3.71 \(\int x^4 \log (d (a+b x+c x^2)^n) \, dx\)

Optimal. Leaf size=207 \[ \frac {b n \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{10 c^5}+\frac {n \sqrt {b^2-4 a c} \left (a^2 c^2-3 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{5 c^5}-\frac {n x \left (2 a^2 c^2-4 a b^2 c+b^4\right )}{5 c^4}+\frac {b n x^2 \left (b^2-3 a c\right )}{10 c^3}-\frac {n x^3 \left (b^2-2 a c\right )}{15 c^2}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25} \]

[Out]

-1/5*(2*a^2*c^2-4*a*b^2*c+b^4)*n*x/c^4+1/10*b*(-3*a*c+b^2)*n*x^2/c^3-1/15*(-2*a*c+b^2)*n*x^3/c^2+1/20*b*n*x^4/
c-2/25*n*x^5+1/10*b*(5*a^2*c^2-5*a*b^2*c+b^4)*n*ln(c*x^2+b*x+a)/c^5+1/5*x^5*ln(d*(c*x^2+b*x+a)^n)+1/5*(a^2*c^2
-3*a*b^2*c+b^4)*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/c^5

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Rubi [A]  time = 0.23, antiderivative size = 207, 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} \[ \frac {b n \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{10 c^5}-\frac {n x \left (2 a^2 c^2-4 a b^2 c+b^4\right )}{5 c^4}+\frac {n \sqrt {b^2-4 a c} \left (a^2 c^2-3 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{5 c^5}-\frac {n x^3 \left (b^2-2 a c\right )}{15 c^2}+\frac {b n x^2 \left (b^2-3 a c\right )}{10 c^3}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

-((b^4 - 4*a*b^2*c + 2*a^2*c^2)*n*x)/(5*c^4) + (b*(b^2 - 3*a*c)*n*x^2)/(10*c^3) - ((b^2 - 2*a*c)*n*x^3)/(15*c^
2) + (b*n*x^4)/(20*c) - (2*n*x^5)/25 + (Sqrt[b^2 - 4*a*c]*(b^4 - 3*a*b^2*c + a^2*c^2)*n*ArcTanh[(b + 2*c*x)/Sq
rt[b^2 - 4*a*c]])/(5*c^5) + (b*(b^4 - 5*a*b^2*c + 5*a^2*c^2)*n*Log[a + b*x + c*x^2])/(10*c^5) + (x^5*Log[d*(a
+ b*x + c*x^2)^n])/5

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 x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx &=\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {1}{5} n \int \frac {x^5 (b+2 c x)}{a+b x+c x^2} \, dx\\ &=\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {1}{5} n \int \left (\frac {b^4-4 a b^2 c+2 a^2 c^2}{c^4}-\frac {b \left (b^2-3 a c\right ) x}{c^3}+\frac {\left (b^2-2 a c\right ) x^2}{c^2}-\frac {b x^3}{c}+2 x^4-\frac {a \left (b^4-4 a b^2 c+2 a^2 c^2\right )+b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) x}{c^4 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n x}{5 c^4}+\frac {b \left (b^2-3 a c\right ) n x^2}{10 c^3}-\frac {\left (b^2-2 a c\right ) n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {n \int \frac {a \left (b^4-4 a b^2 c+2 a^2 c^2\right )+b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) x}{a+b x+c x^2} \, dx}{5 c^4}\\ &=-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n x}{5 c^4}+\frac {b \left (b^2-3 a c\right ) n x^2}{10 c^3}-\frac {\left (b^2-2 a c\right ) n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {\left (\left (b^2-4 a c\right ) \left (b^4-3 a b^2 c+a^2 c^2\right ) n\right ) \int \frac {1}{a+b x+c x^2} \, dx}{10 c^5}+\frac {\left (b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) n\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{10 c^5}\\ &=-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n x}{5 c^4}+\frac {b \left (b^2-3 a c\right ) n x^2}{10 c^3}-\frac {\left (b^2-2 a c\right ) n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) n \log \left (a+b x+c x^2\right )}{10 c^5}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {\left (\left (b^2-4 a c\right ) \left (b^4-3 a b^2 c+a^2 c^2\right ) n\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{5 c^5}\\ &=-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n x}{5 c^4}+\frac {b \left (b^2-3 a c\right ) n x^2}{10 c^3}-\frac {\left (b^2-2 a c\right ) n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {\sqrt {b^2-4 a c} \left (b^4-3 a b^2 c+a^2 c^2\right ) n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{5 c^5}+\frac {b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) n \log \left (a+b x+c x^2\right )}{10 c^5}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 190, normalized size = 0.92 \[ \frac {30 b n \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \log (a+x (b+c x))+60 n \sqrt {b^2-4 a c} \left (a^2 c^2-3 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+c n x \left (-8 c^2 \left (15 a^2-5 a c x^2+3 c^2 x^4\right )-20 b^2 c \left (c x^2-12 a\right )+15 b c^2 x \left (c x^2-6 a\right )-60 b^4+30 b^3 c x\right )+60 c^5 x^5 \log \left (d (a+x (b+c x))^n\right )}{300 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*n*x*(-60*b^4 + 30*b^3*c*x - 20*b^2*c*(-12*a + c*x^2) + 15*b*c^2*x*(-6*a + c*x^2) - 8*c^2*(15*a^2 - 5*a*c*x^
2 + 3*c^2*x^4)) + 60*Sqrt[b^2 - 4*a*c]*(b^4 - 3*a*b^2*c + a^2*c^2)*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]] +
30*b*(b^4 - 5*a*b^2*c + 5*a^2*c^2)*n*Log[a + x*(b + c*x)] + 60*c^5*x^5*Log[d*(a + x*(b + c*x))^n])/(300*c^5)

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fricas [A]  time = 0.47, size = 444, normalized size = 2.14 \[ \left [-\frac {24 \, c^{5} n x^{5} - 60 \, c^{5} x^{5} \log \relax (d) - 15 \, b c^{4} n x^{4} + 20 \, {\left (b^{2} c^{3} - 2 \, a c^{4}\right )} n x^{3} - 30 \, {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} n x^{2} - 30 \, {\left (b^{4} - 3 \, a b^{2} c + a^{2} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} n \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 60 \, {\left (b^{4} c - 4 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} n x - 30 \, {\left (2 \, c^{5} n x^{5} + {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{300 \, c^{5}}, -\frac {24 \, c^{5} n x^{5} - 60 \, c^{5} x^{5} \log \relax (d) - 15 \, b c^{4} n x^{4} + 20 \, {\left (b^{2} c^{3} - 2 \, a c^{4}\right )} n x^{3} - 30 \, {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} n x^{2} - 60 \, {\left (b^{4} - 3 \, a b^{2} c + a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} n \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 60 \, {\left (b^{4} c - 4 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} n x - 30 \, {\left (2 \, c^{5} n x^{5} + {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{300 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/300*(24*c^5*n*x^5 - 60*c^5*x^5*log(d) - 15*b*c^4*n*x^4 + 20*(b^2*c^3 - 2*a*c^4)*n*x^3 - 30*(b^3*c^2 - 3*a*
b*c^3)*n*x^2 - 30*(b^4 - 3*a*b^2*c + a^2*c^2)*sqrt(b^2 - 4*a*c)*n*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqr
t(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 60*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*n*x - 30*(2*c^5*n*x^5 +
(b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*n)*log(c*x^2 + b*x + a))/c^5, -1/300*(24*c^5*n*x^5 - 60*c^5*x^5*log(d) - 15*b*
c^4*n*x^4 + 20*(b^2*c^3 - 2*a*c^4)*n*x^3 - 30*(b^3*c^2 - 3*a*b*c^3)*n*x^2 - 60*(b^4 - 3*a*b^2*c + a^2*c^2)*sqr
t(-b^2 + 4*a*c)*n*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 60*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)
*n*x - 30*(2*c^5*n*x^5 + (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*n)*log(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.20, size = 221, normalized size = 1.07 \[ \frac {1}{5} \, n x^{5} \log \left (c x^{2} + b x + a\right ) - \frac {1}{25} \, {\left (2 \, n - 5 \, \log \relax (d)\right )} x^{5} + \frac {b n x^{4}}{20 \, c} - \frac {{\left (b^{2} n - 2 \, a c n\right )} x^{3}}{15 \, c^{2}} + \frac {{\left (b^{3} n - 3 \, a b c n\right )} x^{2}}{10 \, c^{3}} - \frac {{\left (b^{4} n - 4 \, a b^{2} c n + 2 \, a^{2} c^{2} n\right )} x}{5 \, c^{4}} + \frac {{\left (b^{5} n - 5 \, a b^{3} c n + 5 \, a^{2} b c^{2} n\right )} \log \left (c x^{2} + b x + a\right )}{10 \, c^{5}} - \frac {{\left (b^{6} n - 7 \, a b^{4} c n + 13 \, a^{2} b^{2} c^{2} n - 4 \, a^{3} c^{3} n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{5 \, \sqrt {-b^{2} + 4 \, a c} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*n*x^5*log(c*x^2 + b*x + a) - 1/25*(2*n - 5*log(d))*x^5 + 1/20*b*n*x^4/c - 1/15*(b^2*n - 2*a*c*n)*x^3/c^2 +
1/10*(b^3*n - 3*a*b*c*n)*x^2/c^3 - 1/5*(b^4*n - 4*a*b^2*c*n + 2*a^2*c^2*n)*x/c^4 + 1/10*(b^5*n - 5*a*b^3*c*n
+ 5*a^2*b*c^2*n)*log(c*x^2 + b*x + a)/c^5 - 1/5*(b^6*n - 7*a*b^4*c*n + 13*a^2*b^2*c^2*n - 4*a^3*c^3*n)*arctan(
(2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^5)

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maple [C]  time = 0.81, size = 1621, normalized size = 7.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*ln((c*x^2+b*x+a)^n)+1/10*I*Pi*x^5*csgn(I*d)*csgn(I*d*(c*x^2+b*x+a)^n)^2+1/10*I*Pi*x^5*csgn(I*(c*x^2+b*
x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)^2-1/10*I*Pi*x^5*csgn(I*d*(c*x^2+b*x+a)^n)^3-1/10*I*Pi*x^5*csgn(I*d)*csgn(I*(
c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+1/5*ln(d)*x^5-2/25*n*x^5+1/20*b*n*x^4/c+2/15/c*a*n*x^3-1/15*b^2*n*x^
3/c^2-3/10/c^2*a*b*n*x^2+1/10*b^3*n*x^2/c^3+1/2/c^3*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*c-b^6+2*(-4*a^5*c^5+
25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x+(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*
c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*a^2*b-1/2/c^4*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*c-b^6+2*(-4*a
^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x+(-4*a^5*c^5+25*a^4*b^2*c^4-50*a
^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*a*b^3+1/10/c^5*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*c-b^6
+2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x+(-4*a^5*c^5+25*a^4*b^2*
c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*b^5+1/2/c^3*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*
c-b^6-2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x-(-4*a^5*c^5+25*a^4
*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*a^2*b-1/2/c^4*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7
*a*b^4*c-b^6-2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x-(-4*a^5*c^5
+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*a*b^3+1/10/c^5*n*ln(4*a^3*c^3-13*a^2*b
^2*c^2+7*a*b^4*c-b^6-2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x-(-4
*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*b^5-2/5/c^2*a^2*n*x+4/5/c^3*a*
b^2*n*x-1/5*b^4*n*x/c^4-1/10/c^5*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*c-b^6+2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a
^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*c*x+(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-
10*a*b^8*c+b^10)^(1/2)*b)*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)+1/10
/c^5*n*ln(4*a^3*c^3-13*a^2*b^2*c^2+7*a*b^4*c-b^6-2*(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10
*a*b^8*c+b^10)^(1/2)*c*x-(-4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)*b)*(-
4*a^5*c^5+25*a^4*b^2*c^4-50*a^3*b^4*c^3+35*a^2*b^6*c^2-10*a*b^8*c+b^10)^(1/2)

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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(x^4*log(d*(c*x^2+b*x+a)^n),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 = 0.59, size = 395, normalized size = 1.91 \[ x^2\,\left (\frac {b\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{2\,c}-\frac {a\,b\,n}{10\,c^2}\right )-\frac {2\,n\,x^5}{25}+x\,\left (\frac {a\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{c}-\frac {b\,\left (\frac {b\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{c}-\frac {a\,b\,n}{5\,c^2}\right )}{c}\right )+\frac {x^5\,\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{5}-x^3\,\left (\frac {b^2\,n}{15\,c^2}-\frac {2\,a\,n}{15\,c}\right )+\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^5\,n}{10}+c^2\,\left (\frac {a^2\,n\,\sqrt {b^2-4\,a\,c}}{10}+\frac {a^2\,b\,n}{2}\right )-c\,\left (\frac {a\,b^3\,n}{2}+\frac {3\,a\,b^2\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )+\frac {b^4\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )}{c^5}-\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (c^2\,\left (\frac {a^2\,n\,\sqrt {b^2-4\,a\,c}}{10}-\frac {a^2\,b\,n}{2}\right )-\frac {b^5\,n}{10}+c\,\left (\frac {a\,b^3\,n}{2}-\frac {3\,a\,b^2\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )+\frac {b^4\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )}{c^5}+\frac {b\,n\,x^4}{20\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(d*(a + b*x + c*x^2)^n),x)

[Out]

x^2*((b*((b^2*n)/(5*c^2) - (2*a*n)/(5*c)))/(2*c) - (a*b*n)/(10*c^2)) - (2*n*x^5)/25 + x*((a*((b^2*n)/(5*c^2) -
(2*a*n)/(5*c)))/c - (b*((b*((b^2*n)/(5*c^2) - (2*a*n)/(5*c)))/c - (a*b*n)/(5*c^2)))/c) + (x^5*log(d*(a + b*x
+ c*x^2)^n))/5 - x^3*((b^2*n)/(15*c^2) - (2*a*n)/(15*c)) + (log(b*(b^2 - 4*a*c)^(1/2) - 4*a*c + b^2 + 2*c*x*(b
^2 - 4*a*c)^(1/2))*((b^5*n)/10 + c^2*((a^2*n*(b^2 - 4*a*c)^(1/2))/10 + (a^2*b*n)/2) - c*((a*b^3*n)/2 + (3*a*b^
2*n*(b^2 - 4*a*c)^(1/2))/10) + (b^4*n*(b^2 - 4*a*c)^(1/2))/10))/c^5 - (log(4*a*c + b*(b^2 - 4*a*c)^(1/2) - b^2
+ 2*c*x*(b^2 - 4*a*c)^(1/2))*(c^2*((a^2*n*(b^2 - 4*a*c)^(1/2))/10 - (a^2*b*n)/2) - (b^5*n)/10 + c*((a*b^3*n)/
2 - (3*a*b^2*n*(b^2 - 4*a*c)^(1/2))/10) + (b^4*n*(b^2 - 4*a*c)^(1/2))/10))/c^5 + (b*n*x^4)/(20*c)

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

[Out]

Timed out

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