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

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Rubi [A]  time = 0.0142451, antiderivative size = 39, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0205791, 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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Maple [A]  time = 0.047, size = 38, normalized size = 1. \begin{align*}{\frac{2\,{d}^{2} \left ( 2\,cx+b \right ) ^{3}}{12\,ac-3\,{b}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.55427, size = 304, normalized size = 7.79 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 1.19467, size = 296, normalized size = 7.59 \begin{align*} -\frac{2 \,{\left (2 \,{\left (\frac{2 \,{\left (b^{2} c^{3} d^{2} - 4 \, a c^{4} d^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (b^{3} c^{2} d^{2} - 4 \, a b c^{3} d^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (b^{4} c d^{2} - 4 \, a b^{2} c^{2} d^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{b^{5} d^{2} - 4 \, a b^{3} c d^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} \end{align*}

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]

-1/3*(2*(2*(2*(b^2*c^3*d^2 - 4*a*c^4*d^2)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(b^3*c^2*d^2 - 4*a*b*c^3*
d^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 3*(b^4*c*d^2 - 4*a*b^2*c^2*d^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2
*c^4))*x + (b^5*d^2 - 4*a*b^3*c*d^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)