### 3.2192 $$\int \frac{(d+e x)^3}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=173 $\frac{(2 c d-b e) \left (-2 c e (b d-3 a e)-b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \left (b^2-4 a c\right )^{3/2}}+\frac{e^2 x (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac{(d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{e^3 \log \left (a+b x+c x^2\right )}{2 c^2}$

[Out]

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

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Rubi [A]  time = 0.284977, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {738, 773, 634, 618, 206, 628} $\frac{(2 c d-b e) \left (-2 c e (b d-3 a e)-b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \left (b^2-4 a c\right )^{3/2}}+\frac{e^2 x (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac{(d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{e^3 \log \left (a+b x+c x^2\right )}{2 c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 738

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[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

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

Mathematica [A]  time = 0.340097, size = 201, normalized size = 1.16 $\frac{\frac{2 \left (-2 c \left (a^2 e^3-3 a c d e (d+e x)+c^2 d^3 x\right )+b^2 e^2 (a e-3 c d x)-b c \left (3 a e^2 (d+e x)+c d^2 (d-3 e x)\right )+b^3 e^3 x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac{2 (b e-2 c d) \left (2 c e (b d-3 a e)+b^2 e^2-2 c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+e^3 \log (a+x (b+c x))}{2 c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*(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))))/((b^2 - 4*a*c)*(a + x*(b + c*x))) + (2*(-2*c*d + b*e)*(-2*c^2*d^2 + b^2*e^2 + 2*c
*e*(b*d - 3*a*e))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2) + e^3*Log[a + x*(b + c*x)])/(2*
c^2)

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Maple [B]  time = 0.157, size = 416, normalized size = 2.4 \begin{align*}{\frac{1}{c{x}^{2}+bx+a} \left ({\frac{ \left ( 3\,abc{e}^{3}-6\,{c}^{2}ad{e}^{2}-{b}^{3}{e}^{3}+3\,{b}^{2}cd{e}^{2}-3\,b{c}^{2}{d}^{2}e+2\,{c}^{3}{d}^{3} \right ) x}{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{2\,{a}^{2}c{e}^{3}-a{b}^{2}{e}^{3}+3\,abcd{e}^{2}-6\,a{c}^{2}{d}^{2}e+b{c}^{2}{d}^{3}}{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }} \right ) }+2\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ) a{e}^{3}}{c \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}{e}^{3}}{2\,{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-6\,{\frac{ab{e}^{3}}{c \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+12\,{\frac{ad{e}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-6\,{\frac{b{d}^{2}e}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+4\,{\frac{c{d}^{3}}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}{e}^{3}}{{c}^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*a*b*c*e^3-6*a*c^2*d*e^2-b^3*e^3+3*b^2*c*d*e^2-3*b*c^2*d^2*e+2*c^3*d^3)/c^2/(4*a*c-b^2)*x+(2*a^2*c*e^3-a*b^
2*e^3+3*a*b*c*d*e^2-6*a*c^2*d^2*e+b*c^2*d^3)/(4*a*c-b^2)/c^2)/(c*x^2+b*x+a)+2/c/(4*a*c-b^2)*ln(c*x^2+b*x+a)*a*
e^3-1/2/c^2/(4*a*c-b^2)*ln(c*x^2+b*x+a)*b^2*e^3-6/c/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*
e^3+12/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*d*e^2-6/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a
*c-b^2)^(1/2))*b*d^2*e+4*c/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*d^3+1/c^2/(4*a*c-b^2)^(3/2)*a
rctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*e^3

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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/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94204, size = 2435, normalized size = 14.08 \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/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(b^3*c^2 - 4*a*b*c^3)*d^3 - 12*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e + 6*(a*b^3*c - 4*a^2*b*c^2)*d*e^2 - 2*(a
*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e^3 - (4*a*c^3*d^3 - 6*a*b*c^2*d^2*e + 12*a^2*c^2*d*e^2 + (a*b^3 - 6*a^2*b*c)*
e^3 + (4*c^4*d^3 - 6*b*c^3*d^2*e + 12*a*c^3*d*e^2 + (b^3*c - 6*a*b*c^2)*e^3)*x^2 + (4*b*c^3*d^3 - 6*b^2*c^2*d^
2*e + 12*a*b*c^2*d*e^2 + (b^4 - 6*a*b^2*c)*e^3)*x)*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*(2*(b^2*c^3 - 4*a*c^4)*d^3 - 3*(b^3*c^2 - 4*a*b*c^3)*d^2
*e + 3*(b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*d*e^2 - (b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*e^3)*x - ((b^4*c - 8*a*b^2*c
^2 + 16*a^2*c^3)*e^3*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^3*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e^3)*lo
g(c*x^2 + b*x + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^2 + (b^5*
c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x), -1/2*(2*(b^3*c^2 - 4*a*b*c^3)*d^3 - 12*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e + 6
*(a*b^3*c - 4*a^2*b*c^2)*d*e^2 - 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e^3 - 2*(4*a*c^3*d^3 - 6*a*b*c^2*d^2*e +
12*a^2*c^2*d*e^2 + (a*b^3 - 6*a^2*b*c)*e^3 + (4*c^4*d^3 - 6*b*c^3*d^2*e + 12*a*c^3*d*e^2 + (b^3*c - 6*a*b*c^2)
*e^3)*x^2 + (4*b*c^3*d^3 - 6*b^2*c^2*d^2*e + 12*a*b*c^2*d*e^2 + (b^4 - 6*a*b^2*c)*e^3)*x)*sqrt(-b^2 + 4*a*c)*a
rctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(2*(b^2*c^3 - 4*a*c^4)*d^3 - 3*(b^3*c^2 - 4*a*b*c^3)*
d^2*e + 3*(b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*d*e^2 - (b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*e^3)*x - ((b^4*c - 8*a*b^
2*c^2 + 16*a^2*c^3)*e^3*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^3*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e^3)
*log(c*x^2 + b*x + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^2 + (b
^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x)]

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Sympy [B]  time = 5.31504, size = 1238, normalized size = 7.16 \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/(c*x**2+b*x+a)**2,x)

[Out]

(e**3/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*
c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)))*log(x + (-16*a**2*c**3*(e**3/(2*c**2) - sqrt(-(
4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c**3 - 48*
a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 8*a**2*c*e**3 + 8*a*b**2*c**2*(e**3/(2*c**2) - sqrt(-(4*a*c - b**2)**
3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2
+ 12*a*b**4*c - b**6))) - a*b**2*e**3 - 6*a*b*c*d*e**2 - b**4*c*(e**3/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(b*e
- 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*
b**4*c - b**6))) + 3*b**2*c*d**2*e - 2*b*c**2*d**3)/(6*a*b*c*e**3 - 12*a*c**2*d*e**2 - b**3*e**3 + 6*b*c**2*d*
*2*e - 4*c**3*d**3)) + (e**3/(2*c**2) + sqrt(-(4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c
*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)))*log(x + (-16*a**2*c**3*(
e**3/(2*c**2) + sqrt(-(4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c
**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 8*a**2*c*e**3 + 8*a*b**2*c**2*(e**3/(2*c**2) +
sqrt(-(4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c*
*3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) - a*b**2*e**3 - 6*a*b*c*d*e**2 - b**4*c*(e**3/(2*c**2) + sqrt(-
(4*a*c - b**2)**3)*(b*e - 2*c*d)*(6*a*c*e**2 - b**2*e**2 - 2*b*c*d*e + 2*c**2*d**2)/(2*c**2*(64*a**3*c**3 - 48
*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 3*b**2*c*d**2*e - 2*b*c**2*d**3)/(6*a*b*c*e**3 - 12*a*c**2*d*e**2 -
b**3*e**3 + 6*b*c**2*d**2*e - 4*c**3*d**3)) + (2*a**2*c*e**3 - a*b**2*e**3 + 3*a*b*c*d*e**2 - 6*a*c**2*d**2*e
+ b*c**2*d**3 + x*(3*a*b*c*e**3 - 6*a*c**2*d*e**2 - b**3*e**3 + 3*b**2*c*d*e**2 - 3*b*c**2*d**2*e + 2*c**3*d**
3))/(4*a**2*c**3 - a*b**2*c**2 + x**2*(4*a*c**4 - b**2*c**3) + x*(4*a*b*c**3 - b**3*c**2))

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Giac [A]  time = 1.11513, size = 319, normalized size = 1.84 \begin{align*} -\frac{{\left (4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 12 \, a c^{2} d e^{2} + b^{3} e^{3} - 6 \, a b c e^{3}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{e^{3} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} - \frac{b c^{2} d^{3} - 6 \, a c^{2} d^{2} e + 3 \, a b c d e^{2} - a b^{2} e^{3} + 2 \, a^{2} c e^{3} +{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - 6 \, a c^{2} d e^{2} - b^{3} e^{3} + 3 \, a b c e^{3}\right )} x}{{\left (c x^{2} + b x + a\right )}{\left (b^{2} - 4 \, a c\right )} c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*c^3*d^3 - 6*b*c^2*d^2*e + 12*a*c^2*d*e^2 + b^3*e^3 - 6*a*b*c*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(
(b^2*c^2 - 4*a*c^3)*sqrt(-b^2 + 4*a*c)) + 1/2*e^3*log(c*x^2 + b*x + a)/c^2 - (b*c^2*d^3 - 6*a*c^2*d^2*e + 3*a*
b*c*d*e^2 - a*b^2*e^3 + 2*a^2*c*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - 6*a*c^2*d*e^2 - b^3*e^3 + 3
*a*b*c*e^3)*x)/((c*x^2 + b*x + a)*(b^2 - 4*a*c)*c^2)