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

Optimal. Leaf size=131 \[ \frac{2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 (b+2 c x) \left (4 a f+\frac{b^2 f}{c}-4 b e+8 c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}} \]

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)
^(3/2)) + (2*(8*c*d - 4*b*e + 4*a*f + (b^2*f)/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 = 0.0852556, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1660, 12, 613} \[ \frac{2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 (b+2 c x) \left (4 a f+\frac{b^2 f}{c}-4 b e+8 c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)
^(3/2)) + (2*(8*c*d - 4*b*e + 4*a*f + (b^2*f)/c)*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{d+e x+f x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \int \frac{8 c d-4 b e+4 a f+\frac{b^2 f}{c}}{2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{\left (8 c d-4 b e+4 a f+\frac{b^2 f}{c}\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (8 c d-4 b e+4 a f+\frac{b^2 f}{c}\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.394814, size = 147, normalized size = 1.12 \[ \frac{8 b \left (2 a^2 f+3 a c \left (d-e x+f x^2\right )-2 c^2 x^2 (e x-3 d)\right )+16 c \left (-a^2 e+a c x \left (3 d+f x^2\right )+2 c^2 d x^3\right )-4 b^2 \left (a (e-6 f x)-c x \left (3 d-6 e x+f x^2\right )\right )-2 b^3 (d+3 x (e-f x))}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 185, normalized size = 1.4 \begin{align*}{\frac{16\,a{c}^{2}f{x}^{3}+4\,{b}^{2}cf{x}^{3}-16\,b{c}^{2}e{x}^{3}+32\,{c}^{3}d{x}^{3}+24\,abcf{x}^{2}+6\,{b}^{3}f{x}^{2}-24\,{b}^{2}ce{x}^{2}+48\,b{c}^{2}d{x}^{2}+24\,a{b}^{2}fx-24\,abcex+48\,a{c}^{2}dx-6\,{b}^{3}ex+12\,{b}^{2}cdx+16\,{a}^{2}bf-16\,{a}^{2}ce-4\,a{b}^{2}e+24\,cabd-2\,{b}^{3}d}{48\,{a}^{2}{c}^{2}-24\,ac{b}^{2}+3\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((f*x^2+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 = 23.2333, size = 614, normalized size = 4.69 \begin{align*} \frac{2 \,{\left (8 \, a^{2} b f + 2 \,{\left (8 \, c^{3} d - 4 \, b c^{2} e +{\left (b^{2} c + 4 \, a c^{2}\right )} f\right )} x^{3} + 3 \,{\left (8 \, b c^{2} d - 4 \, b^{2} c e +{\left (b^{3} + 4 \, a b c\right )} f\right )} x^{2} -{\left (b^{3} - 12 \, a b c\right )} d - 2 \,{\left (a b^{2} + 4 \, a^{2} c\right )} e + 3 \,{\left (4 \, a b^{2} f + 2 \,{\left (b^{2} c + 4 \, a c^{2}\right )} d -{\left (b^{3} + 4 \, a b c\right )} e\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(8*a^2*b*f + 2*(8*c^3*d - 4*b*c^2*e + (b^2*c + 4*a*c^2)*f)*x^3 + 3*(8*b*c^2*d - 4*b^2*c*e + (b^3 + 4*a*b*c
)*f)*x^2 - (b^3 - 12*a*b*c)*d - 2*(a*b^2 + 4*a^2*c)*e + 3*(4*a*b^2*f + 2*(b^2*c + 4*a*c^2)*d - (b^3 + 4*a*b*c)
*e)*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(-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((f*x**2+e*x+d)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.36368, size = 356, normalized size = 2.72 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (8 \, c^{3} d + b^{2} c f + 4 \, a c^{2} f - 4 \, b c^{2} e\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (8 \, b c^{2} d + b^{3} f + 4 \, a b c f - 4 \, b^{2} c e\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (2 \, b^{2} c d + 8 \, a c^{2} d + 4 \, a b^{2} f - b^{3} e - 4 \, a b c e\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{b^{3} d - 12 \, a b c d - 8 \, a^{2} b f + 2 \, a b^{2} e + 8 \, a^{2} 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((f*x^2+e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

1/3*(((2*(8*c^3*d + b^2*c*f + 4*a*c^2*f - 4*b*c^2*e)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(8*b*c^2*d + b
^3*f + 4*a*b*c*f - 4*b^2*c*e)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 3*(2*b^2*c*d + 8*a*c^2*d + 4*a*b^2*f -
b^3*e - 4*a*b*c*e)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - (b^3*d - 12*a*b*c*d - 8*a^2*b*f + 2*a*b^2*e + 8*
a^2*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)