2)^n]/(d + e*x),x]

[Out]

-((n*Log[-((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e) - (n*Log
[-((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))]*Log[d + e*x])/e + (Log[d + e*x]*L
og[d*(a + x*(b + c*x))^n])/e - (n*PolyLog[2, (2*c*(d + e*x))/(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)])/e - (n*Poly
Log[2, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/e

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Maple [C]  time = 0.089, size = 493, normalized size = 2.2 \begin{align*}{\frac{\ln \left ( ex+d \right ) \ln \left ( \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) }{e}}-{\frac{n\ln \left ( ex+d \right ) }{e}\ln \left ({ \left ( -2\, \left ( ex+d \right ) c-be+2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) \left ( -be+2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{n\ln \left ( ex+d \right ) }{e}\ln \left ({ \left ( 2\, \left ( ex+d \right ) c+be-2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) \left ( be-2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{n}{e}{\it dilog} \left ({ \left ( -2\, \left ( ex+d \right ) c-be+2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) \left ( -be+2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{n}{e}{\it dilog} \left ({ \left ( 2\, \left ( ex+d \right ) c+be-2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) \left ( be-2\,cd+\sqrt{-4\,c{e}^{2}a+{b}^{2}{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( id \right ){\it csgn} \left ( i \left ( c{x}^{2}+bx+a \right ) ^{n} \right ){\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) }{e}}+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( id \right ) \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{2}}{e}}+{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \,{\it csgn} \left ( i \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{2}}{e}}-{\frac{{\frac{i}{2}}\ln \left ( ex+d \right ) \pi \, \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{3}}{e}}+{\frac{\ln \left ( ex+d \right ) \ln \left ( d \right ) }{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(d*(c*x^2+b*x+a)^n)/(e*x+d),x)

[Out]

ln(e*x+d)/e*ln((c*x^2+b*x+a)^n)-1/e*n*ln(e*x+d)*ln((-2*(e*x+d)*c-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/(-b*e+2
*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))-1/e*n*ln(e*x+d)*ln((2*(e*x+d)*c+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/(b*e-2
*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2)))-1/e*n*dilog((-2*(e*x+d)*c-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/(-b*e+2*c*d+
(-4*a*c*e^2+b^2*e^2)^(1/2)))-1/e*n*dilog((2*(e*x+d)*c+b*e-2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))/(b*e-2*c*d+(-4*a*c
*e^2+b^2*e^2)^(1/2)))-1/2*I*ln(e*x+d)/e*Pi*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+1/2*I*l
n(e*x+d)/e*Pi*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2+1/2*I*ln(e*x+d)/e*Pi*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x
^2+b*x+a)^n)^2-1/2*I*ln(e*x+d)/e*Pi*csgn(I*d*(c*x^2+b*x+a)^n)^3+ln(e*x+d)/e*ln(d)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d),x, algorithm="maxima")

[Out]

integrate(log((c*x^2 + b*x + a)^n*d)/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(d*(c*x**2+b*x+a)**n)/(e*x+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/(e*x+d),x, algorithm="giac")

[Out]

integrate(log((c*x^2 + b*x + a)^n*d)/(e*x + d), x)