3.11.53 \(\int \frac {1}{(b d+2 c d x)^4 \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=79 \[ \frac {4 \sqrt {a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac {2 \sqrt {a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 79, 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 = {693, 682} \begin {gather*} \frac {4 \sqrt {a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac {2 \sqrt {a+b x+c x^2}}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^4*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.76 \begin {gather*} \frac {2 \sqrt {a+x (b+c x)} \left (-4 c \left (a-2 c x^2\right )+3 b^2+8 b c x\right )}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^4*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 62, normalized size = 0.78 \begin {gather*} \frac {2 \sqrt {a+b x+c x^2} \left (-4 a c+3 b^2+8 b c x+8 c^2 x^2\right )}{3 d^4 \left (b^2-4 a c\right )^2 (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^4*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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fricas [B]  time = 0.80, size = 165, normalized size = 2.09 \begin {gather*} \frac {2 \, {\left (8 \, c^{2} x^{2} + 8 \, b c x + 3 \, b^{2} - 4 \, a c\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (8 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{4} x^{3} + 12 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{4} x^{2} + 6 \, {\left (b^{6} c - 8 \, a b^{4} c^{2} + 16 \, a^{2} b^{2} c^{3}\right )} d^{4} x + {\left (b^{7} - 8 \, a b^{5} c + 16 \, a^{2} b^{3} c^{2}\right )} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(8*c^2*x^2 + 8*b*c*x + 3*b^2 - 4*a*c)*sqrt(c*x^2 + b*x + a)/(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^
3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7
- 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)

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giac [A]  time = 0.34, size = 134, normalized size = 1.70 \begin {gather*} \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b c + b^{2} \sqrt {c} - a c^{\frac {3}{2}}\right )}}{3 \, {\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 d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 70, normalized size = 0.89 \begin {gather*} -\frac {2 \left (-8 c^{2} x^{2}-8 b c x +4 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{3 \left (2 c x +b \right )^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(1/(2*c*d*x+b*d)^4/(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 = 0.65, size = 60, normalized size = 0.76 \begin {gather*} \frac {2\,\sqrt {c\,x^2+b\,x+a}\,\left (3\,b^2+8\,b\,c\,x+8\,c^2\,x^2-4\,a\,c\right )}{3\,d^4\,{\left (4\,a\,c-b^2\right )}^2\,{\left (b+2\,c\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{b^{4} \sqrt {a + b x + c x^{2}} + 8 b^{3} c x \sqrt {a + b x + c x^{2}} + 24 b^{2} c^{2} x^{2} \sqrt {a + b x + c x^{2}} + 32 b c^{3} x^{3} \sqrt {a + b x + c x^{2}} + 16 c^{4} x^{4} \sqrt {a + b x + c x^{2}}}\, dx}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(b**4*sqrt(a + b*x + c*x**2) + 8*b**3*c*x*sqrt(a + b*x + c*x**2) + 24*b**2*c**2*x**2*sqrt(a + b*x +
c*x**2) + 32*b*c**3*x**3*sqrt(a + b*x + c*x**2) + 16*c**4*x**4*sqrt(a + b*x + c*x**2)), x)/d**4

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