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

Optimal. Leaf size=165 \[ \frac{n \sqrt{b^2-4 a c} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac{n (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac{n (2 c d-b e) \log (d+e x)}{e \left (a e^2-b d e+c d^2\right )}-\frac{\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.245178, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2525, 800, 634, 618, 206, 628} \[ \frac{n \sqrt{b^2-4 a c} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac{n (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac{n (2 c d-b e) \log (d+e x)}{e \left (a e^2-b d e+c d^2\right )}-\frac{\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.333308, size = 166, normalized size = 1.01 \[ -\frac{n \sqrt{4 a c-b^2} \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{e (b d-a e)-c d^2}+\frac{n (b e-2 c d) \log (d+e x)}{e \left (e (a e-b d)+c d^2\right )}-\frac{n (b e-2 c d) \log (a+x (b+c x))}{2 e \left (e (a e-b d)+c d^2\right )}-\frac{\log \left (d (a+x (b+c x))^n\right )}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.148, size = 6540, normalized size = 39.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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)/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.22275, size = 950, normalized size = 5.76 \begin{align*} \left [\frac{{\left (e^{2} n x + d e n\right )} \sqrt{b^{2} - 4 \, a c} \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 ({\left (2 \, c d e - b e^{2}\right )} n x +{\left (b d e - 2 \, a e^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \,{\left ({\left (2 \, c d e - b e^{2}\right )} n x +{\left (2 \, c d^{2} - b d e\right )} n\right )} \log \left (e x + d\right ) - 2 \,{\left (c d^{2} - b d e + a e^{2}\right )} \log \left (d\right )}{2 \,{\left (c d^{3} e - b d^{2} e^{2} + a d e^{3} +{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x\right )}}, \frac{2 \,{\left (e^{2} n x + d e n\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) +{\left ({\left (2 \, c d e - b e^{2}\right )} n x +{\left (b d e - 2 \, a e^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \,{\left ({\left (2 \, c d e - b e^{2}\right )} n x +{\left (2 \, c d^{2} - b d e\right )} n\right )} \log \left (e x + d\right ) - 2 \,{\left (c d^{2} - b d e + a e^{2}\right )} \log \left (d\right )}{2 \,{\left (c d^{3} e - b d^{2} e^{2} + a d e^{3} +{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(e*x+d)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.39488, size = 383, normalized size = 2.32 \begin{align*} \frac{{\left (2 \, c d n - b n e\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac{{\left (b^{2} n - 4 \, a c n\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{2 \, c d n x e \log \left (x e + d\right ) + c d^{2} n \log \left (c x^{2} + b x + a\right ) - b d n e \log \left (c x^{2} + b x + a\right ) + 2 \, c d^{2} n \log \left (x e + d\right ) - b n x e^{2} \log \left (x e + d\right ) - b d n e \log \left (x e + d\right ) + a n e^{2} \log \left (c x^{2} + b x + a\right ) + c d^{2} \log \left (d\right ) - b d e \log \left (d\right ) + a e^{2} \log \left (d\right )}{c d^{2} x e^{2} + c d^{3} e - b d x e^{3} - b d^{2} e^{2} + a x e^{4} + a d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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