\(\int \frac {1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)} \, dx\) [1004]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (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 = 30, antiderivative size = 17 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2 c e (d+e x)^2} \]

[Out]

-1/2/c/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2 c e (d+e x)^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

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

[In]

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

[Out]

-1/2*1/(c*e*(d + e*x)^2)

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method result size
default \(-\frac {1}{2 c e \left (e x +d \right )^{2}}\) \(16\)
norman \(-\frac {1}{2 c e \left (e x +d \right )^{2}}\) \(16\)
risch \(-\frac {1}{2 c e \left (e x +d \right )^{2}}\) \(16\)
parallelrisch \(-\frac {1}{2 c e \left (e x +d \right )^{2}}\) \(16\)
gosper \(-\frac {1}{2 e c \left (x^{2} e^{2}+2 d e x +d^{2}\right )}\) \(27\)

[In]

int(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2 \, {\left (c e^{3} x^{2} + 2 \, c d e^{2} x + c d^{2} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.82 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=- \frac {1}{2 c d^{2} e + 4 c d e^{2} x + 2 c e^{3} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2 \, {\left (c e^{3} x^{2} + 2 \, c d e^{2} x + c d^{2} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2 \, {\left (e x + d\right )}^{2} c e} \]

[In]

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

[Out]

-1/2/((e*x + d)^2*c*e)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{2\,c\,d^2\,e+4\,c\,d\,e^2\,x+2\,c\,e^3\,x^2} \]

[In]

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

[Out]

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