3.79 \(\int \frac {\log (d (a+b x+c x^2)^n)}{x^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac {b n \left (b^2-3 a c\right ) \log \left (a+b x+c x^2\right )}{6 a^3}+\frac {b n \log (x) \left (b^2-3 a c\right )}{3 a^3}+\frac {n \sqrt {b^2-4 a c} \left (b^2-a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{3 a^3}+\frac {n \left (b^2-2 a c\right )}{3 a^2 x}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}-\frac {b n}{6 a x^2} \]

[Out]

-1/6*b*n/a/x^2+1/3*(-2*a*c+b^2)*n/a^2/x+1/3*b*(-3*a*c+b^2)*n*ln(x)/a^3-1/6*b*(-3*a*c+b^2)*n*ln(c*x^2+b*x+a)/a^
3-1/3*ln(d*(c*x^2+b*x+a)^n)/x^3+1/3*(-a*c+b^2)*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/a^3

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Rubi [A]  time = 0.20, antiderivative size = 149, 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 = {2525, 800, 634, 618, 206, 628} \[ -\frac {b n \left (b^2-3 a c\right ) \log \left (a+b x+c x^2\right )}{6 a^3}+\frac {n \left (b^2-2 a c\right )}{3 a^2 x}+\frac {b n \log (x) \left (b^2-3 a c\right )}{3 a^3}+\frac {n \sqrt {b^2-4 a c} \left (b^2-a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{3 a^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}-\frac {b n}{6 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[d*(a + b*x + c*x^2)^n]/x^4,x]

[Out]

-(b*n)/(6*a*x^2) + ((b^2 - 2*a*c)*n)/(3*a^2*x) + (Sqrt[b^2 - 4*a*c]*(b^2 - a*c)*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2
- 4*a*c]])/(3*a^3) + (b*(b^2 - 3*a*c)*n*Log[x])/(3*a^3) - (b*(b^2 - 3*a*c)*n*Log[a + b*x + c*x^2])/(6*a^3) -
Log[d*(a + b*x + c*x^2)^n]/(3*x^3)

Rule 206

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

Rule 618

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 628

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 634

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 800

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 2525

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^4} \, dx &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}+\frac {1}{3} n \int \frac {b+2 c x}{x^3 \left (a+b x+c x^2\right )} \, dx\\ &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}+\frac {1}{3} n \int \left (\frac {b}{a x^3}+\frac {-b^2+2 a c}{a^2 x^2}+\frac {b^3-3 a b c}{a^3 x}+\frac {-b^4+4 a b^2 c-2 a^2 c^2-b c \left (b^2-3 a c\right ) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=-\frac {b n}{6 a x^2}+\frac {\left (b^2-2 a c\right ) n}{3 a^2 x}+\frac {b \left (b^2-3 a c\right ) n \log (x)}{3 a^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}+\frac {n \int \frac {-b^4+4 a b^2 c-2 a^2 c^2-b c \left (b^2-3 a c\right ) x}{a+b x+c x^2} \, dx}{3 a^3}\\ &=-\frac {b n}{6 a x^2}+\frac {\left (b^2-2 a c\right ) n}{3 a^2 x}+\frac {b \left (b^2-3 a c\right ) n \log (x)}{3 a^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}-\frac {\left (b \left (b^2-3 a c\right ) n\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{6 a^3}-\frac {\left (\left (b^2-4 a c\right ) \left (b^2-a c\right ) n\right ) \int \frac {1}{a+b x+c x^2} \, dx}{6 a^3}\\ &=-\frac {b n}{6 a x^2}+\frac {\left (b^2-2 a c\right ) n}{3 a^2 x}+\frac {b \left (b^2-3 a c\right ) n \log (x)}{3 a^3}-\frac {b \left (b^2-3 a c\right ) n \log \left (a+b x+c x^2\right )}{6 a^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}+\frac {\left (\left (b^2-4 a c\right ) \left (b^2-a c\right ) n\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{3 a^3}\\ &=-\frac {b n}{6 a x^2}+\frac {\left (b^2-2 a c\right ) n}{3 a^2 x}+\frac {\sqrt {b^2-4 a c} \left (b^2-a c\right ) n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{3 a^3}+\frac {b \left (b^2-3 a c\right ) n \log (x)}{3 a^3}-\frac {b \left (b^2-3 a c\right ) n \log \left (a+b x+c x^2\right )}{6 a^3}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 132, normalized size = 0.89 \[ -\frac {\frac {n x \left (a^2 b-2 b x^2 \log (x) \left (b^2-3 a c\right )+b x^2 \left (b^2-3 a c\right ) \log (a+x (b+c x))-2 x^2 \sqrt {b^2-4 a c} \left (b^2-a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )-2 a x \left (b^2-2 a c\right )\right )}{a^3}+2 \log \left (d (a+x (b+c x))^n\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[d*(a + b*x + c*x^2)^n]/x^4,x]

[Out]

-1/6*((n*x*(a^2*b - 2*a*(b^2 - 2*a*c)*x - 2*Sqrt[b^2 - 4*a*c]*(b^2 - a*c)*x^2*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4
*a*c]] - 2*b*(b^2 - 3*a*c)*x^2*Log[x] + b*(b^2 - 3*a*c)*x^2*Log[a + x*(b + c*x)]))/a^3 + 2*Log[d*(a + x*(b + c
*x))^n])/x^3

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fricas [A]  time = 0.53, size = 318, normalized size = 2.13 \[ \left [-\frac {{\left (b^{2} - a c\right )} \sqrt {b^{2} - 4 \, a c} n x^{3} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, {\left (b^{3} - 3 \, a b c\right )} n x^{3} \log \relax (x) + a^{2} b n x - 2 \, {\left (a b^{2} - 2 \, a^{2} c\right )} n x^{2} + 2 \, a^{3} \log \relax (d) + {\left ({\left (b^{3} - 3 \, a b c\right )} n x^{3} + 2 \, a^{3} n\right )} \log \left (c x^{2} + b x + a\right )}{6 \, a^{3} x^{3}}, \frac {2 \, {\left (b^{2} - a c\right )} \sqrt {-b^{2} + 4 \, a c} n x^{3} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (b^{3} - 3 \, a b c\right )} n x^{3} \log \relax (x) - a^{2} b n x + 2 \, {\left (a b^{2} - 2 \, a^{2} c\right )} n x^{2} - 2 \, a^{3} \log \relax (d) - {\left ({\left (b^{3} - 3 \, a b c\right )} n x^{3} + 2 \, a^{3} n\right )} \log \left (c x^{2} + b x + a\right )}{6 \, a^{3} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^4,x, algorithm="fricas")

[Out]

[-1/6*((b^2 - a*c)*sqrt(b^2 - 4*a*c)*n*x^3*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^3 - 3*a*b*c)*n*x^3*log(x) + a^2*b*n*x - 2*(a*b^2 - 2*a^2*c)*n*x^2 + 2*a^3*log(d
) + ((b^3 - 3*a*b*c)*n*x^3 + 2*a^3*n)*log(c*x^2 + b*x + a))/(a^3*x^3), 1/6*(2*(b^2 - a*c)*sqrt(-b^2 + 4*a*c)*n
*x^3*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(b^3 - 3*a*b*c)*n*x^3*log(x) - a^2*b*n*x + 2*(a
*b^2 - 2*a^2*c)*n*x^2 - 2*a^3*log(d) - ((b^3 - 3*a*b*c)*n*x^3 + 2*a^3*n)*log(c*x^2 + b*x + a))/(a^3*x^3)]

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giac [A]  time = 0.25, size = 164, normalized size = 1.10 \[ -\frac {{\left (b^{3} n - 3 \, a b c n\right )} \log \left (c x^{2} + b x + a\right )}{6 \, a^{3}} - \frac {n \log \left (c x^{2} + b x + a\right )}{3 \, x^{3}} + \frac {{\left (b^{3} n - 3 \, a b c n\right )} \log \relax (x)}{3 \, a^{3}} - \frac {{\left (b^{4} n - 5 \, a b^{2} c n + 4 \, a^{2} c^{2} n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt {-b^{2} + 4 \, a c} a^{3}} + \frac {2 \, b^{2} n x^{2} - 4 \, a c n x^{2} - a b n x - 2 \, a^{2} \log \relax (d)}{6 \, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^4,x, algorithm="giac")

[Out]

-1/6*(b^3*n - 3*a*b*c*n)*log(c*x^2 + b*x + a)/a^3 - 1/3*n*log(c*x^2 + b*x + a)/x^3 + 1/3*(b^3*n - 3*a*b*c*n)*l
og(x)/a^3 - 1/3*(b^4*n - 5*a*b^2*c*n + 4*a^2*c^2*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)
*a^3) + 1/6*(2*b^2*n*x^2 - 4*a*c*n*x^2 - a*b*n*x - 2*a^2*log(d))/(a^2*x^3)

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maple [C]  time = 0.71, size = 423, normalized size = 2.84 \[ -\frac {\ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{3 x^{3}}-\frac {6 a b c n \,x^{3} \ln \relax (x )-2 b^{3} n \,x^{3} \ln \relax (x )-2 a^{3} x^{3} \RootOf \left (\textit {\_Z}^{2} a^{3}+c^{3} n^{2}+\left (-3 c b n a +b^{3} n \right ) \textit {\_Z} \right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2} a^{3}+c^{3} n^{2}+\left (-3 c b n a +b^{3} n \right ) \textit {\_Z} \right )^{2} a^{5} b +6 a^{2} b \,c^{3} n^{2}-5 a \,b^{3} c^{2} n^{2}+b^{5} c \,n^{2}+\left (2 a^{4} c^{2} n -4 a^{3} b^{2} c n +a^{2} b^{4} n \right ) \RootOf \left (\textit {\_Z}^{2} a^{3}+c^{3} n^{2}+\left (-3 c b n a +b^{3} n \right ) \textit {\_Z} \right )+\left (4 a^{2} c^{4} n^{2}-4 a \,b^{2} c^{3} n^{2}+b^{4} c^{2} n^{2}+\left (6 a^{5} c -2 a^{4} b^{2}\right ) \RootOf \left (\textit {\_Z}^{2} a^{3}+c^{3} n^{2}+\left (-3 c b n a +b^{3} n \right ) \textit {\_Z} \right )^{2}+\left (-7 a^{3} b \,c^{2} n +2 a^{2} b^{3} c n \right ) \RootOf \left (\textit {\_Z}^{2} a^{3}+c^{3} n^{2}+\left (-3 c b n a +b^{3} n \right ) \textit {\_Z} \right )\right ) x \right )-i \pi \,a^{3} \mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )+i \pi \,a^{3} \mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2}+i \pi \,a^{3} \mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2}-i \pi \,a^{3} \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{3}+4 a^{2} c n \,x^{2}-2 a \,b^{2} n \,x^{2}+a^{2} b n x +2 a^{3} \ln \relax (d )}{6 a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(d*(c*x^2+b*x+a)^n)/x^4,x)

[Out]

-1/3/x^3*ln((c*x^2+b*x+a)^n)-1/6*(-I*Pi*a^3*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
^3*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2+I*Pi*a^3*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2-I*Pi*a^3
*csgn(I*d*(c*x^2+b*x+a)^n)^3+6*ln(x)*a*b*c*n*x^3-2*ln(x)*b^3*n*x^3-2*sum(_R*ln(((6*a^5*c-2*a^4*b^2)*_R^2+(-7*a
^3*b*c^2*n+2*a^2*b^3*c*n)*_R+4*a^2*c^4*n^2-4*a*b^2*c^3*n^2+b^4*c^2*n^2)*x-a^5*b*_R^2+(2*a^4*c^2*n-4*a^3*b^2*c*
n+a^2*b^4*n)*_R+6*a^2*b*c^3*n^2-5*a*b^3*c^2*n^2+b^5*c*n^2),_R=RootOf(a^3*_Z^2+(-3*a*b*c*n+b^3*n)*_Z+c^3*n^2))*
a^3*x^3+4*a^2*c*n*x^2-2*a*b^2*n*x^2+a^2*b*n*x+2*ln(d)*a^3)/a^3/x^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(d*(c*x^2+b*x+a)^n)/x^4,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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.93, size = 505, normalized size = 3.39 \[ \frac {\ln \left (2\,a\,b^4\,\sqrt {b^2-4\,a\,c}-2\,b^6\,x-2\,a\,b^5+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}+13\,a^2\,b^3\,c-20\,a^3\,b\,c^2+4\,a^3\,c^3\,x+2\,a^3\,c^2\,\sqrt {b^2-4\,a\,c}-25\,a^2\,b^2\,c^2\,x+14\,a\,b^4\,c\,x-7\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-10\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+11\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (\frac {b\,c\,n}{2}-\frac {c\,n\,\sqrt {b^2-4\,a\,c}}{6}\right )-\frac {b^3\,n}{6}+\frac {b^2\,n\,\sqrt {b^2-4\,a\,c}}{6}\right )}{a^3}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{3\,x^3}-\frac {\frac {b\,n}{2\,a}+\frac {n\,x\,\left (2\,a\,c-b^2\right )}{a^2}}{3\,x^2}-\frac {\ln \left (2\,a\,b^5+2\,b^6\,x+2\,a\,b^4\,\sqrt {b^2-4\,a\,c}+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}-13\,a^2\,b^3\,c+20\,a^3\,b\,c^2-4\,a^3\,c^3\,x+2\,a^3\,c^2\,\sqrt {b^2-4\,a\,c}+25\,a^2\,b^2\,c^2\,x-14\,a\,b^4\,c\,x-7\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-10\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+11\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3\,n}{6}-a\,\left (\frac {b\,c\,n}{2}+\frac {c\,n\,\sqrt {b^2-4\,a\,c}}{6}\right )+\frac {b^2\,n\,\sqrt {b^2-4\,a\,c}}{6}\right )}{a^3}+\frac {\ln \relax (x)\,\left (b^3\,n-3\,a\,b\,c\,n\right )}{3\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d*(a + b*x + c*x^2)^n)/x^4,x)

[Out]

(log(2*a*b^4*(b^2 - 4*a*c)^(1/2) - 2*b^6*x - 2*a*b^5 + 2*b^5*x*(b^2 - 4*a*c)^(1/2) + 13*a^2*b^3*c - 20*a^3*b*c
^2 + 4*a^3*c^3*x + 2*a^3*c^2*(b^2 - 4*a*c)^(1/2) - 25*a^2*b^2*c^2*x + 14*a*b^4*c*x - 7*a^2*b^2*c*(b^2 - 4*a*c)
^(1/2) - 10*a*b^3*c*x*(b^2 - 4*a*c)^(1/2) + 11*a^2*b*c^2*x*(b^2 - 4*a*c)^(1/2))*(a*((b*c*n)/2 - (c*n*(b^2 - 4*
a*c)^(1/2))/6) - (b^3*n)/6 + (b^2*n*(b^2 - 4*a*c)^(1/2))/6))/a^3 - log(d*(a + b*x + c*x^2)^n)/(3*x^3) - ((b*n)
/(2*a) + (n*x*(2*a*c - b^2))/a^2)/(3*x^2) - (log(2*a*b^5 + 2*b^6*x + 2*a*b^4*(b^2 - 4*a*c)^(1/2) + 2*b^5*x*(b^
2 - 4*a*c)^(1/2) - 13*a^2*b^3*c + 20*a^3*b*c^2 - 4*a^3*c^3*x + 2*a^3*c^2*(b^2 - 4*a*c)^(1/2) + 25*a^2*b^2*c^2*
x - 14*a*b^4*c*x - 7*a^2*b^2*c*(b^2 - 4*a*c)^(1/2) - 10*a*b^3*c*x*(b^2 - 4*a*c)^(1/2) + 11*a^2*b*c^2*x*(b^2 -
4*a*c)^(1/2))*((b^3*n)/6 - a*((b*c*n)/2 + (c*n*(b^2 - 4*a*c)^(1/2))/6) + (b^2*n*(b^2 - 4*a*c)^(1/2))/6))/a^3 +
(log(x)*(b^3*n - 3*a*b*c*n))/(3*a^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(d*(c*x**2+b*x+a)**n)/x**4,x)

[Out]

Timed out

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