3.83 \(\int (d+e x)^3 \log (d (a+b x+c x^2)^n) \, dx\)

Optimal. Leaf size=338 \[ -\frac{n \left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \log \left (a+b x+c x^2\right )}{8 c^4 e}-\frac{e n x^2 \left (-2 c e (a e+2 b d)+b^2 e^2+12 c^2 d^2\right )}{8 c^2}-\frac{n x \left (-2 c^2 d e (4 a e+3 b d)+b c e^2 (3 a e+4 b d)-b^3 e^3+8 c^3 d^3\right )}{4 c^3}+\frac{n \sqrt{b^2-4 a c} (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{4 c^4}+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{e^2 n x^3 (8 c d-b e)}{12 c}-\frac{1}{8} e^3 n x^4 \]

[Out]

-((8*c^3*d^3 - b^3*e^3 + b*c*e^2*(4*b*d + 3*a*e) - 2*c^2*d*e*(3*b*d + 4*a*e))*n*x)/(4*c^3) - (e*(12*c^2*d^2 +
b^2*e^2 - 2*c*e*(2*b*d + a*e))*n*x^2)/(8*c^2) - (e^2*(8*c*d - b*e)*n*x^3)/(12*c) - (e^3*n*x^4)/8 + (Sqrt[b^2 -
4*a*c]*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(4*c
^4) - ((2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b*d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a
*b*d*e + a^2*e^2))*n*Log[a + b*x + c*x^2])/(8*c^4*e) + ((d + e*x)^4*Log[d*(a + b*x + c*x^2)^n])/(4*e)

________________________________________________________________________________________

Rubi [A]  time = 0.515817, antiderivative size = 338, 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} \[ -\frac{n \left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \log \left (a+b x+c x^2\right )}{8 c^4 e}-\frac{e n x^2 \left (-2 c e (a e+2 b d)+b^2 e^2+12 c^2 d^2\right )}{8 c^2}-\frac{n x \left (-2 c^2 d e (4 a e+3 b d)+b c e^2 (3 a e+4 b d)-b^3 e^3+8 c^3 d^3\right )}{4 c^3}+\frac{n \sqrt{b^2-4 a c} (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{4 c^4}+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{e^2 n x^3 (8 c d-b e)}{12 c}-\frac{1}{8} e^3 n x^4 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

-((8*c^3*d^3 - b^3*e^3 + b*c*e^2*(4*b*d + 3*a*e) - 2*c^2*d*e*(3*b*d + 4*a*e))*n*x)/(4*c^3) - (e*(12*c^2*d^2 +
b^2*e^2 - 2*c*e*(2*b*d + a*e))*n*x^2)/(8*c^2) - (e^2*(8*c*d - b*e)*n*x^3)/(12*c) - (e^3*n*x^4)/8 + (Sqrt[b^2 -
4*a*c]*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(4*c
^4) - ((2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b*d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a
*b*d*e + a^2*e^2))*n*Log[a + b*x + c*x^2])/(8*c^4*e) + ((d + e*x)^4*Log[d*(a + b*x + c*x^2)^n])/(4*e)

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 (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx &=\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{n \int \frac{(b+2 c x) (d+e x)^4}{a+b x+c x^2} \, dx}{4 e}\\ &=\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{n \int \left (\frac{e \left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right )}{c^3}+\frac{e^2 \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) x}{c^2}+\frac{e^3 (8 c d-b e) x^2}{c}+2 e^4 x^3+\frac{-4 a b^2 c d e^3+a b^3 e^4-8 a c^2 d e \left (c d^2-a e^2\right )+b c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) x}{c^3 \left (a+b x+c x^2\right )}\right ) \, dx}{4 e}\\ &=-\frac{\left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) n x}{4 c^3}-\frac{e \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) n x^2}{8 c^2}-\frac{e^2 (8 c d-b e) n x^3}{12 c}-\frac{1}{8} e^3 n x^4+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{n \int \frac{-4 a b^2 c d e^3+a b^3 e^4-8 a c^2 d e \left (c d^2-a e^2\right )+b c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) x}{a+b x+c x^2} \, dx}{4 c^3 e}\\ &=-\frac{\left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) n x}{4 c^3}-\frac{e \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) n x^2}{8 c^2}-\frac{e^2 (8 c d-b e) n x^3}{12 c}-\frac{1}{8} e^3 n x^4+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac{\left (\left (b^2-4 a c\right ) (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n\right ) \int \frac{1}{a+b x+c x^2} \, dx}{8 c^4}-\frac{\left (\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) n\right ) \int \frac{b+2 c x}{a+b x+c x^2} \, dx}{8 c^4 e}\\ &=-\frac{\left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) n x}{4 c^3}-\frac{e \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) n x^2}{8 c^2}-\frac{e^2 (8 c d-b e) n x^3}{12 c}-\frac{1}{8} e^3 n x^4-\frac{\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) n \log \left (a+b x+c x^2\right )}{8 c^4 e}+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}+\frac{\left (\left (b^2-4 a c\right ) (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{4 c^4}\\ &=-\frac{\left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) n x}{4 c^3}-\frac{e \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) n x^2}{8 c^2}-\frac{e^2 (8 c d-b e) n x^3}{12 c}-\frac{1}{8} e^3 n x^4+\frac{\sqrt{b^2-4 a c} (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{4 c^4}-\frac{\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) n \log \left (a+b x+c x^2\right )}{8 c^4 e}+\frac{(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}\\ \end{align*}

Mathematica [A]  time = 0.504436, size = 324, normalized size = 0.96 \[ \frac{(d+e x)^4 \log \left (d (a+x (b+c x))^n\right )-\frac{n \left (3 \left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \log (a+x (b+c x))+3 c^2 e^2 x^2 \left (-2 c e (a e+2 b d)+b^2 e^2+12 c^2 d^2\right )+6 c e x \left (-2 c^2 d e (4 a e+3 b d)+b c e^2 (3 a e+4 b d)-b^3 e^3+8 c^3 d^3\right )-6 e \sqrt{b^2-4 a c} (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )+2 c^3 e^3 x^3 (8 c d-b e)+3 c^4 e^4 x^4\right )}{6 c^4}}{4 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*Log[d*(a + b*x + c*x^2)^n],x]

[Out]

(-(n*(6*c*e*(8*c^3*d^3 - b^3*e^3 + b*c*e^2*(4*b*d + 3*a*e) - 2*c^2*d*e*(3*b*d + 4*a*e))*x + 3*c^2*e^2*(12*c^2*
d^2 + b^2*e^2 - 2*c*e*(2*b*d + a*e))*x^2 + 2*c^3*e^3*(8*c*d - b*e)*x^3 + 3*c^4*e^4*x^4 - 6*Sqrt[b^2 - 4*a*c]*e
*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]] + 3*(2*c^4*d^4
+ b^4*e^4 - 4*b^2*c*e^3*(b*d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^2)
)*Log[a + x*(b + c*x)]))/(6*c^4) + (d + e*x)^4*Log[d*(a + x*(b + c*x))^n])/(4*e)

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Maple [C]  time = 0.196, size = 16059, normalized size = 47.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*ln(d*(c*x^2+b*x+a)^n),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((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.4354, size = 1858, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x, algorithm="fricas")

[Out]

[-1/24*(3*c^4*e^3*n*x^4 + 2*(8*c^4*d*e^2 - b*c^3*e^3)*n*x^3 + 3*(12*c^4*d^2*e - 4*b*c^3*d*e^2 + (b^2*c^2 - 2*a
*c^3)*e^3)*n*x^2 - 3*(4*c^3*d^3 - 6*b*c^2*d^2*e + 4*(b^2*c - a*c^2)*d*e^2 - (b^3 - 2*a*b*c)*e^3)*sqrt(b^2 - 4*
a*c)*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)) + 6*(8*c^4*d
^3 - 6*b*c^3*d^2*e + 4*(b^2*c^2 - 2*a*c^3)*d*e^2 - (b^3*c - 3*a*b*c^2)*e^3)*n*x - 3*(2*c^4*e^3*n*x^4 + 8*c^4*d
*e^2*n*x^3 + 12*c^4*d^2*e*n*x^2 + 8*c^4*d^3*n*x + (4*b*c^3*d^3 - 6*(b^2*c^2 - 2*a*c^3)*d^2*e + 4*(b^3*c - 3*a*
b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c + 2*a^2*c^2)*e^3)*n)*log(c*x^2 + b*x + a) - 6*(c^4*e^3*x^4 + 4*c^4*d*e^2*x^3 +
6*c^4*d^2*e*x^2 + 4*c^4*d^3*x)*log(d))/c^4, -1/24*(3*c^4*e^3*n*x^4 + 2*(8*c^4*d*e^2 - b*c^3*e^3)*n*x^3 + 3*(1
2*c^4*d^2*e - 4*b*c^3*d*e^2 + (b^2*c^2 - 2*a*c^3)*e^3)*n*x^2 - 6*(4*c^3*d^3 - 6*b*c^2*d^2*e + 4*(b^2*c - a*c^2
)*d*e^2 - (b^3 - 2*a*b*c)*e^3)*sqrt(-b^2 + 4*a*c)*n*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 6*
(8*c^4*d^3 - 6*b*c^3*d^2*e + 4*(b^2*c^2 - 2*a*c^3)*d*e^2 - (b^3*c - 3*a*b*c^2)*e^3)*n*x - 3*(2*c^4*e^3*n*x^4 +
8*c^4*d*e^2*n*x^3 + 12*c^4*d^2*e*n*x^2 + 8*c^4*d^3*n*x + (4*b*c^3*d^3 - 6*(b^2*c^2 - 2*a*c^3)*d^2*e + 4*(b^3*
c - 3*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c + 2*a^2*c^2)*e^3)*n)*log(c*x^2 + b*x + a) - 6*(c^4*e^3*x^4 + 4*c^4*d*e
^2*x^3 + 6*c^4*d^2*e*x^2 + 4*c^4*d^3*x)*log(d))/c^4]

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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((e*x+d)**3*ln(d*(c*x**2+b*x+a)**n),x)

[Out]

Timed out

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Giac [B]  time = 1.39151, size = 923, normalized size = 2.73 \begin{align*} \frac{b d^{3} n \log \left (c x^{2} + b x + a\right )}{2 \, c} - \frac{{\left (b^{2} d^{3} n - 4 \, a c d^{3} n\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c} - \frac{3 \,{\left (b^{2} d^{2} n e - 2 \, a c d^{2} n e\right )} \log \left (c x^{2} + b x + a\right )}{4 \, c^{2}} + \frac{3 \,{\left (b^{3} d^{2} n e - 4 \, a b c d^{2} n e\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} c^{2}} + \frac{{\left (b^{3} d n e^{2} - 3 \, a b c d n e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} - \frac{{\left (b^{4} d n e^{2} - 5 \, a b^{2} c d n e^{2} + 4 \, a^{2} c^{2} d n e^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{3}} + \frac{6 \, c^{3} n x^{4} e^{3} \log \left (c x^{2} + b x + a\right ) + 24 \, c^{3} d n x^{3} e^{2} \log \left (c x^{2} + b x + a\right ) + 36 \, c^{3} d^{2} n x^{2} e \log \left (c x^{2} + b x + a\right ) - 3 \, c^{3} n x^{4} e^{3} - 16 \, c^{3} d n x^{3} e^{2} - 36 \, c^{3} d^{2} n x^{2} e + 24 \, c^{3} d^{3} n x \log \left (c x^{2} + b x + a\right ) + 6 \, c^{3} x^{4} e^{3} \log \left (d\right ) + 24 \, c^{3} d x^{3} e^{2} \log \left (d\right ) + 36 \, c^{3} d^{2} x^{2} e \log \left (d\right ) - 48 \, c^{3} d^{3} n x + 2 \, b c^{2} n x^{3} e^{3} + 12 \, b c^{2} d n x^{2} e^{2} + 36 \, b c^{2} d^{2} n x e + 24 \, c^{3} d^{3} x \log \left (d\right ) - 3 \, b^{2} c n x^{2} e^{3} + 6 \, a c^{2} n x^{2} e^{3} - 24 \, b^{2} c d n x e^{2} + 48 \, a c^{2} d n x e^{2} + 6 \, b^{3} n x e^{3} - 18 \, a b c n x e^{3}}{24 \, c^{3}} - \frac{{\left (b^{4} n e^{3} - 4 \, a b^{2} c n e^{3} + 2 \, a^{2} c^{2} n e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{8 \, c^{4}} + \frac{{\left (b^{5} n e^{3} - 6 \, a b^{3} c n e^{3} + 8 \, a^{2} b c^{2} n e^{3}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt{-b^{2} + 4 \, a c} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x, algorithm="giac")

[Out]

1/2*b*d^3*n*log(c*x^2 + b*x + a)/c - (b^2*d^3*n - 4*a*c*d^3*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b
^2 + 4*a*c)*c) - 3/4*(b^2*d^2*n*e - 2*a*c*d^2*n*e)*log(c*x^2 + b*x + a)/c^2 + 3/2*(b^3*d^2*n*e - 4*a*b*c*d^2*n
*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^2) + 1/2*(b^3*d*n*e^2 - 3*a*b*c*d*n*e^2)*log(
c*x^2 + b*x + a)/c^3 - (b^4*d*n*e^2 - 5*a*b^2*c*d*n*e^2 + 4*a^2*c^2*d*n*e^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*
a*c))/(sqrt(-b^2 + 4*a*c)*c^3) + 1/24*(6*c^3*n*x^4*e^3*log(c*x^2 + b*x + a) + 24*c^3*d*n*x^3*e^2*log(c*x^2 + b
*x + a) + 36*c^3*d^2*n*x^2*e*log(c*x^2 + b*x + a) - 3*c^3*n*x^4*e^3 - 16*c^3*d*n*x^3*e^2 - 36*c^3*d^2*n*x^2*e
+ 24*c^3*d^3*n*x*log(c*x^2 + b*x + a) + 6*c^3*x^4*e^3*log(d) + 24*c^3*d*x^3*e^2*log(d) + 36*c^3*d^2*x^2*e*log(
d) - 48*c^3*d^3*n*x + 2*b*c^2*n*x^3*e^3 + 12*b*c^2*d*n*x^2*e^2 + 36*b*c^2*d^2*n*x*e + 24*c^3*d^3*x*log(d) - 3*
b^2*c*n*x^2*e^3 + 6*a*c^2*n*x^2*e^3 - 24*b^2*c*d*n*x*e^2 + 48*a*c^2*d*n*x*e^2 + 6*b^3*n*x*e^3 - 18*a*b*c*n*x*e
^3)/c^3 - 1/8*(b^4*n*e^3 - 4*a*b^2*c*n*e^3 + 2*a^2*c^2*n*e^3)*log(c*x^2 + b*x + a)/c^4 + 1/4*(b^5*n*e^3 - 6*a*
b^3*c*n*e^3 + 8*a^2*b*c^2*n*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^4)