3.2396 \(\int \frac {1}{(a+b x+c x^2)^{5/2}} \, dx\)

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]

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

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Rubi [A]  time = 0.01, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {614, 613} \[ \frac {16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

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])

Rule 613

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

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {(8 c) \int \frac {1}{\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)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {16 c (b+2 c 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 = 0.03, size = 57, normalized size = 0.81 \[ \frac {2 (b+2 c x) \left (4 c \left (3 a+2 c x^2\right )-b^2+8 b c x\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

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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fricas [B]  time = 1.20, size = 192, normalized size = 2.74 \[ \frac {2 \, {\left (16 \, c^{3} x^{3} + 24 \, b c^{2} x^{2} - b^{3} + 12 \, a b c + 6 \, {\left (b^{2} c + 4 \, a c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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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giac [B]  time = 0.30, size = 144, normalized size = 2.06 \[ \frac {2 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, c^{3} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, b c^{2}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (b^{2} c + 4 \, a c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {b^{3} - 12 \, a b c}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 + 16*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 - 12*a*b*c)/(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 [A]  time = 0.06, size = 78, normalized size = 1.11 \[ \frac {\frac {32}{3} c^{3} x^{3}+16 b \,c^{2} x^{2}+16 a \,c^{2} x +4 b^{2} c x +8 a b c -\frac {2}{3} b^{3}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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(-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(1/(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 = 1.04, size = 57, normalized size = 0.81 \[ \frac {\left (2\,b+4\,c\,x\right )\,\left (-b^2+8\,b\,c\,x+8\,c^2\,x^2+12\,a\,c\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x + c*x^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \]

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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