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

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 15 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c e (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {27, 12, 32} \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c e (d+e x)} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c e (d+e x)} \]

[In]

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

[Out]

-(1/(c*e*(d + e*x)))

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 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 \(-\frac {1}{c e \left (e x +d \right )}\) \(16\)
parallelrisch \(\frac {x}{c d \left (e x +d \right )}\) \(16\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c e^{2} x + c d e} \]

[In]

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

[Out]

-1/(c*e^2*x + c*d*e)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=- \frac {1}{c d e + c e^{2} x} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c e^{2} x + c d e} \]

[In]

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

[Out]

-1/(c*e^2*x + c*d*e)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{{\left (e x + d\right )} c e} \]

[In]

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

[Out]

-1/((e*x + d)*c*e)

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {1}{c\,x\,e^2+c\,d\,e} \]

[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)