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

Optimal. Leaf size=74 \[ -\frac {8 e (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 (d+e x)^2}{3 \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {768, 636} \begin {gather*} -\frac {8 e (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 (d+e x)^2}{3 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2)/(3*(a + b*x + c*x^2)^(3/2)) - (8*e*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*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 768

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

Rubi steps

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

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Mathematica [A]  time = 0.79, size = 110, normalized size = 1.49 \begin {gather*} \frac {2 \left (8 a^2 e^2+4 b e \left (c x^2 (e x-3 d)-a (d-3 e x)\right )+4 a c \left (d^2+3 e^2 x^2\right )-b^2 \left (d^2+6 d e x-3 e^2 x^2\right )-8 c^2 d e x^3\right )}{3 \left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.61, size = 123, normalized size = 1.66 \begin {gather*} -\frac {2 \left (-8 a^2 e^2+4 a b d e-12 a b e^2 x-4 a c d^2-12 a c e^2 x^2+b^2 d^2+6 b^2 d e x-3 b^2 e^2 x^2+12 b c d e x^2-4 b c e^2 x^3+8 c^2 d e x^3\right )}{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]

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

[Out]

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

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fricas [B]  time = 1.26, size = 194, normalized size = 2.62 \begin {gather*} -\frac {2 \, {\left (4 \, a b d e - 8 \, a^{2} e^{2} + 4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} x^{3} + {\left (b^{2} - 4 \, a c\right )} d^{2} + 3 \, {\left (4 \, b c d e - {\left (b^{2} + 4 \, a c\right )} e^{2}\right )} x^{2} + 6 \, {\left (b^{2} d e - 2 \, a b e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{3} + {\left (b^{4} - 2 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.32, size = 289, normalized size = 3.91 \begin {gather*} -\frac {2 \, {\left ({\left ({\left (\frac {4 \, {\left (2 \, b^{2} c^{2} d e - 8 \, a c^{3} d e - b^{3} c e^{2} + 4 \, a b c^{2} e^{2}\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (4 \, b^{3} c d e - 16 \, a b c^{2} d e - b^{4} e^{2} + 16 \, a^{2} c^{2} e^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {6 \, {\left (b^{4} d e - 4 \, a b^{2} c d e - 2 \, a b^{3} e^{2} + 8 \, a^{2} b c e^{2}\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2} + 4 \, a b^{3} d e - 16 \, a^{2} b c d e - 8 \, a^{2} b^{2} e^{2} + 32 \, a^{3} c e^{2}}{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((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/3*(((4*(2*b^2*c^2*d*e - 8*a*c^3*d*e - b^3*c*e^2 + 4*a*b*c^2*e^2)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(4*b^
3*c*d*e - 16*a*b*c^2*d*e - b^4*e^2 + 16*a^2*c^2*e^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 6*(b^4*d*e - 4*a*b^2*
c*d*e - 2*a*b^3*e^2 + 8*a^2*b*c*e^2)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + (b^4*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2
*d^2 + 4*a*b^3*d*e - 16*a^2*b*c*d*e - 8*a^2*b^2*e^2 + 32*a^3*c*e^2)/(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 = 123, normalized size = 1.66 \begin {gather*} -\frac {2 \left (4 b c \,e^{2} x^{3}-8 c^{2} d e \,x^{3}+12 a c \,e^{2} x^{2}+3 b^{2} e^{2} x^{2}-12 b c d e \,x^{2}+12 a b \,e^{2} x -6 b^{2} d e x +8 a^{2} e^{2}-4 a b d e +4 a c \,d^{2}-b^{2} d^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (4 a c -b^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((2*c*x+b)*(e*x+d)^2/(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 = 2.27, size = 122, normalized size = 1.65 \begin {gather*} -\frac {2\,\left (8\,a^2\,e^2-4\,a\,b\,d\,e+12\,a\,b\,e^2\,x+4\,a\,c\,d^2+12\,a\,c\,e^2\,x^2-b^2\,d^2-6\,b^2\,d\,e\,x+3\,b^2\,e^2\,x^2-12\,b\,c\,d\,e\,x^2+4\,b\,c\,e^2\,x^3-8\,c^2\,d\,e\,x^3\right )}{3\,\left (4\,a\,c-b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(8*a^2*e^2 - b^2*d^2 + 3*b^2*e^2*x^2 + 4*a*c*d^2 + 12*a*b*e^2*x - 6*b^2*d*e*x + 12*a*c*e^2*x^2 + 4*b*c*e^2
*x^3 - 8*c^2*d*e*x^3 - 4*a*b*d*e - 12*b*c*d*e*x^2))/(3*(4*a*c - b^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((2*c*x+b)*(e*x+d)**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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