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]

(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

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Rubi [A]  time = 0.112237, antiderivative size = 109, normalized size of antiderivative = 1., 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 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 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*}

Mathematica [A]  time = 0.0870621, 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]

-(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]))/(4*c^2)

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Maple [C]  time = 0.093, size = 510, normalized size = 4.7 \begin{align*}{\frac{{x}^{2}\ln \left ( \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) }{2}}+{\frac{i}{4}} \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{2}{\it csgn} \left ( i \left ( c{x}^{2}+bx+a \right ) ^{n} \right ){x}^{2}\pi -{\frac{i}{4}}\pi \,{x}^{2}{\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 ) -{\frac{i}{4}}\pi \,{x}^{2} \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{3}+{\frac{i}{4}} \left ({\it csgn} \left ( id \left ( c{x}^{2}+bx+a \right ) ^{n} \right ) \right ) ^{2}{\it csgn} \left ( id \right ){x}^{2}\pi +{\frac{\ln \left ( d \right ){x}^{2}}{2}}-{\frac{n{x}^{2}}{2}}+{\frac{an}{2\,c}\ln \left ( 2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) }-{\frac{{b}^{2}n}{4\,{c}^{2}}\ln \left ( 2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) }+{\frac{an}{2\,c}\ln \left ( -2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}-\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) }-{\frac{{b}^{2}n}{4\,{c}^{2}}\ln \left ( -2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}-\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) }+{\frac{bnx}{2\,c}}-{\frac{n}{4\,{c}^{2}}\ln \left ( 2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}}+{\frac{n}{4\,{c}^{2}}\ln \left ( -2\,\sqrt{-4\,a{b}^{2}c+{b}^{4}}cx-4\,abc+{b}^{3}-\sqrt{-4\,a{b}^{2}c+{b}^{4}}b \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}} \end{align*}

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*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)*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*d*(c*x^2+b*x+a)^n)-1/4*I*Pi*x^2*csgn(I*d*(c*x^2+b*x+a)^n)^3+1/4*I*csgn(I*
d*(c*x^2+b*x+a)^n)^2*csgn(I*d)*x^2*Pi+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., 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(x*log(d*(c*x^2+b*x+a)^n),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94003, size = 567, normalized size = 5.2 \begin{align*} \left [-\frac{2 \, c^{2} n x^{2} - 2 \, c^{2} x^{2} \log \left (d\right ) - 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 \left (d\right ) - 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 ] \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.33988, size = 153, normalized size = 1.4 \begin{align*} \frac{1}{2} \, n x^{2} \log \left (c x^{2} + b x + a\right ) - \frac{1}{2} \,{\left (n - \log \left (d\right )\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}} \end{align*}

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)