3.1085 \(\int \frac {1}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac {1}{4 c e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {607} \[ -\frac {1}{4 c e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-5/2),x]

[Out]

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

Rule 607

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 25, normalized size = 0.61 \[ -\frac {d+e x}{4 e \left (c (d+e x)^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-5/2),x]

[Out]

-1/4*(d + e*x)/(e*(c*(d + e*x)^2)^(5/2))

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fricas [B]  time = 1.02, size = 97, normalized size = 2.37 \[ -\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \, {\left (c^{3} e^{6} x^{5} + 5 \, c^{3} d e^{5} x^{4} + 10 \, c^{3} d^{2} e^{4} x^{3} + 10 \, c^{3} d^{3} e^{3} x^{2} + 5 \, c^{3} d^{4} e^{2} x + c^{3} d^{5} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(c^3*e^6*x^5 + 5*c^3*d*e^5*x^4 + 10*c^3*d^2*e^4*x^3 + 10*c^3*d^3*e^3*
x^2 + 5*c^3*d^4*e^2*x + c^3*d^5*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 33, normalized size = 0.80 \[ -\frac {e x +d}{4 \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x)

[Out]

-1/4*(e*x+d)/e/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)

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maxima [A]  time = 1.35, size = 17, normalized size = 0.41 \[ -\frac {1}{4 \, c^{\frac {5}{2}} e^{5} {\left (x + \frac {d}{e}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-1/4/(c^(5/2)*e^5*(x + d/e)^4)

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mupad [B]  time = 0.50, size = 37, normalized size = 0.90 \[ -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{4\,c^3\,e\,{\left (d+e\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(5/2),x)

[Out]

-(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)/(4*c^3*e*(d + e*x)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Integral((c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(-5/2), x)

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