3.11.10 \(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^4} \, dx\)

Optimal. Leaf size=39 \[ \frac {2 \left (a+b x+c x^2\right )^{3/2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \]

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {682} \begin {gather*} \frac {2 \left (a+b x+c x^2\right )^{3/2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 682

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.97 \begin {gather*} \frac {2 (a+x (b+c x))^{3/2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.49, size = 39, normalized size = 1.00 \begin {gather*} \frac {2 \left (a+b x+c x^2\right )^{3/2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.71, size = 98, normalized size = 2.51 \begin {gather*} \frac {2 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{3 \, {\left (8 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{4} x^{3} + 12 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{4} x^{2} + 6 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} d^{4} x + {\left (b^{5} - 4 \, a b^{3} c\right )} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^4,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.33, size = 205, normalized size = 5.26 \begin {gather*} \frac {12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} c^{\frac {5}{2}} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b c^{2} + 18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{2} c^{\frac {3}{2}} + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{3} c + b^{4} \sqrt {c} - 2 \, a b^{2} c^{\frac {3}{2}} + 4 \, a^{2} c^{\frac {5}{2}}}{12 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b \sqrt {c} + b^{2} - 2 \, a c\right )}^{3} c^{2} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^4,x, algorithm="giac")

[Out]

1/12*(12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*c^(5/2) + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c^2 + 18*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*c^(3/2) + 6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^3*c + b^4*sqrt(c) -
2*a*b^2*c^(3/2) + 4*a^2*c^(5/2))/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))*b*sqrt(c) + b^2 - 2*a*c)^3*c^2*d^4)

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maple [A]  time = 0.06, size = 38, normalized size = 0.97 \begin {gather*} -\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 \left (2 c x +b \right )^{3} \left (4 a c -b^{2}\right ) d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^4,x)

[Out]

-2/3*(c*x^2+b*x+a)^(3/2)/(2*c*x+b)^3/d^4/(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((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^4,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 = 0.75, size = 111, normalized size = 2.85 \begin {gather*} \frac {2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x\,\left (18\,b^4\,c\,d^4-72\,a\,b^2\,c^2\,d^4\right )-x^3\,\left (96\,a\,c^4\,d^4-24\,b^2\,c^3\,d^4\right )+3\,b^5\,d^4+x^2\,\left (36\,b^3\,c^2\,d^4-144\,a\,b\,c^3\,d^4\right )-12\,a\,b^3\,c\,d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^4,x)

[Out]

(2*(a + b*x + c*x^2)^(3/2))/(x*(18*b^4*c*d^4 - 72*a*b^2*c^2*d^4) - x^3*(96*a*c^4*d^4 - 24*b^2*c^3*d^4) + 3*b^5
*d^4 + x^2*(36*b^3*c^2*d^4 - 144*a*b*c^3*d^4) - 12*a*b^3*c*d^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**4,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x)/d*
*4

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