\(\int \frac {d+e x}{(c d^2+2 c d e x+c e^2 x^2)^2} \, dx\) [1013]

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

Optimal result

Integrand size = 28, antiderivative size = 17 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 c^2 e (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 12, 32} \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 c^2 e (d+e x)^2} \]

[In]

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

[Out]

-1/2*1/(c^2*e*(d + e*x)^2)

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 c^2 e (d+e x)^2} \]

[In]

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

[Out]

-1/2*1/(c^2*e*(d + e*x)^2)

Maple [A] (verified)

Time = 2.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method result size
gosper \(-\frac {1}{2 c^{2} e \left (e x +d \right )^{2}}\) \(16\)
default \(-\frac {1}{2 c^{2} e \left (e x +d \right )^{2}}\) \(16\)
risch \(-\frac {1}{2 c^{2} e \left (e x +d \right )^{2}}\) \(16\)
parallelrisch \(-\frac {1}{2 c^{2} e \left (e x +d \right )^{2}}\) \(16\)
norman \(\frac {-\frac {d}{2 e c}-\frac {x}{2 c}}{c \left (e x +d \right )^{3}}\) \(28\)

[In]

int((e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (15) = 30\).

Time = 0.37 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.94 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 \, {\left (c^{2} e^{3} x^{2} + 2 \, c^{2} d e^{2} x + c^{2} d^{2} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (15) = 30\).

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.12 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=- \frac {1}{2 c^{2} d^{2} e + 4 c^{2} d e^{2} x + 2 c^{2} e^{3} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} c e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2 \, {\left (c d^{2} + {\left (e x^{2} + 2 \, d x\right )} c e\right )} c e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^2} \, dx=-\frac {1}{2\,c^2\,d^2\,e+4\,c^2\,d\,e^2\,x+2\,c^2\,e^3\,x^2} \]

[In]

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

[Out]

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