3.1254 \(\int \frac {(b d+2 c d x)^2}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=39 \[ -\frac {2 d^2 (b+2 c x)^3}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {682} \[ -\frac {2 d^2 (b+2 c x)^3}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 682

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.97 \[ -\frac {2 d^2 (b+2 c x)^3}{3 \left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.33, size = 146, normalized size = 3.74 \[ -\frac {2 \, {\left (8 \, c^{3} d^{2} x^{3} + 12 \, b c^{2} d^{2} x^{2} + 6 \, b^{2} c d^{2} x + b^{3} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{3} + {\left (b^{4} - 2 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.27, size = 195, normalized size = 5.00 \[ -\frac {2 \, {\left (2 \, {\left (2 \, {\left (\frac {2 \, {\left (b^{2} c^{3} d^{2} - 4 \, a c^{4} d^{2}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (b^{3} c^{2} d^{2} - 4 \, a b c^{3} d^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (b^{4} c d^{2} - 4 \, a b^{2} c^{2} d^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {b^{5} d^{2} - 4 \, a b^{3} c d^{2}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(2*(2*(2*(b^2*c^3*d^2 - 4*a*c^4*d^2)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(b^3*c^2*d^2 - 4*a*b*c^3*d^2)/(
b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 3*(b^4*c*d^2 - 4*a*b^2*c^2*d^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (b^5*d^
2 - 4*a*b^3*c*d^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 [A]  time = 0.05, size = 38, normalized size = 0.97 \[ \frac {2 \left (2 c x +b \right )^{3} d^{2}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((2*c*d*x+b*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 = 0.64, size = 67, normalized size = 1.72 \[ \frac {2\,b^3\,d^2+12\,b^2\,c\,d^2\,x+24\,b\,c^2\,d^2\,x^2+16\,c^3\,d^2\,x^3}{\left (12\,a\,c-3\,b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*d^2 + 16*c^3*d^2*x^3 + 24*b*c^2*d^2*x^2 + 12*b^2*c*d^2*x)/((12*a*c - 3*b^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((2*c*d*x+b*d)**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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