3.1591 \(\int \frac{(b+2 c x) (d+e x)}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=59 \[ -\frac{4 e (b+2 c x)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 (d+e x)}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0222422, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {768, 613} \[ -\frac{4 e (b+2 c x)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 (d+e x)}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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]

Rubi steps

\begin{align*} \int \frac{(b+2 c x) (d+e x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac{1}{3} (2 e) \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 (d+e x)}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 e (b+2 c x)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.186567, size = 64, normalized size = 1.08 \[ -\frac{2 \left (2 b e \left (a+3 c x^2\right )+4 c \left (c e x^3-a d\right )+b^2 (d+3 e x)\right )}{3 \left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 67, normalized size = 1.1 \begin{align*}{\frac{8\,{c}^{2}e{x}^{3}+12\,bce{x}^{2}+6\,{b}^{2}ex+4\,abe-8\,acd+2\,{b}^{2}d}{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*x+b)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(4*c^2*e*x^3+6*b*c*e*x^2+3*b^2*e*x+2*a*b*e-4*a*c*d+b^2*d)/(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*x+b)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.43269, size = 313, normalized size = 5.31 \begin{align*} -\frac{2 \,{\left (4 \, c^{2} e x^{3} + 6 \, b c e x^{2} + 3 \, b^{2} e x + 2 \, a b e +{\left (b^{2} - 4 \, a c\right )} d\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*x+b)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(4*c^2*e*x^3 + 6*b*c*e*x^2 + 3*b^2*e*x + 2*a*b*e + (b^2 - 4*a*c)*d)*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*x+b)*(e*x+d)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.35618, size = 309, normalized size = 5.24 \begin{align*} -\frac{{\left (2 \,{\left (\frac{2 \,{\left (b^{2} c^{2} e - 4 \, a c^{3} e\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (b^{3} c e - 4 \, a b c^{2} e\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (b^{4} e - 4 \, a b^{2} c e\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d + 2 \, a b^{3} e - 8 \, a^{2} b c e}{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*x+b)*(e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

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