3.1.78 \(\int \frac {\log (d (a+b x+c x^2)^n)}{x^3} \, dx\) [78]

Optimal. Leaf size=121 \[ -\frac {b n}{2 a x}-\frac {b \sqrt {b^2-4 a c} n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 a^2}-\frac {\left (b^2-2 a c\right ) n \log (x)}{2 a^2}+\frac {\left (b^2-2 a c\right ) n \log \left (a+b x+c x^2\right )}{4 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2605, 814, 648, 632, 212, 642} \begin {gather*} \frac {n \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{4 a^2}-\frac {n \log (x) \left (b^2-2 a c\right )}{2 a^2}-\frac {b n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}-\frac {b n}{2 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[d*(a + b*x + c*x^2)^n]/x^3,x]

[Out]

-1/2*(b*n)/(a*x) - (b*Sqrt[b^2 - 4*a*c]*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(2*a^2) - ((b^2 - 2*a*c)*n*L
og[x])/(2*a^2) + ((b^2 - 2*a*c)*n*Log[a + b*x + c*x^2])/(4*a^2) - Log[d*(a + b*x + c*x^2)^n]/(2*x^2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2605

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

Rubi steps

\begin {align*} \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^3} \, dx &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}+\frac {1}{2} n \int \frac {b+2 c x}{x^2 \left (a+b x+c x^2\right )} \, dx\\ &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}+\frac {1}{2} n \int \left (\frac {b}{a x^2}+\frac {-b^2+2 a c}{a^2 x}+\frac {b \left (b^2-3 a c\right )+c \left (b^2-2 a c\right ) x}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=-\frac {b n}{2 a x}-\frac {\left (b^2-2 a c\right ) n \log (x)}{2 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}+\frac {n \int \frac {b \left (b^2-3 a c\right )+c \left (b^2-2 a c\right ) x}{a+b x+c x^2} \, dx}{2 a^2}\\ &=-\frac {b n}{2 a x}-\frac {\left (b^2-2 a c\right ) n \log (x)}{2 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}+\frac {\left (b \left (b^2-4 a c\right ) n\right ) \int \frac {1}{a+b x+c x^2} \, dx}{4 a^2}+\frac {\left (\left (b^2-2 a c\right ) n\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{4 a^2}\\ &=-\frac {b n}{2 a x}-\frac {\left (b^2-2 a c\right ) n \log (x)}{2 a^2}+\frac {\left (b^2-2 a c\right ) n \log \left (a+b x+c x^2\right )}{4 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}-\frac {\left (b \left (b^2-4 a c\right ) n\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{2 a^2}\\ &=-\frac {b n}{2 a x}-\frac {b \sqrt {b^2-4 a c} n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 a^2}-\frac {\left (b^2-2 a c\right ) n \log (x)}{2 a^2}+\frac {\left (b^2-2 a c\right ) n \log \left (a+b x+c x^2\right )}{4 a^2}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 105, normalized size = 0.87 \begin {gather*} -\frac {\frac {n x \left (2 a b+2 b \sqrt {b^2-4 a c} x \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+2 \left (b^2-2 a c\right ) x \log (x)-\left (b^2-2 a c\right ) x \log (a+x (b+c x))\right )}{a^2}+2 \log \left (d (a+x (b+c x))^n\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[d*(a + b*x + c*x^2)^n]/x^3,x]

[Out]

-1/4*((n*x*(2*a*b + 2*b*Sqrt[b^2 - 4*a*c]*x*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]] + 2*(b^2 - 2*a*c)*x*Log[x]
- (b^2 - 2*a*c)*x*Log[a + x*(b + c*x)]))/a^2 + 2*Log[d*(a + x*(b + c*x))^n])/x^2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.06, size = 1178, normalized size = 9.74

method	result	size
risch	\(-\frac {\ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{2 x^{2}}-\frac {-i \pi \,a^{2} \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 )+i \pi \,a^{2} \mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2}+i \pi \,a^{2} \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}-i \pi \,a^{2} \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{3}+2 n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}-6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}+8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}+5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) c a \,x^{2}-n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}-6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}+8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}+5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) b^{2} x^{2}+2 n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}+6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}-8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}-5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) c a \,x^{2}-n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}+6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}-8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}-5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) b^{2} x^{2}-4 n \ln \left (x \right ) c a \,x^{2}+2 n \ln \left (x \right ) b^{2} x^{2}-n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}-6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}+8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}+5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c -2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}\, x^{2}+n \ln \left (\left (-16 a^{2} b^{2} c^{2}+12 a \,b^{4} c -2 b^{6}+6 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} c^{2}-8 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{2} c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, b^{4}\right ) x -12 a^{3} b \,c^{2}+11 b^{3} c \,a^{2}-2 a \,b^{5}-5 \sqrt {-4 a \,b^{2} c +b^{4}}\, a^{2} b c +2 \sqrt {-4 a \,b^{2} c +b^{4}}\, a \,b^{3}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}\, x^{2}+2 a b n x +2 \ln \left (d \right ) a^{2}}{4 a^{2} x^{2}}\)	\(1178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(d*(c*x^2+b*x+a)^n)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/x^2*ln((c*x^2+b*x+a)^n)-1/4*(-I*Pi*a^2*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+I*Pi*a
^2*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2+I*Pi*a^2*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*a^2
*csgn(I*d*(c*x^2+b*x+a)^n)^3+2*n*ln((-16*a^2*b^2*c^2+12*a*b^4*c-2*b^6-6*(-4*a*b^2*c+b^4)^(1/2)*a^2*c^2+8*(-4*a
*b^2*c+b^4)^(1/2)*a*b^2*c-2*(-4*a*b^2*c+b^4)^(1/2)*b^4)*x-12*a^3*b*c^2+11*b^3*c*a^2-2*a*b^5+5*(-4*a*b^2*c+b^4)
^(1/2)*a^2*b*c-2*(-4*a*b^2*c+b^4)^(1/2)*a*b^3)*c*a*x^2-n*ln((-16*a^2*b^2*c^2+12*a*b^4*c-2*b^6-6*(-4*a*b^2*c+b^
4)^(1/2)*a^2*c^2+8*(-4*a*b^2*c+b^4)^(1/2)*a*b^2*c-2*(-4*a*b^2*c+b^4)^(1/2)*b^4)*x-12*a^3*b*c^2+11*b^3*c*a^2-2*
a*b^5+5*(-4*a*b^2*c+b^4)^(1/2)*a^2*b*c-2*(-4*a*b^2*c+b^4)^(1/2)*a*b^3)*b^2*x^2+2*n*ln((-16*a^2*b^2*c^2+12*a*b^
4*c-2*b^6+6*(-4*a*b^2*c+b^4)^(1/2)*a^2*c^2-8*(-4*a*b^2*c+b^4)^(1/2)*a*b^2*c+2*(-4*a*b^2*c+b^4)^(1/2)*b^4)*x-12
*a^3*b*c^2+11*b^3*c*a^2-2*a*b^5-5*(-4*a*b^2*c+b^4)^(1/2)*a^2*b*c+2*(-4*a*b^2*c+b^4)^(1/2)*a*b^3)*c*a*x^2-n*ln(
(-16*a^2*b^2*c^2+12*a*b^4*c-2*b^6+6*(-4*a*b^2*c+b^4)^(1/2)*a^2*c^2-8*(-4*a*b^2*c+b^4)^(1/2)*a*b^2*c+2*(-4*a*b^
2*c+b^4)^(1/2)*b^4)*x-12*a^3*b*c^2+11*b^3*c*a^2-2*a*b^5-5*(-4*a*b^2*c+b^4)^(1/2)*a^2*b*c+2*(-4*a*b^2*c+b^4)^(1
/2)*a*b^3)*b^2*x^2-4*n*ln(x)*c*a*x^2+2*n*ln(x)*b^2*x^2-n*ln((-16*a^2*b^2*c^2+12*a*b^4*c-2*b^6-6*(-4*a*b^2*c+b^
4)^(1/2)*a^2*c^2+8*(-4*a*b^2*c+b^4)^(1/2)*a*b^2*c-2*(-4*a*b^2*c+b^4)^(1/2)*b^4)*x-12*a^3*b*c^2+11*b^3*c*a^2-2*
a*b^5+5*(-4*a*b^2*c+b^4)^(1/2)*a^2*b*c-2*(-4*a*b^2*c+b^4)^(1/2)*a*b^3)*(-4*a*b^2*c+b^4)^(1/2)*x^2+n*ln((-16*a^
2*b^2*c^2+12*a*b^4*c-2*b^6+6*(-4*a*b^2*c+b^4)^(1/2)*a^2*c^2-8*(-4*a*b^2*c+b^4)^(1/2)*a*b^2*c+2*(-4*a*b^2*c+b^4
)^(1/2)*b^4)*x-12*a^3*b*c^2+11*b^3*c*a^2-2*a*b^5-5*(-4*a*b^2*c+b^4)^(1/2)*a^2*b*c+2*(-4*a*b^2*c+b^4)^(1/2)*a*b
^3)*(-4*a*b^2*c+b^4)^(1/2)*x^2+2*a*b*n*x+2*ln(d)*a^2)/a^2/x^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.37, size = 261, normalized size = 2.16 \begin {gather*} \left [\frac {\sqrt {b^{2} - 4 \, a c} b n x^{2} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, {\left (b^{2} - 2 \, a c\right )} n x^{2} \log \left (x\right ) - 2 \, a b n x - 2 \, a^{2} \log \left (d\right ) + {\left ({\left (b^{2} - 2 \, a c\right )} n x^{2} - 2 \, a^{2} n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, a^{2} x^{2}}, -\frac {2 \, \sqrt {-b^{2} + 4 \, a c} b n x^{2} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (b^{2} - 2 \, a c\right )} n x^{2} \log \left (x\right ) + 2 \, a b n x + 2 \, a^{2} \log \left (d\right ) - {\left ({\left (b^{2} - 2 \, a c\right )} n x^{2} - 2 \, a^{2} n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^3,x, algorithm="fricas")

[Out]

[1/4*(sqrt(b^2 - 4*a*c)*b*n*x^2*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2
+ b*x + a)) - 2*(b^2 - 2*a*c)*n*x^2*log(x) - 2*a*b*n*x - 2*a^2*log(d) + ((b^2 - 2*a*c)*n*x^2 - 2*a^2*n)*log(c
*x^2 + b*x + a))/(a^2*x^2), -1/4*(2*sqrt(-b^2 + 4*a*c)*b*n*x^2*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4
*a*c)) + 2*(b^2 - 2*a*c)*n*x^2*log(x) + 2*a*b*n*x + 2*a^2*log(d) - ((b^2 - 2*a*c)*n*x^2 - 2*a^2*n)*log(c*x^2 +
b*x + a))/(a^2*x^2)]

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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(d*(c*x**2+b*x+a)**n)/x**3,x)

[Out]

Timed out

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Giac [A]
time = 3.15, size = 129, normalized size = 1.07 \begin {gather*} \frac {{\left (b^{2} n - 2 \, a c n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, a^{2}} - \frac {n \log \left (c x^{2} + b x + a\right )}{2 \, x^{2}} - \frac {{\left (b^{2} n - 2 \, a c n\right )} \log \left (x\right )}{2 \, a^{2}} + \frac {{\left (b^{3} n - 4 \, a b c n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} - \frac {b n x + a \log \left (d\right )}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^3,x, algorithm="giac")

[Out]

1/4*(b^2*n - 2*a*c*n)*log(c*x^2 + b*x + a)/a^2 - 1/2*n*log(c*x^2 + b*x + a)/x^2 - 1/2*(b^2*n - 2*a*c*n)*log(x)
/a^2 + 1/2*(b^3*n - 4*a*b*c*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^2) - 1/2*(b*n*x +
a*log(d))/(a*x^2)

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Mupad [B]
time = 0.76, size = 474, normalized size = 3.92 \begin {gather*} \frac {\ln \left (\frac {b^3\,c^2\,n^2-2\,a\,b\,c^3\,n^2}{4\,a^2}+\frac {\left (b^2\,n-2\,a\,c\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {\left (\frac {x\,\left (24\,a^3\,c^2-8\,a^2\,b^2\,c\right )}{4\,a^2}-a\,b\,c\right )\,\left (b^2\,n-2\,a\,c\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,a^2}-\frac {2\,a\,b^3\,c\,n-6\,a^2\,b\,c^2\,n}{4\,a^2}+\frac {x\,\left (12\,a^2\,c^3\,n-4\,a\,b^2\,c^2\,n\right )}{4\,a^2}\right )}{4\,a^2}+\frac {b^2\,c^3\,n^2\,x}{4\,a^2}\right )\,\left (b^2\,n-2\,a\,c\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,a^2}-\frac {\ln \left (x\right )\,\left (b^2\,n-2\,a\,c\,n\right )}{2\,a^2}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{2\,x^2}-\frac {\ln \left (\frac {b^3\,c^2\,n^2-2\,a\,b\,c^3\,n^2}{4\,a^2}+\frac {\left (2\,a\,c\,n-b^2\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {2\,a\,b^3\,c\,n-6\,a^2\,b\,c^2\,n}{4\,a^2}+\frac {\left (\frac {x\,\left (24\,a^3\,c^2-8\,a^2\,b^2\,c\right )}{4\,a^2}-a\,b\,c\right )\,\left (2\,a\,c\,n-b^2\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,a^2}-\frac {x\,\left (12\,a^2\,c^3\,n-4\,a\,b^2\,c^2\,n\right )}{4\,a^2}\right )}{4\,a^2}+\frac {b^2\,c^3\,n^2\,x}{4\,a^2}\right )\,\left (2\,a\,c\,n-b^2\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,a^2}-\frac {b\,n}{2\,a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d*(a + b*x + c*x^2)^n)/x^3,x)

[Out]

(log((b^3*c^2*n^2 - 2*a*b*c^3*n^2)/(4*a^2) + ((b^2*n - 2*a*c*n + b*n*(b^2 - 4*a*c)^(1/2))*((((x*(24*a^3*c^2 -
8*a^2*b^2*c))/(4*a^2) - a*b*c)*(b^2*n - 2*a*c*n + b*n*(b^2 - 4*a*c)^(1/2)))/(4*a^2) - (2*a*b^3*c*n - 6*a^2*b*c
^2*n)/(4*a^2) + (x*(12*a^2*c^3*n - 4*a*b^2*c^2*n))/(4*a^2)))/(4*a^2) + (b^2*c^3*n^2*x)/(4*a^2))*(b^2*n - 2*a*c
*n + b*n*(b^2 - 4*a*c)^(1/2)))/(4*a^2) - (log(x)*(b^2*n - 2*a*c*n))/(2*a^2) - log(d*(a + b*x + c*x^2)^n)/(2*x^
2) - (log((b^3*c^2*n^2 - 2*a*b*c^3*n^2)/(4*a^2) + ((2*a*c*n - b^2*n + b*n*(b^2 - 4*a*c)^(1/2))*((2*a*b^3*c*n -
6*a^2*b*c^2*n)/(4*a^2) + (((x*(24*a^3*c^2 - 8*a^2*b^2*c))/(4*a^2) - a*b*c)*(2*a*c*n - b^2*n + b*n*(b^2 - 4*a*
c)^(1/2)))/(4*a^2) - (x*(12*a^2*c^3*n - 4*a*b^2*c^2*n))/(4*a^2)))/(4*a^2) + (b^2*c^3*n^2*x)/(4*a^2))*(2*a*c*n
- b^2*n + b*n*(b^2 - 4*a*c)^(1/2)))/(4*a^2) - (b*n)/(2*a*x)

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