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

Optimal. Leaf size=91 \[ \frac {8 (b+2 c x) (2 c d-b e)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 91, 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) (2 c d-b e)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (-2 a e+x (2 c d-b e)+b d)}{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[(d + e*x)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (8*(2*c*d - b*e)*(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 {d+e x}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {(4 (2 c d-b e)) \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {8 (2 c d-b e) (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 = 0.12, size = 110, normalized size = 1.21 \begin {gather*} -\frac {2 \left (8 c \left (a^2 e-3 a c d x-2 c^2 d x^3\right )+2 b^2 (a e+3 c x (2 e x-d))+4 b c \left (2 c x^2 (e x-3 d)-3 a (d-e x)\right )+b^3 (d+3 e 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[(d + e*x)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b^3*(d + 3*e*x) + 8*c*(a^2*e - 3*a*c*d*x - 2*c^2*d*x^3) + 4*b*c*(-3*a*(d - e*x) + 2*c*x^2*(-3*d + e*x)) +
2*b^2*(a*e + 3*c*x*(-d + 2*e*x))))/(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 = 1.35 \begin {gather*} -\frac {2 \left (8 a^2 c e+2 a b^2 e-12 a b c d+12 a b c e x-24 a c^2 d x+b^3 d+3 b^3 e x-6 b^2 c d x+12 b^2 c e x^2-24 b c^2 d x^2+8 b c^2 e x^3-16 c^3 d 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[(d + e*x)/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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fricas [B]  time = 1.09, size = 247, normalized size = 2.71 \begin {gather*} \frac {2 \, {\left (8 \, {\left (2 \, c^{3} d - b c^{2} e\right )} x^{3} + 12 \, {\left (2 \, b c^{2} d - b^{2} c e\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 (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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(8*(2*c^3*d - b*c^2*e)*x^3 + 12*(2*b*c^2*d - b^2*c*e)*x^2 - (b^3 - 12*a*b*c)*d - 2*(a*b^2 + 4*a^2*c)*e + 3
*(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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giac [B]  time = 0.31, size = 203, normalized size = 2.23 \begin {gather*} \frac {2 \, {\left ({\left (4 \, {\left (\frac {2 \, {\left (2 \, c^{3} d - b c^{2} e\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (2 \, b c^{2} d - b^{2} c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (2 \, b^{2} c d + 8 \, a c^{2} d - b^{3} e - 4 \, a b c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {b^{3} d - 12 \, a b c d + 2 \, a b^{2} e + 8 \, a^{2} c e}{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((e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*((4*(2*(2*c^3*d - b*c^2*e)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(2*b*c^2*d - b^2*c*e)/(b^4 - 8*a*b^2*c + 1
6*a^2*c^2))*x + 3*(2*b^2*c*d + 8*a*c^2*d - b^3*e - 4*a*b*c*e)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x - (b^3*d - 12*
a*b*c*d + 2*a*b^2*e + 8*a^2*c*e)/(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 = 0.04, size = 131, normalized size = 1.44 \begin {gather*} -\frac {2 \left (8 b \,c^{2} e \,x^{3}-16 c^{3} d \,x^{3}+12 b^{2} c e \,x^{2}-24 b \,c^{2} d \,x^{2}+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} c e +2 a \,b^{2} e -12 a b c d +b^{3} d \right )}{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((e*x+d)/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3/(c*x^2+b*x+a)^(3/2)*(8*b*c^2*e*x^3-16*c^3*d*x^3+12*b^2*c*e*x^2-24*b*c^2*d*x^2+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*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 = 0.00, 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((e*x+d)/(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 = 1.30, size = 121, normalized size = 1.33 \begin {gather*} -\frac {2\,\left (8\,e\,a^2\,c+2\,e\,a\,b^2+12\,e\,a\,b\,c\,x-12\,d\,a\,b\,c-24\,d\,a\,c^2\,x+3\,e\,b^3\,x+d\,b^3+12\,e\,b^2\,c\,x^2-6\,d\,b^2\,c\,x+8\,e\,b\,c^2\,x^3-24\,d\,b\,c^2\,x^2-16\,d\,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((d + e*x)/(a + b*x + c*x^2)^(5/2),x)

[Out]

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

[Out]

Timed out

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