3.2203 \(\int \frac {(d+e x)^3}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=158 \[ -\frac {6 (2 c d-b e) \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (d+e x) (2 c d-b e) (-2 a e+x (2 c d-b e)+b d)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )} \]

[Out]

-1/2*(2*c*x+b)*(e*x+d)^3/(-4*a*c+b^2)/(c*x^2+b*x+a)^2+3/2*(-b*e+2*c*d)*(e*x+d)*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*
a*c+b^2)^2/(c*x^2+b*x+a)-6*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)
^(5/2)

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {728, 722, 618, 206} \[ -\frac {6 (2 c d-b e) \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (d+e x) (2 c d-b e) (-2 a e+x (2 c d-b e)+b d)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b + 2*c*x)*(d + e*x)^3)/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*(2*c*d - b*e)*(d + e*x)*(b*d - 2*a*e + (
2*c*d - b*e)*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) - (6*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)*ArcTanh[(b +
2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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 722

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*(2*p + 3)*(c*d
^2 - b*d*e + a*e^2))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ
[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rule 728

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*(b + 2*
c*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[(m*(2*c*d - b*e))/((p + 1)*(b^2 - 4*a*c)),
Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {(3 (2 c d-b e)) \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c d-b e) (d+e x) (b d-2 a e+(2 c d-b e) x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {\left (3 (2 c d-b e) \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c d-b e) (d+e x) (b d-2 a e+(2 c d-b e) x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\left (6 (2 c d-b e) \left (c d^2-b d e+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2}\\ &=-\frac {(b+2 c x) (d+e x)^3}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c d-b e) (d+e x) (b d-2 a e+(2 c d-b e) x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {6 (2 c d-b e) \left (c d^2-b d e+a e^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 308, normalized size = 1.95 \[ \frac {1}{2} \left (\frac {2 c \left (a^2 e^3-3 a c d e (d+e x)+c^2 d^3 x\right )+b^2 e^2 (3 c d x-a e)+b c \left (3 a e^2 (d+e x)+c d^2 (d-3 e x)\right )-b^3 e^3 x}{c^2 \left (4 a c-b^2\right ) (a+x (b+c x))^2}+\frac {4 c^2 \left (-4 a^2 e^3+3 a c d e^2 x+3 c^2 d^3 x\right )+b^2 c e \left (5 a e^2-9 c d^2+6 c d e x\right )+6 b c^2 \left (a e^2 (d-e x)+c d^2 (d-3 e x)\right )+b^4 \left (-e^3\right )+3 b^3 c d e^2}{c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac {12 (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^3*c*d*e^2 - b^4*e^3 + b^2*c*e*(-9*c*d^2 + 5*a*e^2 + 6*c*d*e*x) + 4*c^2*(-4*a^2*e^3 + 3*c^2*d^3*x + 3*a*c
*d*e^2*x) + 6*b*c^2*(c*d^2*(d - 3*e*x) + a*e^2*(d - e*x)))/(c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))) + (-(b^3*e^
3*x) + b^2*e^2*(-(a*e) + 3*c*d*x) + 2*c*(a^2*e^3 + c^2*d^3*x - 3*a*c*d*e*(d + e*x)) + b*c*(c*d^2*(d - 3*e*x) +
3*a*e^2*(d + e*x)))/(c^2*(-b^2 + 4*a*c)*(a + x*(b + c*x))^2) + (12*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*A
rcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/2

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fricas [B]  time = 1.06, size = 2084, normalized size = 13.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((b^5*c - 14*a*b^3*c^2 + 40*a^2*b*c^3)*d^3 + 3*(a*b^4*c + 4*a^2*b^2*c^2 - 32*a^3*c^3)*d^2*e - 18*(a^2*b^
3*c - 4*a^3*b*c^2)*d*e^2 + (a^2*b^4 + 4*a^3*b^2*c - 32*a^4*c^2)*e^3 - 6*(2*(b^2*c^4 - 4*a*c^5)*d^3 - 3*(b^3*c^
3 - 4*a*b*c^4)*d^2*e + (b^4*c^2 - 2*a*b^2*c^3 - 8*a^2*c^4)*d*e^2 - (a*b^3*c^2 - 4*a^2*b*c^3)*e^3)*x^3 - (18*(b
^3*c^3 - 4*a*b*c^4)*d^3 - 27*(b^4*c^2 - 4*a*b^2*c^3)*d^2*e + 9*(b^5*c - 2*a*b^3*c^2 - 8*a^2*b*c^3)*d*e^2 - (b^
6 - 3*a*b^4*c + 12*a^2*b^2*c^2 - 64*a^3*c^3)*e^3)*x^2 + 6*(2*a^2*c^3*d^3 - 3*a^2*b*c^2*d^2*e - a^3*b*c*e^3 + (
2*c^5*d^3 - 3*b*c^4*d^2*e - a*b*c^3*e^3 + (b^2*c^3 + 2*a*c^4)*d*e^2)*x^4 + (a^2*b^2*c + 2*a^3*c^2)*d*e^2 + 2*(
2*b*c^4*d^3 - 3*b^2*c^3*d^2*e - a*b^2*c^2*e^3 + (b^3*c^2 + 2*a*b*c^3)*d*e^2)*x^3 + (2*(b^2*c^3 + 2*a*c^4)*d^3
- 3*(b^3*c^2 + 2*a*b*c^3)*d^2*e + (b^4*c + 4*a*b^2*c^2 + 4*a^2*c^3)*d*e^2 - (a*b^3*c + 2*a^2*b*c^2)*e^3)*x^2 +
2*(2*a*b*c^3*d^3 - 3*a*b^2*c^2*d^2*e - a^2*b^2*c*e^3 + (a*b^3*c + 2*a^2*b*c^2)*d*e^2)*x)*sqrt(b^2 - 4*a*c)*lo
g((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*(2*(b^4*c^2 + a*b
^2*c^3 - 20*a^2*c^4)*d^3 - 3*(b^5*c + a*b^3*c^2 - 20*a^2*b*c^3)*d^2*e + 3*(5*a*b^4*c - 22*a^2*b^2*c^2 + 8*a^3*
c^3)*d*e^2 - (a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)*e^3)*x)/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4*b^2*c^3 - 64*a^5*
c^4 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^4 + 2*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4
- 64*a^3*b*c^5)*x^3 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)*x^2 + 2*(a*b^7*c
- 12*a^2*b^5*c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x), -1/2*((b^5*c - 14*a*b^3*c^2 + 40*a^2*b*c^3)*d^3 + 3*(a*b
^4*c + 4*a^2*b^2*c^2 - 32*a^3*c^3)*d^2*e - 18*(a^2*b^3*c - 4*a^3*b*c^2)*d*e^2 + (a^2*b^4 + 4*a^3*b^2*c - 32*a^
4*c^2)*e^3 - 6*(2*(b^2*c^4 - 4*a*c^5)*d^3 - 3*(b^3*c^3 - 4*a*b*c^4)*d^2*e + (b^4*c^2 - 2*a*b^2*c^3 - 8*a^2*c^4
)*d*e^2 - (a*b^3*c^2 - 4*a^2*b*c^3)*e^3)*x^3 - (18*(b^3*c^3 - 4*a*b*c^4)*d^3 - 27*(b^4*c^2 - 4*a*b^2*c^3)*d^2*
e + 9*(b^5*c - 2*a*b^3*c^2 - 8*a^2*b*c^3)*d*e^2 - (b^6 - 3*a*b^4*c + 12*a^2*b^2*c^2 - 64*a^3*c^3)*e^3)*x^2 + 1
2*(2*a^2*c^3*d^3 - 3*a^2*b*c^2*d^2*e - a^3*b*c*e^3 + (2*c^5*d^3 - 3*b*c^4*d^2*e - a*b*c^3*e^3 + (b^2*c^3 + 2*a
*c^4)*d*e^2)*x^4 + (a^2*b^2*c + 2*a^3*c^2)*d*e^2 + 2*(2*b*c^4*d^3 - 3*b^2*c^3*d^2*e - a*b^2*c^2*e^3 + (b^3*c^2
+ 2*a*b*c^3)*d*e^2)*x^3 + (2*(b^2*c^3 + 2*a*c^4)*d^3 - 3*(b^3*c^2 + 2*a*b*c^3)*d^2*e + (b^4*c + 4*a*b^2*c^2 +
4*a^2*c^3)*d*e^2 - (a*b^3*c + 2*a^2*b*c^2)*e^3)*x^2 + 2*(2*a*b*c^3*d^3 - 3*a*b^2*c^2*d^2*e - a^2*b^2*c*e^3 +
(a*b^3*c + 2*a^2*b*c^2)*d*e^2)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 2
*(2*(b^4*c^2 + a*b^2*c^3 - 20*a^2*c^4)*d^3 - 3*(b^5*c + a*b^3*c^2 - 20*a^2*b*c^3)*d^2*e + 3*(5*a*b^4*c - 22*a^
2*b^2*c^2 + 8*a^3*c^3)*d*e^2 - (a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)*e^3)*x)/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4
*b^2*c^3 - 64*a^5*c^4 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^4 + 2*(b^7*c^2 - 12*a*b^5*c^3
+ 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*x^3 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)
*x^2 + 2*(a*b^7*c - 12*a^2*b^5*c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x)]

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giac [B]  time = 0.17, size = 446, normalized size = 2.82 \[ \frac {6 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2} + 2 \, a c d e^{2} - a b e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {12 \, c^{4} d^{3} x^{3} - 18 \, b c^{3} d^{2} x^{3} e + 18 \, b c^{3} d^{3} x^{2} + 6 \, b^{2} c^{2} d x^{3} e^{2} + 12 \, a c^{3} d x^{3} e^{2} - 27 \, b^{2} c^{2} d^{2} x^{2} e + 4 \, b^{2} c^{2} d^{3} x + 20 \, a c^{3} d^{3} x - 6 \, a b c^{2} x^{3} e^{3} + 9 \, b^{3} c d x^{2} e^{2} + 18 \, a b c^{2} d x^{2} e^{2} - 6 \, b^{3} c d^{2} x e - 30 \, a b c^{2} d^{2} x e - b^{3} c d^{3} + 10 \, a b c^{2} d^{3} - b^{4} x^{2} e^{3} - a b^{2} c x^{2} e^{3} - 16 \, a^{2} c^{2} x^{2} e^{3} + 30 \, a b^{2} c d x e^{2} - 12 \, a^{2} c^{2} d x e^{2} - 3 \, a b^{2} c d^{2} e - 24 \, a^{2} c^{2} d^{2} e - 2 \, a b^{3} x e^{3} - 10 \, a^{2} b c x e^{3} + 18 \, a^{2} b c d e^{2} - a^{2} b^{2} e^{3} - 8 \, a^{3} c e^{3}}{2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2 + 2*a*c*d*e^2 - a*b*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4 -
8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) + 1/2*(12*c^4*d^3*x^3 - 18*b*c^3*d^2*x^3*e + 18*b*c^3*d^3*x^2 + 6
*b^2*c^2*d*x^3*e^2 + 12*a*c^3*d*x^3*e^2 - 27*b^2*c^2*d^2*x^2*e + 4*b^2*c^2*d^3*x + 20*a*c^3*d^3*x - 6*a*b*c^2*
x^3*e^3 + 9*b^3*c*d*x^2*e^2 + 18*a*b*c^2*d*x^2*e^2 - 6*b^3*c*d^2*x*e - 30*a*b*c^2*d^2*x*e - b^3*c*d^3 + 10*a*b
*c^2*d^3 - b^4*x^2*e^3 - a*b^2*c*x^2*e^3 - 16*a^2*c^2*x^2*e^3 + 30*a*b^2*c*d*x*e^2 - 12*a^2*c^2*d*x*e^2 - 3*a*
b^2*c*d^2*e - 24*a^2*c^2*d^2*e - 2*a*b^3*x*e^3 - 10*a^2*b*c*x*e^3 + 18*a^2*b*c*d*e^2 - a^2*b^2*e^3 - 8*a^3*c*e
^3)/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*(c*x^2 + b*x + a)^2)

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maple [B]  time = 0.06, size = 695, normalized size = 4.40 \[ -\frac {6 a b \,e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {12 a c d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {6 b^{2} d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}-\frac {18 b c \,d^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {12 c^{2} d^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {-\frac {3 \left (a b \,e^{3}-2 a c d \,e^{2}-b^{2} d \,e^{2}+3 b c \,d^{2} e -2 c^{2} d^{3}\right ) c \,x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {\left (16 a^{2} c^{2} e^{3}+a \,b^{2} c \,e^{3}-18 a b \,c^{2} d \,e^{2}+b^{4} e^{3}-9 b^{3} c d \,e^{2}+27 b^{2} c^{2} d^{2} e -18 b \,c^{3} d^{3}\right ) x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}-\frac {\left (5 a^{2} b c \,e^{3}+6 a^{2} c^{2} d \,e^{2}+a \,b^{3} e^{3}-15 a \,b^{2} c d \,e^{2}+15 a b \,c^{2} d^{2} e -10 a \,c^{3} d^{3}+3 b^{3} c \,d^{2} e -2 b^{2} c^{2} d^{3}\right ) x}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}-\frac {8 a^{3} c \,e^{3}+a^{2} b^{2} e^{3}-18 a^{2} b c d \,e^{2}+24 a^{2} c^{2} d^{2} e +3 a \,b^{2} c \,d^{2} e -10 a b \,c^{2} d^{3}+b^{3} c \,d^{3}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}}{\left (c \,x^{2}+b x +a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x+a)^3,x)

[Out]

(-3*c*(a*b*e^3-2*a*c*d*e^2-b^2*d*e^2+3*b*c*d^2*e-2*c^2*d^3)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3-1/2*(16*a^2*c^2*e^3
+a*b^2*c*e^3-18*a*b*c^2*d*e^2+b^4*e^3-9*b^3*c*d*e^2+27*b^2*c^2*d^2*e-18*b*c^3*d^3)/c/(16*a^2*c^2-8*a*b^2*c+b^4
)*x^2-1/c*(5*a^2*b*c*e^3+6*a^2*c^2*d*e^2+a*b^3*e^3-15*a*b^2*c*d*e^2+15*a*b*c^2*d^2*e-10*a*c^3*d^3+3*b^3*c*d^2*
e-2*b^2*c^2*d^3)/(16*a^2*c^2-8*a*b^2*c+b^4)*x-1/2/c*(8*a^3*c*e^3+a^2*b^2*e^3-18*a^2*b*c*d*e^2+24*a^2*c^2*d^2*e
+3*a*b^2*c*d^2*e-10*a*b*c^2*d^3+b^3*c*d^3)/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2-6/(16*a^2*c^2-8*a*b^2*c
+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*e^3+12/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^
(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*c*d*e^2+6/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c
*x+b)/(4*a*c-b^2)^(1/2))*b^2*d*e^2-18/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2
)^(1/2))*b*c*d^2*e+12/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^2*d^3

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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((e*x+d)^3/(c*x^2+b*x+a)^3,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 = 1.36, size = 636, normalized size = 4.03 \[ \frac {6\,\mathrm {atan}\left (\frac {\left (\frac {3\,\left (b\,e-2\,c\,d\right )\,\left (16\,a^2\,b\,c^2-8\,a\,b^3\,c+b^5\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {6\,c\,x\,\left (b\,e-2\,c\,d\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{3\,b^2\,d\,e^2-9\,b\,c\,d^2\,e-3\,a\,b\,e^3+6\,c^2\,d^3+6\,a\,c\,d\,e^2}\right )\,\left (b\,e-2\,c\,d\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {8\,a^3\,c\,e^3+a^2\,b^2\,e^3-18\,a^2\,b\,c\,d\,e^2+24\,a^2\,c^2\,d^2\,e+3\,a\,b^2\,c\,d^2\,e-10\,a\,b\,c^2\,d^3+b^3\,c\,d^3}{2\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (16\,a^2\,c^2\,e^3+a\,b^2\,c\,e^3-18\,a\,b\,c^2\,d\,e^2+b^4\,e^3-9\,b^3\,c\,d\,e^2+27\,b^2\,c^2\,d^2\,e-18\,b\,c^3\,d^3\right )}{2\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {3\,c\,x^3\,\left (b^2\,d\,e^2-3\,b\,c\,d^2\,e-a\,b\,e^3+2\,c^2\,d^3+2\,a\,c\,d\,e^2\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {x\,\left (5\,a^2\,b\,c\,e^3+6\,a^2\,c^2\,d\,e^2+a\,b^3\,e^3-15\,a\,b^2\,c\,d\,e^2+15\,a\,b\,c^2\,d^2\,e-10\,a\,c^3\,d^3+3\,b^3\,c\,d^2\,e-2\,b^2\,c^2\,d^3\right )}{c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + b*x + c*x^2)^3,x)

[Out]

(6*atan((((3*(b*e - 2*c*d)*(b^5 + 16*a^2*b*c^2 - 8*a*b^3*c)*(a*e^2 + c*d^2 - b*d*e))/((4*a*c - b^2)^(5/2)*(b^4
+ 16*a^2*c^2 - 8*a*b^2*c)) + (6*c*x*(b*e - 2*c*d)*(a*e^2 + c*d^2 - b*d*e))/(4*a*c - b^2)^(5/2))*(b^4 + 16*a^2
*c^2 - 8*a*b^2*c))/(6*c^2*d^3 + 3*b^2*d*e^2 - 3*a*b*e^3 + 6*a*c*d*e^2 - 9*b*c*d^2*e))*(b*e - 2*c*d)*(a*e^2 + c
*d^2 - b*d*e))/(4*a*c - b^2)^(5/2) - ((8*a^3*c*e^3 + b^3*c*d^3 + a^2*b^2*e^3 + 24*a^2*c^2*d^2*e - 10*a*b*c^2*d
^3 + 3*a*b^2*c*d^2*e - 18*a^2*b*c*d*e^2)/(2*c*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(b^4*e^3 - 18*b*c^3*d^3 +
16*a^2*c^2*e^3 + 27*b^2*c^2*d^2*e + a*b^2*c*e^3 - 9*b^3*c*d*e^2 - 18*a*b*c^2*d*e^2))/(2*c*(b^4 + 16*a^2*c^2 -
8*a*b^2*c)) - (3*c*x^3*(2*c^2*d^3 + b^2*d*e^2 - a*b*e^3 + 2*a*c*d*e^2 - 3*b*c*d^2*e))/(b^4 + 16*a^2*c^2 - 8*a
*b^2*c) + (x*(a*b^3*e^3 - 10*a*c^3*d^3 - 2*b^2*c^2*d^3 + 6*a^2*c^2*d*e^2 + 5*a^2*b*c*e^3 + 3*b^3*c*d^2*e + 15*
a*b*c^2*d^2*e - 15*a*b^2*c*d*e^2))/(c*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^2*(2*a*c + b^2) + a^2 + c^2*x^4 + 2*
a*b*x + 2*b*c*x^3)

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sympy [B]  time = 10.91, size = 1180, normalized size = 7.47 \[ 3 \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (x + \frac {- 192 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) + 144 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) - 36 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) + 3 a b^{2} e^{3} - 6 a b c d e^{2} + 3 b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) - 3 b^{3} d e^{2} + 9 b^{2} c d^{2} e - 6 b c^{2} d^{3}}{6 a b c e^{3} - 12 a c^{2} d e^{2} - 6 b^{2} c d e^{2} + 18 b c^{2} d^{2} e - 12 c^{3} d^{3}} \right )} - 3 \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (x + \frac {192 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) - 144 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) + 36 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) + 3 a b^{2} e^{3} - 6 a b c d e^{2} - 3 b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) - 3 b^{3} d e^{2} + 9 b^{2} c d^{2} e - 6 b c^{2} d^{3}}{6 a b c e^{3} - 12 a c^{2} d e^{2} - 6 b^{2} c d e^{2} + 18 b c^{2} d^{2} e - 12 c^{3} d^{3}} \right )} + \frac {- 8 a^{3} c e^{3} - a^{2} b^{2} e^{3} + 18 a^{2} b c d e^{2} - 24 a^{2} c^{2} d^{2} e - 3 a b^{2} c d^{2} e + 10 a b c^{2} d^{3} - b^{3} c d^{3} + x^{3} \left (- 6 a b c^{2} e^{3} + 12 a c^{3} d e^{2} + 6 b^{2} c^{2} d e^{2} - 18 b c^{3} d^{2} e + 12 c^{4} d^{3}\right ) + x^{2} \left (- 16 a^{2} c^{2} e^{3} - a b^{2} c e^{3} + 18 a b c^{2} d e^{2} - b^{4} e^{3} + 9 b^{3} c d e^{2} - 27 b^{2} c^{2} d^{2} e + 18 b c^{3} d^{3}\right ) + x \left (- 10 a^{2} b c e^{3} - 12 a^{2} c^{2} d e^{2} - 2 a b^{3} e^{3} + 30 a b^{2} c d e^{2} - 30 a b c^{2} d^{2} e + 20 a c^{3} d^{3} - 6 b^{3} c d^{2} e + 4 b^{2} c^{2} d^{3}\right )}{32 a^{4} c^{3} - 16 a^{3} b^{2} c^{2} + 2 a^{2} b^{4} c + x^{4} \left (32 a^{2} c^{5} - 16 a b^{2} c^{4} + 2 b^{4} c^{3}\right ) + x^{3} \left (64 a^{2} b c^{4} - 32 a b^{3} c^{3} + 4 b^{5} c^{2}\right ) + x^{2} \left (64 a^{3} c^{4} - 12 a b^{4} c^{2} + 2 b^{6} c\right ) + x \left (64 a^{3} b c^{3} - 32 a^{2} b^{3} c^{2} + 4 a b^{5} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x+a)**3,x)

[Out]

3*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2)*log(x + (-192*a**3*c**3*sqrt(-1/(4*a*c -
b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2) + 144*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d
)*(a*e**2 - b*d*e + c*d**2) - 36*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2) +
3*a*b**2*e**3 - 6*a*b*c*d*e**2 + 3*b**6*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2) -
3*b**3*d*e**2 + 9*b**2*c*d**2*e - 6*b*c**2*d**3)/(6*a*b*c*e**3 - 12*a*c**2*d*e**2 - 6*b**2*c*d*e**2 + 18*b*c**
2*d**2*e - 12*c**3*d**3)) - 3*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2)*log(x + (192*
a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2) - 144*a**2*b**2*c**2*sqrt(-1/(4*a
*c - b**2)**5)*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2) + 36*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*
(a*e**2 - b*d*e + c*d**2) + 3*a*b**2*e**3 - 6*a*b*c*d*e**2 - 3*b**6*sqrt(-1/(4*a*c - b**2)**5)*(b*e - 2*c*d)*(
a*e**2 - b*d*e + c*d**2) - 3*b**3*d*e**2 + 9*b**2*c*d**2*e - 6*b*c**2*d**3)/(6*a*b*c*e**3 - 12*a*c**2*d*e**2 -
6*b**2*c*d*e**2 + 18*b*c**2*d**2*e - 12*c**3*d**3)) + (-8*a**3*c*e**3 - a**2*b**2*e**3 + 18*a**2*b*c*d*e**2 -
24*a**2*c**2*d**2*e - 3*a*b**2*c*d**2*e + 10*a*b*c**2*d**3 - b**3*c*d**3 + x**3*(-6*a*b*c**2*e**3 + 12*a*c**3
*d*e**2 + 6*b**2*c**2*d*e**2 - 18*b*c**3*d**2*e + 12*c**4*d**3) + x**2*(-16*a**2*c**2*e**3 - a*b**2*c*e**3 + 1
8*a*b*c**2*d*e**2 - b**4*e**3 + 9*b**3*c*d*e**2 - 27*b**2*c**2*d**2*e + 18*b*c**3*d**3) + x*(-10*a**2*b*c*e**3
- 12*a**2*c**2*d*e**2 - 2*a*b**3*e**3 + 30*a*b**2*c*d*e**2 - 30*a*b*c**2*d**2*e + 20*a*c**3*d**3 - 6*b**3*c*d
**2*e + 4*b**2*c**2*d**3))/(32*a**4*c**3 - 16*a**3*b**2*c**2 + 2*a**2*b**4*c + x**4*(32*a**2*c**5 - 16*a*b**2*
c**4 + 2*b**4*c**3) + x**3*(64*a**2*b*c**4 - 32*a*b**3*c**3 + 4*b**5*c**2) + x**2*(64*a**3*c**4 - 12*a*b**4*c*
*2 + 2*b**6*c) + x*(64*a**3*b*c**3 - 32*a**2*b**3*c**2 + 4*a*b**5*c))

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