\(\int \frac {1}{(d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)} \, dx\) [1005]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 c e (d+e x)^3} \]

[Out]

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

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} \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 c e (d+e x)^3} \]

[In]

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

[Out]

-1/3*1/(c*e*(d + e*x)^3)

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)^4} \, dx \\ & = \frac {\int \frac {1}{(d+e x)^4} \, dx}{c} \\ & = -\frac {1}{3 c e (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 c e (d+e x)^3} \]

[In]

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

[Out]

-1/3*1/(c*e*(d + e*x)^3)

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {1}{3 c e \left (e x +d \right )^{3}}\)	\(16\)
norman	\(-\frac {1}{3 c e \left (e x +d \right )^{3}}\)	\(16\)
gosper	\(-\frac {1}{3 \left (e x +d \right ) e c \left (x^{2} e^{2}+2 d e x +d^{2}\right )}\)	\(34\)
risch	\(-\frac {1}{3 \left (e x +d \right ) e c \left (x^{2} e^{2}+2 d e x +d^{2}\right )}\)	\(34\)
parallelrisch	\(-\frac {1}{3 \left (e x +d \right ) e c \left (x^{2} e^{2}+2 d e x +d^{2}\right )}\)	\(34\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 \, {\left (c e^{4} x^{3} + 3 \, c d e^{3} x^{2} + 3 \, c d^{2} e^{2} x + c d^{3} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (14) = 28\).

Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.59 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=- \frac {1}{3 c d^{3} e + 9 c d^{2} e^{2} x + 9 c d e^{3} x^{2} + 3 c e^{4} x^{3}} \]

[In]

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

[Out]

-1/(3*c*d**3*e + 9*c*d**2*e**2*x + 9*c*d*e**3*x**2 + 3*c*e**4*x**3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (15) = 30\).

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 \, {\left (c e^{4} x^{3} + 3 \, c d e^{3} x^{2} + 3 \, c d^{2} e^{2} x + c d^{3} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3 \, {\left (e x + d\right )}^{3} c e} \]

[In]

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

[Out]

-1/3/((e*x + d)^3*c*e)

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.41 \[ \int \frac {1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{3\,c\,d^3\,e+9\,c\,d^2\,e^2\,x+9\,c\,d\,e^3\,x^2+3\,c\,e^4\,x^3} \]

[In]

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

[Out]

-1/(3*c*e^4*x^3 + 3*c*d^3*e + 9*c*d^2*e^2*x + 9*c*d*e^3*x^2)