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

Optimal. Leaf size=94 \[ \frac {8 (2 a+b x) (A b-2 a B)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 x^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {804, 636} \begin {gather*} \frac {8 (2 a+b x) (A b-2 a B)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 x^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*(A*b - 2*a*B - (b*B - 2*A*c)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (8*(A*b - 2*a*B)*(2*a + b
*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

Rule 636

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

Rule 804

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(b*f - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(m
*(b*(e*f + d*g) - 2*(c*d*f + a*e*g)))/((p + 1)*(b^2 - 4*a*c)), 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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 110, normalized size = 1.17 \begin {gather*} \frac {2 \left (-16 a^3 B+8 a^2 (A b-3 B x (b+c x))+2 a x \left (A \left (6 b^2+6 b c x+4 c^2 x^2\right )-3 b B x (b+2 c x)\right )+b^2 x^2 (3 A b+2 A c x+b B x)\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.18, size = 134, normalized size = 1.43 \begin {gather*} -\frac {2 \left (16 a^3 B-8 a^2 A b+24 a^2 b B x+24 a^2 B c x^2-12 a A b^2 x-12 a A b c x^2-8 a A c^2 x^3+6 a b^2 B x^2+12 a b B c x^3-3 A b^3 x^2-2 A b^2 c x^3-b^3 B x^3\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-8*a^2*A*b + 16*a^3*B - 12*a*A*b^2*x + 24*a^2*b*B*x - 3*A*b^3*x^2 + 6*a*b^2*B*x^2 - 12*a*A*b*c*x^2 + 24*a
^2*B*c*x^2 - b^3*B*x^3 - 2*A*b^2*c*x^3 + 12*a*b*B*c*x^3 - 8*a*A*c^2*x^3))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)
^(3/2))

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fricas [B]  time = 1.16, size = 248, normalized size = 2.64 \begin {gather*} -\frac {2 \, {\left (16 \, B a^{3} - 8 \, A a^{2} b - {\left (B b^{3} + 8 \, A a c^{2} - 2 \, {\left (6 \, B a b - A b^{2}\right )} c\right )} x^{3} + 3 \, {\left (2 \, B a b^{2} - A b^{3} + 4 \, {\left (2 \, B a^{2} - A a b\right )} c\right )} x^{2} + 12 \, {\left (2 \, B a^{2} b - A a b^{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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(16*B*a^3 - 8*A*a^2*b - (B*b^3 + 8*A*a*c^2 - 2*(6*B*a*b - A*b^2)*c)*x^3 + 3*(2*B*a*b^2 - A*b^3 + 4*(2*B*a
^2 - A*a*b)*c)*x^2 + 12*(2*B*a^2*b - A*a*b^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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giac [B]  time = 0.26, size = 195, normalized size = 2.07 \begin {gather*} \frac {2 \, {\left ({\left ({\left (\frac {{\left (B b^{3} - 12 \, B a b c + 2 \, A b^{2} c + 8 \, A a c^{2}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} - \frac {3 \, {\left (2 \, B a b^{2} - A b^{3} + 8 \, B a^{2} c - 4 \, A a b c\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {12 \, {\left (2 \, B a^{2} b - A a b^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {8 \, {\left (2 \, B a^{3} - A a^{2} b\right )}}{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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((((B*b^3 - 12*B*a*b*c + 2*A*b^2*c + 8*A*a*c^2)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) - 3*(2*B*a*b^2 - A*b^3 +
8*B*a^2*c - 4*A*a*b*c)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x - 12*(2*B*a^2*b - A*a*b^2)/(b^4 - 8*a*b^2*c + 16*a^2*
c^2))*x - 8*(2*B*a^3 - A*a^2*b)/(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 = 0.06, size = 141, normalized size = 1.50 \begin {gather*} \frac {\frac {16}{3} A a \,c^{2} x^{3}+\frac {4}{3} A \,b^{2} c \,x^{3}-8 B a b c \,x^{3}+\frac {2}{3} B \,b^{3} x^{3}+8 A a b c \,x^{2}+2 A \,b^{3} x^{2}-16 B \,a^{2} c \,x^{2}-4 B a \,b^{2} x^{2}+8 A a \,b^{2} x -16 B \,a^{2} b x +\frac {16}{3} A \,a^{2} b -\frac {32}{3} B \,a^{3}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(8*A*a*c^2*x^3+2*A*b^2*c*x^3-12*B*a*b*c*x^3+B*b^3*x^3+12*A*a*b*c*x^2+3*A*b^3*x^2-24*B*
a^2*c*x^2-6*B*a*b^2*x^2+12*A*a*b^2*x-24*B*a^2*b*x+8*A*a^2*b-16*B*a^3)/(16*a^2*c^2-8*a*b^2*c+b^4)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)/(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 = 1.73, size = 131, normalized size = 1.39 \begin {gather*} \frac {2\,\left (-16\,B\,a^3-24\,B\,a^2\,b\,x+8\,A\,a^2\,b-24\,B\,a^2\,c\,x^2-6\,B\,a\,b^2\,x^2+12\,A\,a\,b^2\,x-12\,B\,a\,b\,c\,x^3+12\,A\,a\,b\,c\,x^2+8\,A\,a\,c^2\,x^3+B\,b^3\,x^3+3\,A\,b^3\,x^2+2\,A\,b^2\,c\,x^3\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(3*A*b^3*x^2 - 16*B*a^3 + B*b^3*x^3 + 8*A*a^2*b + 12*A*a*b^2*x - 24*B*a^2*b*x - 6*B*a*b^2*x^2 + 8*A*a*c^2*x
^3 - 24*B*a^2*c*x^2 + 2*A*b^2*c*x^3 + 12*A*a*b*c*x^2 - 12*B*a*b*c*x^3))/(3*(4*a*c - b^2)^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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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