### 3.2382 $$\int \frac{(d+e x)^4}{(a+b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=286 $\frac{e \sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e^2 \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{7/2}}-\frac{2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (d+e x)^2 \sqrt{a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}$

[Out]

(-2*(d + e*x)^3*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (2*e*(2*c*d - b*e)*(d
+ e*x)^2*Sqrt[a + b*x + c*x^2])/(c*(b^2 - 4*a*c)) + (e*(32*c^3*d^3 - 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e)
- 8*c^2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*x)*Sqrt[a + b*x + c*x^2]
)/(4*c^3*(b^2 - 4*a*c)) + (3*e^2*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]
*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))

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Rubi [A]  time = 0.317787, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {738, 832, 779, 621, 206} $\frac{e \sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e^2 \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{7/2}}-\frac{2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (d+e x)^2 \sqrt{a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^4/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^3*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (2*e*(2*c*d - b*e)*(d
+ e*x)^2*Sqrt[a + b*x + c*x^2])/(c*(b^2 - 4*a*c)) + (e*(32*c^3*d^3 - 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e)
- 8*c^2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*x)*Sqrt[a + b*x + c*x^2]
)/(4*c^3*(b^2 - 4*a*c)) + (3*e^2*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]
*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))

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 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 779

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], 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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 \int \frac{(d+e x)^2 (-3 e (b d-2 a e)-3 e (2 c d-b e) x)}{\sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac{2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (2 c d-b e) (d+e x)^2 \sqrt{a+b x+c x^2}}{c \left (b^2-4 a c\right )}-\frac{2 \int \frac{(d+e x) \left (-\frac{3}{2} e \left (b^2 d e-20 a c d e+4 b \left (c d^2+a e^2\right )\right )-\frac{3}{2} e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{3 c \left (b^2-4 a c\right )}\\ &=-\frac{2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (2 c d-b e) (d+e x)^2 \sqrt{a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac{e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{\left (3 e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac{2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (2 c d-b e) (d+e x)^2 \sqrt{a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac{e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{\left (3 e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{4 c^3}\\ &=-\frac{2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 e (2 c d-b e) (d+e x)^2 \sqrt{a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac{e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.451202, size = 316, normalized size = 1.1 $\frac{2 \sqrt{c} \left (4 b c \left (-13 a^2 e^4+a c e^2 \left (12 d^2+40 d e x-5 e^2 x^2\right )+2 c^2 d^3 (d-4 e x)\right )+8 c^2 \left (a^2 e^3 (16 d+3 e x)+a c e \left (-12 d^2 e x-8 d^3+8 d e^2 x^2+e^3 x^3\right )+2 c^2 d^4 x\right )-2 b^2 c e^2 \left (a e (24 d+31 e x)+c x \left (-24 d^2+8 d e x+e^2 x^2\right )\right )+b^3 e^3 (15 a e+c x (5 e x-48 d))+15 b^4 e^4 x\right )-3 e^2 \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^4/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[c]*(15*b^4*e^4*x + b^3*e^3*(15*a*e + c*x*(-48*d + 5*e*x)) + 4*b*c*(-13*a^2*e^4 + 2*c^2*d^3*(d - 4*e*x)
+ a*c*e^2*(12*d^2 + 40*d*e*x - 5*e^2*x^2)) - 2*b^2*c*e^2*(a*e*(24*d + 31*e*x) + c*x*(-24*d^2 + 8*d*e*x + e^2*
x^2)) + 8*c^2*(2*c^2*d^4*x + a^2*e^3*(16*d + 3*e*x) + a*c*e*(-8*d^3 - 12*d^2*e*x + 8*d*e^2*x^2 + e^3*x^3))) -
3*(b^2 - 4*a*c)*e^2*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(
2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.054, size = 913, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+b*x+a)^(3/2),x)

[Out]

3*d^2*e^2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-8*d^3*e*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4*d^3*e*b^2/c/(4
*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+16*d*e^3*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4*d^3*e/c/(c*x^2+b*x+a)^(1/2)+3
/2*e^4*a/c^2*x/(c*x^2+b*x+a)^(1/2)+15/16*e^4*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-13/4*e^4*b/c^3*a/(c*x^2+b
*x+a)^(1/2)-6*d^2*e^2*x/c/(c*x^2+b*x+a)^(1/2)+3*d^2*e^2*b/c^2/(c*x^2+b*x+a)^(1/2)-6*d*e^3*b/c^(5/2)*ln((1/2*b+
c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+4*d*e^3*x^2/c/(c*x^2+b*x+a)^(1/2)-3*d*e^3*b^2/c^3/(c*x^2+b*x+a)^(1/2)+8*d*e^
3*a/c^2/(c*x^2+b*x+a)^(1/2)-5/4*e^4*b/c^2*x^2/(c*x^2+b*x+a)^(1/2)-15/8*e^4*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)-13/2*
e^4*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-6*d*e^3*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+8*d*e^3*a/c^
2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+6*d^2*e^2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+2*d^4*(2*c*x+b)/(4*a*c
-b^2)/(c*x^2+b*x+a)^(1/2)+1/2*e^4*x^3/c/(c*x^2+b*x+a)^(1/2)+15/16*e^4*b^3/c^4/(c*x^2+b*x+a)^(1/2)+15/8*e^4*b^2
/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/2*e^4*a/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1
/2))+6*d^2*e^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+15/8*e^4*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^
(1/2)*x-13/4*e^4*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+6*d*e^3*b/c^2*x/(c*x^2+b*x+a)^(1/2)-3*d*e^3*b^4/c^3
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 6.10687, size = 2488, normalized size = 8.7 \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(16*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 16*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (5*a*b^4 - 24*a^2*b^2*c + 1
6*a^3*c^2)*e^4 + (16*(b^2*c^3 - 4*a*c^4)*d^2*e^2 - 16*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (5*b^4*c - 24*a*b^2*c^2 +
16*a^2*c^3)*e^4)*x^2 + (16*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 - 16*(b^4*c - 4*a*b^2*c^2)*d*e^3 + (5*b^5 - 24*a*b^3*
c + 16*a^2*b*c^2)*e^4)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c)
- 4*a*c) + 4*(8*b*c^4*d^4 - 64*a*c^4*d^3*e + 48*a*b*c^3*d^2*e^2 - 2*(b^2*c^3 - 4*a*c^4)*e^4*x^3 - 16*(3*a*b^2
*c^2 - 8*a^2*c^3)*d*e^3 + (15*a*b^3*c - 52*a^2*b*c^2)*e^4 - (16*(b^2*c^3 - 4*a*c^4)*d*e^3 - 5*(b^3*c^2 - 4*a*b
*c^3)*e^4)*x^2 + (16*c^5*d^4 - 32*b*c^4*d^3*e + 48*(b^2*c^3 - 2*a*c^4)*d^2*e^2 - 16*(3*b^3*c^2 - 10*a*b*c^3)*d
*e^3 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5
- 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*(3*(16*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 16*(a*b^3*c - 4*a^2*
b*c^2)*d*e^3 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*e^4 + (16*(b^2*c^3 - 4*a*c^4)*d^2*e^2 - 16*(b^3*c^2 - 4*a
*b*c^3)*d*e^3 + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*e^4)*x^2 + (16*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 - 16*(b^4*c
- 4*a*b^2*c^2)*d*e^3 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*e^4)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*
(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(8*b*c^4*d^4 - 64*a*c^4*d^3*e + 48*a*b*c^3*d^2*e^2 - 2*(b^2*
c^3 - 4*a*c^4)*e^4*x^3 - 16*(3*a*b^2*c^2 - 8*a^2*c^3)*d*e^3 + (15*a*b^3*c - 52*a^2*b*c^2)*e^4 - (16*(b^2*c^3 -
4*a*c^4)*d*e^3 - 5*(b^3*c^2 - 4*a*b*c^3)*e^4)*x^2 + (16*c^5*d^4 - 32*b*c^4*d^3*e + 48*(b^2*c^3 - 2*a*c^4)*d^2
*e^2 - 16*(3*b^3*c^2 - 10*a*b*c^3)*d*e^3 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*e^4)*x)*sqrt(c*x^2 + b*x + a
))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{4}}{\left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x)**4/(a + b*x + c*x**2)**(3/2), x)

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Giac [A]  time = 1.17574, size = 509, normalized size = 1.78 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (b^{2} c^{2} e^{4} - 4 \, a c^{3} e^{4}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac{16 \, b^{2} c^{2} d e^{3} - 64 \, a c^{3} d e^{3} - 5 \, b^{3} c e^{4} + 20 \, a b c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{16 \, c^{4} d^{4} - 32 \, b c^{3} d^{3} e + 48 \, b^{2} c^{2} d^{2} e^{2} - 96 \, a c^{3} d^{2} e^{2} - 48 \, b^{3} c d e^{3} + 160 \, a b c^{2} d e^{3} + 15 \, b^{4} e^{4} - 62 \, a b^{2} c e^{4} + 24 \, a^{2} c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{8 \, b c^{3} d^{4} - 64 \, a c^{3} d^{3} e + 48 \, a b c^{2} d^{2} e^{2} - 48 \, a b^{2} c d e^{3} + 128 \, a^{2} c^{2} d e^{3} + 15 \, a b^{3} e^{4} - 52 \, a^{2} b c e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt{c x^{2} + b x + a}} - \frac{3 \,{\left (16 \, c^{2} d^{2} e^{2} - 16 \, b c d e^{3} + 5 \, b^{2} e^{4} - 4 \, a c e^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{7}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/4*(((2*(b^2*c^2*e^4 - 4*a*c^3*e^4)*x/(b^2*c^3 - 4*a*c^4) + (16*b^2*c^2*d*e^3 - 64*a*c^3*d*e^3 - 5*b^3*c*e^4
+ 20*a*b*c^2*e^4)/(b^2*c^3 - 4*a*c^4))*x - (16*c^4*d^4 - 32*b*c^3*d^3*e + 48*b^2*c^2*d^2*e^2 - 96*a*c^3*d^2*e^
2 - 48*b^3*c*d*e^3 + 160*a*b*c^2*d*e^3 + 15*b^4*e^4 - 62*a*b^2*c*e^4 + 24*a^2*c^2*e^4)/(b^2*c^3 - 4*a*c^4))*x
- (8*b*c^3*d^4 - 64*a*c^3*d^3*e + 48*a*b*c^2*d^2*e^2 - 48*a*b^2*c*d*e^3 + 128*a^2*c^2*d*e^3 + 15*a*b^3*e^4 - 5
2*a^2*b*c*e^4)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 3/8*(16*c^2*d^2*e^2 - 16*b*c*d*e^3 + 5*b^2*e^4 - 4
*a*c*e^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)