3.17.45 \(\int \frac {1}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=62 \[ -\frac {2 \left (a e^2+c d^2+2 c d e x\right )}{\left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {613} \begin {gather*} -\frac {2 \left (a e^2+c d^2+2 c d e x\right )}{\left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c*d^2 + a*e^2 + 2*c*d*e*x))/((c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.79 \begin {gather*} -\frac {2 \left (a e^2+c d (d+2 e x)\right )}{\left (c d^2-a e^2\right )^2 \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a*e^2 + c*d*(d + 2*e*x)))/((c*d^2 - a*e^2)^2*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [A]  time = 0.58, size = 79, normalized size = 1.27 \begin {gather*} -\frac {2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{(d+e x) \left (a e^2-c d^2\right )^2 (a e+c d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(-3/2),x]

[Out]

(-2*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/((-(c*d^2) + a*e^2)^2*(a*e + c*d*
x)*(d + e*x))

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fricas [B]  time = 1.04, size = 153, normalized size = 2.47 \begin {gather*} -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

-2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^
3*d*e^5 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6
)*x)

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giac [A]  time = 0.50, size = 99, normalized size = 1.60 \begin {gather*} -\frac {2 \, {\left (\frac {2 \, c d x e}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {c d^{2} + a e^{2}}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}}\right )}}{\sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 86, normalized size = 1.39 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (e x +d \right ) \left (2 c d e x +a \,e^{2}+c \,d^{2}\right )}{\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2),x)

[Out]

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

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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(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [B]  time = 0.61, size = 75, normalized size = 1.21 \begin {gather*} -\frac {\frac {c\,d^2}{2}+c\,x\,d\,e+\frac {a\,e^2}{2}}{\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{4}-a\,c\,d^2\,e^2\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a d e + c d e x^{2} + x \left (a e^{2} + c d^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral((a*d*e + c*d*e*x**2 + x*(a*e**2 + c*d**2))**(-3/2), x)

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