3.2385 \(\int \frac{1}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=70 \[ \frac{16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b + 2*c*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (16*c*(b + 2*c*x))/
(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.0395237, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143 \[ \frac{16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x + c*x^2)^(-5/2),x]

[Out]

(-2*(b + 2*c*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (16*c*(b + 2*c*x))/
(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

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Rubi in Sympy [A]  time = 4.98096, size = 68, normalized size = 0.97 \[ \frac{8 c \left (2 b + 4 c x\right )}{3 \left (- 4 a c + b^{2}\right )^{2} \sqrt{a + b x + c x^{2}}} - \frac{2 \left (b + 2 c x\right )}{3 \left (- 4 a c + b^{2}\right ) \left (a + b x + c x^{2}\right )^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/(c*x**2+b*x+a)**(5/2),x)

[Out]

8*c*(2*b + 4*c*x)/(3*(-4*a*c + b**2)**2*sqrt(a + b*x + c*x**2)) - 2*(b + 2*c*x)/
(3*(-4*a*c + b**2)*(a + b*x + c*x**2)**(3/2))

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Mathematica [A]  time = 0.0561161, size = 55, normalized size = 0.79 \[ -\frac{2 (b+2 c x) \left (-4 c \left (3 a+2 c x^2\right )+b^2-8 b c x\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [A]  time = 0.006, size = 78, normalized size = 1.1 \[{\frac{32\,{c}^{3}{x}^{3}+48\,b{c}^{2}{x}^{2}+48\,a{c}^{2}x+12\,{b}^{2}cx+24\,abc-2\,{b}^{3}}{48\,{a}^{2}{c}^{2}-24\,ac{b}^{2}+3\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/(c*x^2+b*x+a)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(-5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.29035, size = 259, normalized size = 3.7 \[ \frac{2 \,{\left (16 \, c^{3} x^{3} + 24 \, b c^{2} x^{2} - b^{3} + 12 \, a b c + 6 \,{\left (b^{2} c + 4 \, a c^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(-5/2),x, algorithm="fricas")

[Out]

2/3*(16*c^3*x^3 + 24*b*c^2*x^2 - b^3 + 12*a*b*c + 6*(b^2*c + 4*a*c^2)*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral((a + b*x + c*x**2)**(-5/2), x)

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GIAC/XCAS [A]  time = 0.216935, size = 194, normalized size = 2.77 \[ \frac{2 \,{\left (2 \,{\left (4 \,{\left (\frac{2 \, c^{3} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac{3 \, b c^{2}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac{3 \,{\left (b^{2} c + 4 \, a c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac{b^{3} - 12 \, a b c}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(-5/2),x, algorithm="giac")

[Out]

2/3*(2*(4*(2*c^3*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*b*c^2/(b^4 - 8*a*b^2*c + 1
6*a^2*c^2))*x + 3*(b^2*c + 4*a*c^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x - (b^3 - 1
2*a*b*c)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)