3.11.85 \(\int \frac {1}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1085]

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 = 0.00, 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 = {621} \begin {gather*} -\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 {gather*}

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

Rule 621

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

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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Maple [A]
time = 0.54, size = 33, normalized size = 0.80

method	result	size
risch	\(-\frac {1}{4 c^{2} \left (e x +d \right )^{3} \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(27\)
gosper	\(-\frac {e x +d}{4 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(33\)
default	\(-\frac {e x +d}{4 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(33\)
trager	\(\frac {\left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 c^{3} d^{4} \left (e x +d \right )^{5}}\)	\(68\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 15, normalized size = 0.37 \begin {gather*} -\frac {e^{\left (-5\right )}}{4 \, {\left (d e^{\left (-1\right )} + x\right )}^{4} c^{\frac {5}{2}}} \end {gather*}

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*e^(-5)/((d*e^(-1) + x)^4*c^(5/2))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (37) = 74\).
time = 1.96, size = 93, normalized size = 2.27 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{4 \, {\left (c^{3} x^{5} e^{6} + 5 \, c^{3} d x^{4} e^{5} + 10 \, c^{3} d^{2} x^{3} e^{4} + 10 \, c^{3} d^{3} x^{2} e^{3} + 5 \, c^{3} d^{4} x e^{2} + c^{3} d^{5} e\right )}} \end {gather*}

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

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

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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Giac [A]
time = 1.07, size = 24, normalized size = 0.59 \begin {gather*} -\frac {e^{\left (-1\right )}}{4 \, {\left (x e + d\right )}^{4} c^{\frac {5}{2}} \mathrm {sgn}\left (x e + d\right )} \end {gather*}

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]

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

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Mupad [B]
time = 0.50, size = 37, normalized size = 0.90 \begin {gather*} -\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} \end {gather*}

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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