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

Optimal. Leaf size=228 \[ -\frac {n \text {Li}_2\left (\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{e}-\frac {n \text {Li}_2\left (\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e}-\frac {n \log (d+e x) \log \left (-\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )}{e}-\frac {n \log (d+e x) \log \left (-\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )}{e}+\frac {\log (d+e x) \log \left (d \left (a+b x+c x^2\right )^n\right )}{e} \]

[Out]

ln(e*x+d)*ln(d*(c*x^2+b*x+a)^n)/e-n*ln(e*x+d)*ln(-e*(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2
))))/e-n*ln(e*x+d)*ln(-e*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))/e-n*polylog(2,2*c*(e*x
+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))/e-n*polylog(2,2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))/e

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Rubi [A]  time = 0.41, antiderivative size = 228, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.30, size = 226, normalized size = 0.99 \[ -\frac {n \text {Li}_2\left (\frac {2 c (d+e x)}{2 c d-b e+\sqrt {b^2-4 a c} e}\right )}{e}-\frac {n \text {Li}_2\left (\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e}-\frac {n \log (d+e x) \log \left (-\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )}{e}-\frac {n \log (d+e x) \log \left (-\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )}{e}+\frac {\log (d+e x) \log \left (d (a+x (b+c x))^n\right )}{e} \]

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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{e x + d}, x\right ) \]

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

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)

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maple [C]  time = 0.37, size = 493, normalized size = 2.16 \[ -\frac {i \pi \,\mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right ) \ln \left (e x +d \right )}{2 e}+\frac {i \pi \,\mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2} \ln \left (e x +d \right )}{2 e}+\frac {i \pi \,\mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2} \ln \left (e x +d \right )}{2 e}-\frac {i \pi \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{3} \ln \left (e x +d \right )}{2 e}-\frac {n \ln \left (\frac {-b e +2 c d -2 \left (e x +d \right ) c +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}{-b e +2 c d +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}\right ) \ln \left (e x +d \right )}{e}-\frac {n \ln \left (\frac {b e -2 c d +2 \left (e x +d \right ) c +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}{b e -2 c d +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}\right ) \ln \left (e x +d \right )}{e}-\frac {n \dilog \left (\frac {-b e +2 c d -2 \left (e x +d \right ) c +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}{-b e +2 c d +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}\right )}{e}-\frac {n \dilog \left (\frac {b e -2 c d +2 \left (e x +d \right ) c +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}{b e -2 c d +\sqrt {-4 c \,e^{2} a +b^{2} e^{2}}}\right )}{e}+\frac {\ln \relax (d ) \ln \left (e x +d \right )}{e}+\frac {\ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right ) \ln \left (e x +d \right )}{e} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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