3.1253 \(\int \frac {(b d+2 c d x)^3}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=52 \[ -\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}}-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*d^3*(2*c*x+b)^2/(c*x^2+b*x+a)^(3/2)-16/3*c*d^3/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {686, 629} \[ -\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}}-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^3*(b + 2*c*x)^2)/(3*(a + b*x + c*x^2)^(3/2)) - (16*c*d^3)/(3*Sqrt[a + b*x + c*x^2])

Rule 629

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

Rule 686

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

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac {1}{3} \left (8 c d^2\right ) \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.81 \[ -\frac {2 d^3 \left (4 c \left (2 a+3 c x^2\right )+b^2+12 b c x\right )}{3 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 1.54, size = 83, normalized size = 1.60 \[ -\frac {2 \, {\left (12 \, c^{2} d^{3} x^{2} + 12 \, b c d^{3} x + {\left (b^{2} + 8 \, a c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 74, normalized size = 1.42 \[ -\frac {2 \, {\left (b^{2} d^{3} - 4 \, a c d^{3} + 12 \, {\left (c d x^{2} + b d x + a d\right )} c d^{2}\right )} d}{3 \, {\left (c d x^{2} + b d x + a d\right )} \sqrt {\frac {c d x^{2} + b d x + a d}{d}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 39, normalized size = 0.75 \[ -\frac {2 \left (12 c^{2} x^{2}+12 b c x +8 a c +b^{2}\right ) d^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3/(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 = 0.70, size = 62, normalized size = 1.19 \[ -\frac {2\,b^2\,d^3+24\,c\,d^3\,\left (c\,x^2+b\,x+a\right )-8\,a\,c\,d^3}{\sqrt {c\,x^2+b\,x+a}\,\left (3\,c\,x^2+3\,b\,x+3\,a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x)

[Out]

-(2*b^2*d^3 + 24*c*d^3*(a + b*x + c*x^2) - 8*a*c*d^3)/((a + b*x + c*x^2)^(1/2)*(3*a + 3*b*x + 3*c*x^2))

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sympy [B]  time = 1.75, size = 264, normalized size = 5.08 \[ - \frac {16 a c d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {2 b^{2} d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 b c d^{3} x}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 c^{2} d^{3} x^{2}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(5/2),x)

[Out]

-16*a*c*d**3/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2)) - 2
*b**2*d**3/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2)) - 24*
b*c*d**3*x/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2)) - 24*
c**2*d**3*x**2/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2))

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