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

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

[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)

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Rubi [A]  time = 0.199732, antiderivative size = 149, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.366032, 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]

-((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])/(6*x^3)

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Maple [C]  time = 0.085, size = 423, normalized size = 2.8 \begin{align*} -{\frac{\ln \left ( \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) }{3\,{x}^{3}}}-{\frac{-i\pi \,{a}^{3}{\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 ) +i\pi \,{a}^{3}{\it csgn} \left ( id \right ) \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{2}+i\pi \,{a}^{3}{\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}-i\pi \,{a}^{3} \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{3}+6\,\ln \left ( x \right ) abcn{x}^{3}-2\,\ln \left ( x \right ){b}^{3}n{x}^{3}-2\,\sum _{{\it \_R}={\it RootOf} \left ({a}^{3}{{\it \_Z}}^{2}+ \left ( -3\,cbna+{b}^{3}n \right ){\it \_Z}+{c}^{3}{n}^{2} \right ) }{\it \_R}\,\ln \left ( \left ( \left ( 6\,{a}^{5}c-2\,{a}^{4}{b}^{2} \right ){{\it \_R}}^{2}+ \left ( -7\,{a}^{3}b{c}^{2}n+2\,{a}^{2}{b}^{3}cn \right ){\it \_R}+4\,{a}^{2}{c}^{4}{n}^{2}-4\,a{b}^{2}{c}^{3}{n}^{2}+{b}^{4}{c}^{2}{n}^{2} \right ) x-{a}^{5}b{{\it \_R}}^{2}+ \left ( 2\,{a}^{4}{c}^{2}n-4\,{a}^{3}{b}^{2}cn+{a}^{2}{b}^{4}n \right ){\it \_R}+6\,{a}^{2}b{c}^{3}{n}^{2}-5\,a{b}^{3}{c}^{2}{n}^{2}+{b}^{5}c{n}^{2} \right ){a}^{3}{x}^{3}+4\,{a}^{2}cn{x}^{2}-2\,a{b}^{2}n{x}^{2}+{a}^{2}bnx+2\,\ln \left ( d \right ){a}^{3}}{6\,{a}^{3}{x}^{3}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.74687, size = 741, normalized size = 4.97 \begin{align*} \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 \left (x\right ) + a^{2} b n x - 2 \,{\left (a b^{2} - 2 \, a^{2} c\right )} n x^{2} + 2 \, a^{3} \log \left (d\right ) +{\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 \left (x\right ) - a^{2} b n x + 2 \,{\left (a b^{2} - 2 \, a^{2} c\right )} n x^{2} - 2 \, a^{3} \log \left (d\right ) -{\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 ] \end{align*}

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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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)/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.20834, size = 221, normalized size = 1.48 \begin{align*} -\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 \left (x\right )}{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 \left (d\right )}{6 \, a^{2} x^{3}} \end{align*}

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)