3.11.3 \(\int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx\) [1003]

Optimal. Leaf size=15 \[ -\frac {1}{c e (d+e x)} \]

[Out]

-1/c/e/(e*x+d)

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {27, 12, 32} \begin {gather*} -\frac {1}{c e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-1),x]

[Out]

-(1/(c*e*(d + e*x)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} -\frac {1}{c e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-1),x]

[Out]

-(1/(c*e*(d + e*x)))

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Maple [A]
time = 0.53, size = 16, normalized size = 1.07

method	result	size
gosper	\(-\frac {1}{c e \left (e x +d \right )}\)	\(16\)
default	\(-\frac {1}{c e \left (e x +d \right )}\)	\(16\)
norman	\(\frac {x}{c d \left (e x +d \right )}\)	\(16\)
risch	\(-\frac {1}{c e \left (e x +d \right )}\)	\(16\)

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),x,method=_RETURNVERBOSE)

[Out]

-1/c/e/(e*x+d)

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Maxima [A]
time = 0.27, size = 15, normalized size = 1.00 \begin {gather*} -\frac {1}{c x e^{2} + c d e} \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),x, algorithm="maxima")

[Out]

-1/(c*x*e^2 + c*d*e)

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Fricas [A]
time = 2.16, size = 15, normalized size = 1.00 \begin {gather*} -\frac {1}{c x e^{2} + c d e} \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),x, algorithm="fricas")

[Out]

-1/(c*x*e^2 + c*d*e)

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Sympy [A]
time = 0.05, size = 14, normalized size = 0.93 \begin {gather*} - \frac {1}{c d e + c e^{2} x} \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),x)

[Out]

-1/(c*d*e + c*e**2*x)

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Giac [A]
time = 1.40, size = 15, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (-1\right )}}{{\left (x e + d\right )} c} \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),x, algorithm="giac")

[Out]

-e^(-1)/((x*e + d)*c)

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Mupad [B]
time = 0.39, size = 15, normalized size = 1.00 \begin {gather*} -\frac {1}{c\,x\,e^2+c\,d\,e} \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),x)

[Out]

-1/(c*d*e + c*e^2*x)

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