3.1235 \(\int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=59 \[ \frac {4}{3} d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}+\frac {2}{3} d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2} \]

[Out]

4/3*(-4*a*c+b^2)*d^3*(c*x^2+b*x+a)^(1/2)+2/3*d^3*(2*c*x+b)^2*(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 59, 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 = {692, 629} \[ \frac {4}{3} d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}+\frac {2}{3} d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 692

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.73 \[ \frac {2}{3} d^3 \sqrt {a+x (b+c x)} \left (4 c \left (c x^2-2 a\right )+3 b^2+4 b c x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.32, size = 48, normalized size = 0.81 \[ \frac {2}{3} \, {\left (4 \, c^{2} d^{3} x^{2} + 4 \, b c d^{3} x + {\left (3 \, b^{2} - 8 \, a c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 58, normalized size = 0.98 \[ 2 \, \sqrt {c x^{2} + b x + a} b^{2} d^{3} + \frac {8}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} c d^{3} - 8 \, \sqrt {c x^{2} + b x + a} a c d^{3} \]

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 41, normalized size = 0.69 \[ -\frac {2 \left (-4 c^{2} x^{2}-4 b c x +8 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{3}}{3} \]

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)^(1/2),x)

[Out]

-2/3*d^3*(-4*c^2*x^2-4*b*c*x+8*a*c-3*b^2)*(c*x^2+b*x+a)^(1/2)

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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)^(1/2),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 = 0.58, size = 48, normalized size = 0.81 \[ \left (\frac {8\,c^2\,d^3\,x^2}{3}-\frac {2\,d^3\,\left (8\,a\,c-3\,b^2\right )}{3}+\frac {8\,b\,c\,d^3\,x}{3}\right )\,\sqrt {c\,x^2+b\,x+a} \]

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)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.37, size = 97, normalized size = 1.64 \[ - \frac {16 a c d^{3} \sqrt {a + b x + c x^{2}}}{3} + 2 b^{2} d^{3} \sqrt {a + b x + c x^{2}} + \frac {8 b c d^{3} x \sqrt {a + b x + c x^{2}}}{3} + \frac {8 c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{3} \]

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)**(1/2),x)

[Out]

-16*a*c*d**3*sqrt(a + b*x + c*x**2)/3 + 2*b**2*d**3*sqrt(a + b*x + c*x**2) + 8*b*c*d**3*x*sqrt(a + b*x + c*x**
2)/3 + 8*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/3

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