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

Optimal. Leaf size=59 \[ \frac {4}{15} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2} \]

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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} \begin {gather*} \frac {4}{15} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2} \end {gather*}

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*(a + b*x + c*x^2)^(3/2))/15 + (2*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2))/5

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 (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx &=\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {1}{5} \left (2 \left (b^2-4 a c\right ) d^2\right ) \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx\\ &=\frac {4}{15} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 44, normalized size = 0.75 \begin {gather*} \frac {2}{15} d^3 (a+x (b+c x))^{3/2} \left (4 c \left (3 c x^2-2 a\right )+5 b^2+12 b c x\right ) \end {gather*}

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*(a + x*(b + c*x))^(3/2)*(5*b^2 + 12*b*c*x + 4*c*(-2*a + 3*c*x^2)))/15

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IntegrateAlgebraic [A]  time = 0.50, size = 102, normalized size = 1.73 \begin {gather*} -\frac {2}{15} \sqrt {a+b x+c x^2} \left (8 a^2 c d^3-5 a b^2 d^3-4 a b c d^3 x-4 a c^2 d^3 x^2-5 b^3 d^3 x-17 b^2 c d^3 x^2-24 b c^2 d^3 x^3-12 c^3 d^3 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x + c*x^2]*(-5*a*b^2*d^3 + 8*a^2*c*d^3 - 5*b^3*d^3*x - 4*a*b*c*d^3*x - 17*b^2*c*d^3*x^2 - 4*a*c
^2*d^3*x^2 - 24*b*c^2*d^3*x^3 - 12*c^3*d^3*x^4))/15

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fricas [A]  time = 0.42, size = 91, normalized size = 1.54 \begin {gather*} \frac {2}{15} \, {\left (12 \, c^{3} d^{3} x^{4} + 24 \, b c^{2} d^{3} x^{3} + {\left (17 \, b^{2} c + 4 \, a c^{2}\right )} d^{3} x^{2} + {\left (5 \, b^{3} + 4 \, a b c\right )} d^{3} x + {\left (5 \, a b^{2} - 8 \, a^{2} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \end {gather*}

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]

2/15*(12*c^3*d^3*x^4 + 24*b*c^2*d^3*x^3 + (17*b^2*c + 4*a*c^2)*d^3*x^2 + (5*b^3 + 4*a*b*c)*d^3*x + (5*a*b^2 -
8*a^2*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 \begin {gather*} \frac {2}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} b^{2} d^{3} + \frac {8}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} c d^{3} - \frac {8}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} a c d^{3} \end {gather*}

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]

2/3*(c*x^2 + b*x + a)^(3/2)*b^2*d^3 + 8/5*(c*x^2 + b*x + a)^(5/2)*c*d^3 - 8/3*(c*x^2 + b*x + a)^(3/2)*a*c*d^3

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maple [A]  time = 0.05, size = 41, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (-12 c^{2} x^{2}-12 b c x +8 a c -5 b^{2}\right ) d^{3}}{15} \end {gather*}

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]

-2/15*(c*x^2+b*x+a)^(3/2)*(-12*c^2*x^2-12*b*c*x+8*a*c-5*b^2)*d^3

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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*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 positive, negative or zero?

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mupad [B]  time = 0.58, size = 89, normalized size = 1.51 \begin {gather*} \sqrt {c\,x^2+b\,x+a}\,\left (\frac {8\,c^3\,d^3\,x^4}{5}-\frac {2\,a\,d^3\,\left (8\,a\,c-5\,b^2\right )}{15}+\frac {2\,c\,d^3\,x^2\,\left (17\,b^2+4\,a\,c\right )}{15}+\frac {16\,b\,c^2\,d^3\,x^3}{5}+\frac {2\,b\,d^3\,x\,\left (5\,b^2+4\,a\,c\right )}{15}\right ) \end {gather*}

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]

(a + b*x + c*x^2)^(1/2)*((8*c^3*d^3*x^4)/5 - (2*a*d^3*(8*a*c - 5*b^2))/15 + (2*c*d^3*x^2*(4*a*c + 17*b^2))/15
+ (16*b*c^2*d^3*x^3)/5 + (2*b*d^3*x*(4*a*c + 5*b^2))/15)

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sympy [B]  time = 0.40, size = 216, normalized size = 3.66 \begin {gather*} - \frac {16 a^{2} c d^{3} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 a b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{3} + \frac {8 a b c d^{3} x \sqrt {a + b x + c x^{2}}}{15} + \frac {8 a c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{3} + \frac {34 b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {16 b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {8 c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{5} \end {gather*}

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**2*c*d**3*sqrt(a + b*x + c*x**2)/15 + 2*a*b**2*d**3*sqrt(a + b*x + c*x**2)/3 + 8*a*b*c*d**3*x*sqrt(a + b
*x + c*x**2)/15 + 8*a*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/15 + 2*b**3*d**3*x*sqrt(a + b*x + c*x**2)/3 + 34*b
**2*c*d**3*x**2*sqrt(a + b*x + c*x**2)/15 + 16*b*c**2*d**3*x**3*sqrt(a + b*x + c*x**2)/5 + 8*c**3*d**3*x**4*sq
rt(a + b*x + c*x**2)/5

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