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

Optimal. Leaf size=90 \[ -\frac {8 (b+2 c x) (b B-2 A c)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {638, 613} \begin {gather*} -\frac {8 (b+2 c x) (b B-2 A c)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {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[(A + B*x)/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {A+B x}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {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 (b B-2 A c)) \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {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 (b B-2 A c) (b+2 c 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 = 0.10, size = 99, normalized size = 1.10 \begin {gather*} -\frac {2 \left (B \left (8 a^2 c+2 a b (b+6 c x)+b x \left (3 b^2+12 b c x+8 c^2 x^2\right )\right )+A (b+2 c x) \left (-4 c \left (3 a+2 c x^2\right )+b^2-8 b c x\right )\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[(A + B*x)/(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)) + B*(8*a^2*c + 2*a*b*(b + 6*c*x) + b*x*(3*b^2 + 12*b*
c*x + 8*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 = 0.88, size = 123, normalized size = 1.37 \begin {gather*} -\frac {2 \left (8 a^2 B c-12 a A b c-24 a A c^2 x+2 a b^2 B+12 a b B c x+A b^3-6 A b^2 c x-24 A b c^2 x^2-16 A c^3 x^3+3 b^3 B x+12 b^2 B c x^2+8 b B c^2 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[(A + B*x)/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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fricas [B]  time = 1.14, size = 245, normalized size = 2.72 \begin {gather*} -\frac {2 \, {\left (2 \, B a b^{2} + A b^{3} + 8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{3} + 12 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{2} + 4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c + 3 \, {\left (B b^{3} - 8 \, A a c^{2} + 2 \, {\left (2 \, B a b - A b^{2}\right )} c\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((B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(2*B*a*b^2 + A*b^3 + 8*(B*b*c^2 - 2*A*c^3)*x^3 + 12*(B*b^2*c - 2*A*b*c^2)*x^2 + 4*(2*B*a^2 - 3*A*a*b)*c +
3*(B*b^3 - 8*A*a*c^2 + 2*(2*B*a*b - A*b^2)*c)*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 = 0.31, size = 193, normalized size = 2.14 \begin {gather*} -\frac {2 \, {\left ({\left (4 \, {\left (\frac {2 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (B b^{3} + 4 \, B a b c - 2 \, A b^{2} c - 8 \, A a c^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {2 \, B a b^{2} + A b^{3} + 8 \, B a^{2} c - 12 \, A a b c}{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((B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/3*((4*(2*(B*b*c^2 - 2*A*c^3)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(B*b^2*c - 2*A*b*c^2)/(b^4 - 8*a*b^2*c +
16*a^2*c^2))*x + 3*(B*b^3 + 4*B*a*b*c - 2*A*b^2*c - 8*A*a*c^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (2*B*a*b^2
+ A*b^3 + 8*B*a^2*c - 12*A*a*b*c)/(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 = 132, normalized size = 1.47 \begin {gather*} \frac {\frac {32}{3} A \,c^{3} x^{3}-\frac {16}{3} B b \,c^{2} x^{3}+16 A b \,c^{2} x^{2}-8 B \,b^{2} c \,x^{2}+16 A a \,c^{2} x +4 A \,b^{2} c x -8 B a b c x -2 B \,b^{3} x +8 A a b c -\frac {2}{3} A \,b^{3}-\frac {16}{3} B \,a^{2} c -\frac {4}{3} B a \,b^{2}}{\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((B*x+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*B*b*c^2*x^3+24*A*b*c^2*x^2-12*B*b^2*c*x^2+24*A*a*c^2*x+6*A*b^2*c*x-12*
B*a*b*c*x-3*B*b^3*x+12*A*a*b*c-A*b^3-8*B*a^2*c-2*B*a*b^2)/(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((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 = 1.52, size = 121, normalized size = 1.34 \begin {gather*} -\frac {2\,\left (8\,B\,a^2\,c+2\,B\,a\,b^2+12\,B\,a\,b\,c\,x-12\,A\,a\,b\,c-24\,A\,a\,c^2\,x+3\,B\,b^3\,x+A\,b^3+12\,B\,b^2\,c\,x^2-6\,A\,b^2\,c\,x+8\,B\,b\,c^2\,x^3-24\,A\,b\,c^2\,x^2-16\,A\,c^3\,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((A + B*x)/(a + b*x + c*x^2)^(5/2),x)

[Out]

-(2*(A*b^3 - 16*A*c^3*x^3 + 2*B*a*b^2 + 8*B*a^2*c + 3*B*b^3*x - 24*A*a*c^2*x - 6*A*b^2*c*x - 24*A*b*c^2*x^2 +
12*B*b^2*c*x^2 + 8*B*b*c^2*x^3 - 12*A*a*b*c + 12*B*a*b*c*x))/(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((B*x+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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