3.2394 \(\int \frac {(d+e x)^2}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac {8 (2 c d-b e) (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x) (d+e x)^2}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*(2*c*x+b)*(e*x+d)^2/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)+8/3*(-b*e+2*c*d)*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+
b^2)^2/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {728, 636} \[ \frac {8 (2 c d-b e) (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x) (d+e x)^2}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

Rule 636

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

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)^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b+2 c x) (d+e x)^2}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {(4 (2 c d-b e)) \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {2 (b+2 c x) (d+e x)^2}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {8 (2 c d-b e) (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 167, normalized size = 1.70 \[ \frac {2 \left (4 b \left (2 a^2 e^2+3 a c (d-e x)^2+2 c^2 d x^2 (3 d-2 e x)\right )+8 c \left (-2 a^2 d e+a c x \left (3 d^2+e^2 x^2\right )+2 c^2 d^2 x^3\right )+b^2 \left (2 c x \left (3 d^2-12 d e x+e^2 x^2\right )-4 a e (d-3 e x)\right )-\left (b^3 \left (d^2+6 d e x-3 e^2 x^2\right )\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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fricas [B]  time = 2.99, size = 305, normalized size = 3.11 \[ \frac {2 \, {\left (8 \, a^{2} b e^{2} + 2 \, {\left (8 \, c^{3} d^{2} - 8 \, b c^{2} d e + {\left (b^{2} c + 4 \, a c^{2}\right )} e^{2}\right )} x^{3} - {\left (b^{3} - 12 \, a b c\right )} d^{2} - 4 \, {\left (a b^{2} + 4 \, a^{2} c\right )} d e + 3 \, {\left (8 \, b c^{2} d^{2} - 8 \, b^{2} c d e + {\left (b^{3} + 4 \, a b c\right )} e^{2}\right )} x^{2} + 6 \, {\left (2 \, a b^{2} e^{2} + {\left (b^{2} c + 4 \, a c^{2}\right )} d^{2} - {\left (b^{3} + 4 \, a b c\right )} d e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(8*a^2*b*e^2 + 2*(8*c^3*d^2 - 8*b*c^2*d*e + (b^2*c + 4*a*c^2)*e^2)*x^3 - (b^3 - 12*a*b*c)*d^2 - 4*(a*b^2 +
4*a^2*c)*d*e + 3*(8*b*c^2*d^2 - 8*b^2*c*d*e + (b^3 + 4*a*b*c)*e^2)*x^2 + 6*(2*a*b^2*e^2 + (b^2*c + 4*a*c^2)*d
^2 - (b^3 + 4*a*b*c)*d*e)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^
3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b
^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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giac [B]  time = 0.32, size = 263, normalized size = 2.68 \[ \frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (8 \, c^{3} d^{2} - 8 \, b c^{2} d e + b^{2} c e^{2} + 4 \, a c^{2} e^{2}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (8 \, b c^{2} d^{2} - 8 \, b^{2} c d e + b^{3} e^{2} + 4 \, a b c e^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {6 \, {\left (b^{2} c d^{2} + 4 \, a c^{2} d^{2} - b^{3} d e - 4 \, a b c d e + 2 \, a b^{2} e^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {b^{3} d^{2} - 12 \, a b c d^{2} + 4 \, a b^{2} d e + 16 \, a^{2} c d e - 8 \, a^{2} b e^{2}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*(((2*(8*c^3*d^2 - 8*b*c^2*d*e + b^2*c*e^2 + 4*a*c^2*e^2)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(8*b*c^2*d^2
- 8*b^2*c*d*e + b^3*e^2 + 4*a*b*c*e^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 6*(b^2*c*d^2 + 4*a*c^2*d^2 - b^3*d
*e - 4*a*b*c*d*e + 2*a*b^2*e^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x - (b^3*d^2 - 12*a*b*c*d^2 + 4*a*b^2*d*e + 16
*a^2*c*d*e - 8*a^2*b*e^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)

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maple [B]  time = 0.06, size = 215, normalized size = 2.19 \[ \frac {\frac {16}{3} a \,c^{2} e^{2} x^{3}+\frac {4}{3} b^{2} c \,e^{2} x^{3}-\frac {32}{3} b \,c^{2} d e \,x^{3}+\frac {32}{3} c^{3} d^{2} x^{3}+8 a b c \,e^{2} x^{2}+2 b^{3} e^{2} x^{2}-16 b^{2} c d e \,x^{2}+16 b \,c^{2} d^{2} x^{2}+8 a \,b^{2} e^{2} x -16 a b c d e x +16 a \,c^{2} d^{2} x -4 b^{3} d e x +4 b^{2} c \,d^{2} x +\frac {16}{3} a^{2} b \,e^{2}-\frac {32}{3} a^{2} c d e -\frac {8}{3} a \,b^{2} d e +8 a b c \,d^{2}-\frac {2}{3} b^{3} d^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(8*a*c^2*e^2*x^3+2*b^2*c*e^2*x^3-16*b*c^2*d*e*x^3+16*c^3*d^2*x^3+12*a*b*c*e^2*x^2+3*b^
3*e^2*x^2-24*b^2*c*d*e*x^2+24*b*c^2*d^2*x^2+12*a*b^2*e^2*x-24*a*b*c*d*e*x+24*a*c^2*d^2*x-6*b^3*d*e*x+6*b^2*c*d
^2*x+8*a^2*b*e^2-16*a^2*c*d*e-4*a*b^2*d*e+12*a*b*c*d^2-b^3*d^2)/(16*a^2*c^2-8*a*b^2*c+b^4)

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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)^2/(c*x^2+b*x+a)^(5/2),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 zero or nonzero?

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mupad [B]  time = 1.42, size = 321, normalized size = 3.28 \[ \frac {2\,b^3\,e^2\,\left (c\,x^2+b\,x+a\right )-2\,b^3\,c\,d^2-2\,b^4\,e^2\,x-2\,a\,b^3\,e^2-16\,a^2\,c^2\,e^2\,x-4\,b^2\,c^2\,d^2\,x+8\,a\,b\,c^2\,d^2+8\,a^2\,b\,c\,e^2-32\,a^2\,c^2\,d\,e+16\,a\,c^3\,d^2\,x+16\,b\,c^2\,d^2\,\left (c\,x^2+b\,x+a\right )+32\,c^3\,d^2\,x\,\left (c\,x^2+b\,x+a\right )+12\,a\,b^2\,c\,e^2\,x+16\,a\,c^2\,e^2\,x\,\left (c\,x^2+b\,x+a\right )+4\,b^2\,c\,e^2\,x\,\left (c\,x^2+b\,x+a\right )+8\,a\,b^2\,c\,d\,e+4\,b^3\,c\,d\,e\,x+8\,a\,b\,c\,e^2\,\left (c\,x^2+b\,x+a\right )-16\,b^2\,c\,d\,e\,\left (c\,x^2+b\,x+a\right )-16\,a\,b\,c^2\,d\,e\,x-32\,b\,c^2\,d\,e\,x\,\left (c\,x^2+b\,x+a\right )}{\left (48\,a^2\,c^3-24\,a\,b^2\,c^2+3\,b^4\,c\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + b*x + c*x^2)^(5/2),x)

[Out]

(2*b^3*e^2*(a + b*x + c*x^2) - 2*b^3*c*d^2 - 2*b^4*e^2*x - 2*a*b^3*e^2 - 16*a^2*c^2*e^2*x - 4*b^2*c^2*d^2*x +
8*a*b*c^2*d^2 + 8*a^2*b*c*e^2 - 32*a^2*c^2*d*e + 16*a*c^3*d^2*x + 16*b*c^2*d^2*(a + b*x + c*x^2) + 32*c^3*d^2*
x*(a + b*x + c*x^2) + 12*a*b^2*c*e^2*x + 16*a*c^2*e^2*x*(a + b*x + c*x^2) + 4*b^2*c*e^2*x*(a + b*x + c*x^2) +
8*a*b^2*c*d*e + 4*b^3*c*d*e*x + 8*a*b*c*e^2*(a + b*x + c*x^2) - 16*b^2*c*d*e*(a + b*x + c*x^2) - 16*a*b*c^2*d*
e*x - 32*b*c^2*d*e*x*(a + b*x + c*x^2))/((3*b^4*c + 48*a^2*c^3 - 24*a*b^2*c^2)*(a + b*x + c*x^2)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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