### 3.2182 $$\int \frac{(d+e x)^4}{a+b x+c x^2} \, dx$$

Optimal. Leaf size=243 $-\frac{\left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^4 \sqrt{b^2-4 a c}}+\frac{e (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^4}+\frac{e^2 x \left (-c e (a e+4 b d)+b^2 e^2+6 c^2 d^2\right )}{c^3}+\frac{e^3 x^2 (4 c d-b e)}{2 c^2}+\frac{e^4 x^3}{3 c}$

[Out]

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

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Rubi [A]  time = 0.439051, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {701, 634, 618, 206, 628} $-\frac{\left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^4 \sqrt{b^2-4 a c}}+\frac{e (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^4}+\frac{e^2 x \left (-c e (a e+4 b d)+b^2 e^2+6 c^2 d^2\right )}{c^3}+\frac{e^3 x^2 (4 c d-b e)}{2 c^2}+\frac{e^4 x^3}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^4/(a + b*x + c*x^2),x]

[Out]

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

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

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

Mathematica [A]  time = 0.192607, size = 240, normalized size = 0.99 $\frac{\frac{6 \left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+6 c e^2 x \left (-c e (a e+4 b d)+b^2 e^2+6 c^2 d^2\right )+3 e (2 c d-b e) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))+3 c^2 e^3 x^2 (4 c d-b e)+2 c^3 e^4 x^3}{6 c^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^4/(a + b*x + c*x^2),x]

[Out]

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

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Maple [B]  time = 0.154, size = 595, normalized size = 2.5 \begin{align*}{\frac{{e}^{4}{x}^{3}}{3\,c}}-{\frac{{e}^{4}{x}^{2}b}{2\,{c}^{2}}}+2\,{\frac{d{e}^{3}{x}^{2}}{c}}-{\frac{{e}^{4}ax}{{c}^{2}}}+{\frac{{b}^{2}{e}^{4}x}{{c}^{3}}}-4\,{\frac{{e}^{3}bdx}{{c}^{2}}}+6\,{\frac{{d}^{2}{e}^{2}x}{c}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ) ab{e}^{4}}{{c}^{3}}}-2\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ) ad{e}^{3}}{{c}^{2}}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{3}{e}^{4}}{2\,{c}^{4}}}+2\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}d{e}^{3}}{{c}^{3}}}-3\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ) b{d}^{2}{e}^{2}}{{c}^{2}}}+2\,{\frac{\ln \left ( c{x}^{2}+bx+a \right ){d}^{3}e}{c}}+2\,{\frac{{a}^{2}{e}^{4}}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-4\,{\frac{a{b}^{2}{e}^{4}}{{c}^{3}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+12\,{\frac{d{e}^{3}ab}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-12\,{\frac{a{d}^{2}{e}^{2}}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{d}^{4}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{4}{e}^{4}}{{c}^{4}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-4\,{\frac{{b}^{3}d{e}^{3}}{{c}^{3}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+6\,{\frac{{b}^{2}{d}^{2}{e}^{2}}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-4\,{\frac{b{d}^{3}e}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+b*x+a),x)

[Out]

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

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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)^4/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.20638, size = 1693, normalized size = 6.97 \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)^4/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 6.10659, size = 1554, normalized size = 6.4 \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)**4/(c*x**2+b*x+a),x)

[Out]

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

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Giac [A]  time = 1.12097, size = 358, normalized size = 1.47 \begin{align*} \frac{2 \, c^{2} x^{3} e^{4} + 12 \, c^{2} d x^{2} e^{3} + 36 \, c^{2} d^{2} x e^{2} - 3 \, b c x^{2} e^{4} - 24 \, b c d x e^{3} + 6 \, b^{2} x e^{4} - 6 \, a c x e^{4}}{6 \, c^{3}} + \frac{{\left (4 \, c^{3} d^{3} e - 6 \, b c^{2} d^{2} e^{2} + 4 \, b^{2} c d e^{3} - 4 \, a c^{2} d e^{3} - b^{3} e^{4} + 2 \, a b c e^{4}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} + \frac{{\left (2 \, c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 12 \, a c^{3} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + 12 \, a b c^{2} d e^{3} + b^{4} e^{4} - 4 \, a b^{2} c e^{4} + 2 \, a^{2} c^{2} e^{4}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\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)^4/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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