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

Optimal. Leaf size=109 \[ -\frac {n \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{4 c^2}-\frac {b n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 c^2}+\frac {1}{2} x^2 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {b n x}{2 c}-\frac {n x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2525, 800, 634, 618, 206, 628} \[ -\frac {n \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{4 c^2}-\frac {b n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{2 c^2}+\frac {1}{2} x^2 \log \left (d \left (a+b x+c x^2\right )^n\right )+\frac {b n x}{2 c}-\frac {n x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 94, normalized size = 0.86 \[ -\frac {n \left (b^2-2 a c\right ) \log (a+x (b+c x))+2 b n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )-2 c x \left (c x \log \left (d (a+x (b+c x))^n\right )+n (b-c x)\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 245, normalized size = 2.25 \[ \left [-\frac {2 \, c^{2} n x^{2} - 2 \, c^{2} x^{2} \log \relax (d) - 2 \, b c n x - \sqrt {b^{2} - 4 \, a c} b n \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - {\left (2 \, c^{2} n x^{2} - {\left (b^{2} - 2 \, a c\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, c^{2}}, -\frac {2 \, c^{2} n x^{2} - 2 \, c^{2} x^{2} \log \relax (d) - 2 \, b c n x + 2 \, \sqrt {-b^{2} + 4 \, a c} b n \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left (2 \, c^{2} n x^{2} - {\left (b^{2} - 2 \, a c\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 113, normalized size = 1.04 \[ \frac {1}{2} \, n x^{2} \log \left (c x^{2} + b x + a\right ) - \frac {1}{2} \, {\left (n - \log \relax (d)\right )} x^{2} + \frac {b n x}{2 \, c} - \frac {{\left (b^{2} n - 2 \, a c n\right )} \log \left (c x^{2} + b x + a\right )}{4 \, c^{2}} + \frac {{\left (b^{3} n - 4 \, a b c n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.68, size = 510, normalized size = 4.68 \[ -\frac {i \pi \,x^{2} \mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i d \right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2}}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{2}}{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )^{3}}{4}-\frac {n \,x^{2}}{2}+\frac {x^{2} \ln \relax (d )}{2}+\frac {x^{2} \ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{2}+\frac {a n \ln \left (-4 a b c +b^{3}-2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x -\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{2 c}+\frac {a n \ln \left (-4 a b c +b^{3}+2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x +\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{2 c}-\frac {b^{2} n \ln \left (-4 a b c +b^{3}-2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x -\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{4 c^{2}}-\frac {b^{2} n \ln \left (-4 a b c +b^{3}+2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x +\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{4 c^{2}}+\frac {b n x}{2 c}+\frac {\sqrt {-4 a \,b^{2} c +b^{4}}\, n \ln \left (-4 a b c +b^{3}-2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x -\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{4 c^{2}}-\frac {\sqrt {-4 a \,b^{2} c +b^{4}}\, n \ln \left (-4 a b c +b^{3}+2 \sqrt {-4 a \,b^{2} c +b^{4}}\, c x +\sqrt {-4 a \,b^{2} c +b^{4}}\, b \right )}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*ln((c*x^2+b*x+a)^n)-1/4*I*Pi*x^2*csgn(I*d)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)+1/4*I*csg
n(I*d*(c*x^2+b*x+a)^n)^2*csgn(I*d)*x^2*Pi+1/4*I*csgn(I*d*(c*x^2+b*x+a)^n)^2*csgn(I*(c*x^2+b*x+a)^n)*x^2*Pi-1/4
*I*Pi*x^2*csgn(I*d*(c*x^2+b*x+a)^n)^3+1/2*ln(d)*x^2-1/2*n*x^2+1/2/c*n*ln(-2*(-4*a*b^2*c+b^4)^(1/2)*c*x-4*a*b*c
+b^3-(-4*a*b^2*c+b^4)^(1/2)*b)*a-1/4/c^2*n*ln(-2*(-4*a*b^2*c+b^4)^(1/2)*c*x-4*a*b*c+b^3-(-4*a*b^2*c+b^4)^(1/2)
*b)*b^2+1/2/c*n*ln(2*(-4*a*b^2*c+b^4)^(1/2)*c*x-4*a*b*c+b^3+(-4*a*b^2*c+b^4)^(1/2)*b)*a-1/4/c^2*n*ln(2*(-4*a*b
^2*c+b^4)^(1/2)*c*x-4*a*b*c+b^3+(-4*a*b^2*c+b^4)^(1/2)*b)*b^2+1/2*b*n*x/c+1/4/c^2*n*ln(-2*(-4*a*b^2*c+b^4)^(1/
2)*c*x-4*a*b*c+b^3-(-4*a*b^2*c+b^4)^(1/2)*b)*(-4*a*b^2*c+b^4)^(1/2)-1/4/c^2*n*ln(2*(-4*a*b^2*c+b^4)^(1/2)*c*x-
4*a*b*c+b^3+(-4*a*b^2*c+b^4)^(1/2)*b)*(-4*a*b^2*c+b^4)^(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*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 = 0.55, size = 166, normalized size = 1.52 \[ \frac {x^2\,\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{2}-\frac {n\,x^2}{2}-\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,n-2\,a\,c\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,c^2}+\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (2\,a\,c\,n-b^2\,n+b\,n\,\sqrt {b^2-4\,a\,c}\right )}{4\,c^2}+\frac {b\,n\,x}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(d*(a + b*x + c*x^2)^n),x)

[Out]

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

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sympy [A]  time = 165.46, size = 369, normalized size = 3.39 \[ \begin {cases} - \frac {b^{2} n \log {\left (\frac {b^{2}}{4 c} + b x + c x^{2} \right )}}{8 c^{2}} + \frac {b n x}{2 c} + \frac {n x^{2} \log {\left (\frac {b^{2}}{4 c} + b x + c x^{2} \right )}}{2} - \frac {n x^{2}}{2} + \frac {x^{2} \log {\relax (d )}}{2} & \text {for}\: a = \frac {b^{2}}{4 c} \\- \frac {a^{2} n \log {\left (a + b x \right )}}{2 b^{2}} + \frac {a n x}{2 b} + \frac {n x^{2} \log {\left (a + b x \right )}}{2} - \frac {n x^{2}}{4} + \frac {x^{2} \log {\relax (d )}}{2} & \text {for}\: c = 0 \\- \frac {a b n \log {\left (a + b x + c x^{2} \right )}}{c \sqrt {- 4 a c + b^{2}}} + \frac {2 a b n \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{c \sqrt {- 4 a c + b^{2}}} + \frac {a n \log {\left (a + b x + c x^{2} \right )}}{2 c} + \frac {b^{3} n \log {\left (a + b x + c x^{2} \right )}}{4 c^{2} \sqrt {- 4 a c + b^{2}}} - \frac {b^{3} n \log {\left (\frac {b}{2 c} + x + \frac {\sqrt {- 4 a c + b^{2}}}{2 c} \right )}}{2 c^{2} \sqrt {- 4 a c + b^{2}}} - \frac {b^{2} n \log {\left (a + b x + c x^{2} \right )}}{4 c^{2}} + \frac {b n x}{2 c} + \frac {n x^{2} \log {\left (a + b x + c x^{2} \right )}}{2} - \frac {n x^{2}}{2} + \frac {x^{2} \log {\relax (d )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-b**2*n*log(b**2/(4*c) + b*x + c*x**2)/(8*c**2) + b*n*x/(2*c) + n*x**2*log(b**2/(4*c) + b*x + c*x**
2)/2 - n*x**2/2 + x**2*log(d)/2, Eq(a, b**2/(4*c))), (-a**2*n*log(a + b*x)/(2*b**2) + a*n*x/(2*b) + n*x**2*log
(a + b*x)/2 - n*x**2/4 + x**2*log(d)/2, Eq(c, 0)), (-a*b*n*log(a + b*x + c*x**2)/(c*sqrt(-4*a*c + b**2)) + 2*a
*b*n*log(b/(2*c) + x + sqrt(-4*a*c + b**2)/(2*c))/(c*sqrt(-4*a*c + b**2)) + a*n*log(a + b*x + c*x**2)/(2*c) +
b**3*n*log(a + b*x + c*x**2)/(4*c**2*sqrt(-4*a*c + b**2)) - b**3*n*log(b/(2*c) + x + sqrt(-4*a*c + b**2)/(2*c)
)/(2*c**2*sqrt(-4*a*c + b**2)) - b**2*n*log(a + b*x + c*x**2)/(4*c**2) + b*n*x/(2*c) + n*x**2*log(a + b*x + c*
x**2)/2 - n*x**2/2 + x**2*log(d)/2, True))

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