3.183 \(\int \frac{A+C x^2}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac{2 (b+2 c x) \left (4 a C+8 A c+\frac{b^2 C}{c}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (2*(8*A
*c + 4*a*C + (b^2*C)/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 = 0.093144, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1660, 12, 613} \[ \frac{2 (b+2 c x) \left (4 a C+8 A c+\frac{b^2 C}{c}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 \left (x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (2*(8*A
*c + 4*a*C + (b^2*C)/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{A+C x^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\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 A c+4 a C+\frac{b^2 C}{c}}{2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac{2 \left (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\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 A c+4 a C+\frac{b^2 C}{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 (b c \left (A+\frac{a C}{c}\right )+\left (2 A c^2+\left (b^2-2 a c\right ) C\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 A c+4 a C+\frac{b^2 C}{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 = 0.889565, size = 107, normalized size = 0.94 \[ \frac{2 C \left (8 a^2 b+4 a x \left (3 b^2+3 b c x+2 c^2 x^2\right )+b^2 x^2 (3 b+2 c x)\right )-2 A (b+2 c x) \left (-4 c \left (3 a+2 c x^2\right )+b^2-8 b c x\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 137, normalized size = 1.2 \begin{align*}{\frac{32\,A{x}^{3}{c}^{3}+16\,Ca{c}^{2}{x}^{3}+4\,C{b}^{2}c{x}^{3}+48\,A{x}^{2}b{c}^{2}+24\,Cabc{x}^{2}+6\,C{b}^{3}{x}^{2}+48\,Aa{c}^{2}x+12\,A{b}^{2}cx+24\,Ca{b}^{2}x+24\,Aabc-2\,A{b}^{3}+16\,C{a}^{2}b}{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((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(16*A*c^3*x^3+8*C*a*c^2*x^3+2*C*b^2*c*x^3+24*A*b*c^2*x^2+12*C*a*b*c*x^2+3*C*b^3*x^2+24
*A*a*c^2*x+6*A*b^2*c*x+12*C*a*b^2*x+12*A*a*b*c-A*b^3+8*C*a^2*b)/(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((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 9.07228, size = 520, normalized size = 4.56 \begin{align*} \frac{2 \,{\left (8 \, C a^{2} b - A b^{3} + 12 \, A a b c + 2 \,{\left (C b^{2} c + 4 \, C a c^{2} + 8 \, A c^{3}\right )} x^{3} + 3 \,{\left (C b^{3} + 4 \, C a b c + 8 \, A b c^{2}\right )} x^{2} + 6 \,{\left (2 \, C a b^{2} + A b^{2} c + 4 \, A 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(8*C*a^2*b - A*b^3 + 12*A*a*b*c + 2*(C*b^2*c + 4*C*a*c^2 + 8*A*c^3)*x^3 + 3*(C*b^3 + 4*C*a*b*c + 8*A*b*c^2
)*x^2 + 6*(2*C*a*b^2 + A*b^2*c + 4*A*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + C x^{2}}{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Integral((A + C*x**2)/(a + b*x + c*x**2)**(5/2), x)

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Giac [B]  time = 1.24251, size = 293, normalized size = 2.57 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (C b^{2} c + 4 \, C a c^{2} + 8 \, A c^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (C b^{3} + 4 \, C a b c + 8 \, A b c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{6 \,{\left (2 \, C a b^{2} + A b^{2} c + 4 \, A a c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{8 \, C a^{2} b - A b^{3} + 12 \, A a b c}{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((C*x^2+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

1/3*(((2*(C*b^2*c + 4*C*a*c^2 + 8*A*c^3)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(C*b^3 + 4*C*a*b*c + 8*A*b
*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(2*C*a*b^2 + A*b^2*c + 4*A*a*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 1
6*a^2*c^4))*x + (8*C*a^2*b - A*b^3 + 12*A*a*b*c)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)