\(\int \frac {1}{(c d^2+2 c d e x+c e^2 x^2)^3} \, dx\) [1026]

Optimal result

Integrand size = 22, antiderivative size = 17 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 c^3 e (d+e x)^5} \]

[Out]

-1/5/c^3/e/(e*x+d)^5

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {27, 12, 32} \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 c^3 e (d+e x)^5} \]

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-3),x]

[Out]

-1/5*1/(c^3*e*(d + e*x)^5)

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^3 (d+e x)^6} \, dx \\ & = \frac {\int \frac {1}{(d+e x)^6} \, dx}{c^3} \\ & = -\frac {1}{5 c^3 e (d+e x)^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 c^3 e (d+e x)^5} \]

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(-3),x]

[Out]

-1/5*1/(c^3*e*(d + e*x)^5)

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {1}{5 c^{3} e \left (e x +d \right )^{5}}\)	\(16\)
norman	\(-\frac {1}{5 c^{3} e \left (e x +d \right )^{5}}\)	\(16\)
risch	\(-\frac {1}{5 c^{3} e \left (e x +d \right )^{5}}\)	\(16\)
gosper	\(-\frac {1}{5 \left (e x +d \right ) \left (x^{2} e^{2}+2 d e x +d^{2}\right )^{2} e \,c^{3}}\)	\(34\)
parallelrisch	\(-\frac {1}{5 \left (e x +d \right ) \left (x^{2} e^{2}+2 d e x +d^{2}\right )^{2} e \,c^{3}}\)	\(34\)

[In]

int(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x,method=_RETURNVERBOSE)

[Out]

-1/5/c^3/e/(e*x+d)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (15) = 30\).

Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 4.41 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (c^{3} e^{6} x^{5} + 5 \, c^{3} d e^{5} x^{4} + 10 \, c^{3} d^{2} e^{4} x^{3} + 10 \, c^{3} d^{3} e^{3} x^{2} + 5 \, c^{3} d^{4} e^{2} x + c^{3} d^{5} e\right )}} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="fricas")

[Out]

-1/5/(c^3*e^6*x^5 + 5*c^3*d*e^5*x^4 + 10*c^3*d^2*e^4*x^3 + 10*c^3*d^3*e^3*x^2 + 5*c^3*d^4*e^2*x + c^3*d^5*e)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (15) = 30\).

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.82 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=- \frac {1}{5 c^{3} d^{5} e + 25 c^{3} d^{4} e^{2} x + 50 c^{3} d^{3} e^{3} x^{2} + 50 c^{3} d^{2} e^{4} x^{3} + 25 c^{3} d e^{5} x^{4} + 5 c^{3} e^{6} x^{5}} \]

[In]

integrate(1/(c*e**2*x**2+2*c*d*e*x+c*d**2)**3,x)

[Out]

-1/(5*c**3*d**5*e + 25*c**3*d**4*e**2*x + 50*c**3*d**3*e**3*x**2 + 50*c**3*d**2*e**4*x**3 + 25*c**3*d*e**5*x**
4 + 5*c**3*e**6*x**5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (15) = 30\).

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 4.41 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (c^{3} e^{6} x^{5} + 5 \, c^{3} d e^{5} x^{4} + 10 \, c^{3} d^{2} e^{4} x^{3} + 10 \, c^{3} d^{3} e^{3} x^{2} + 5 \, c^{3} d^{4} e^{2} x + c^{3} d^{5} e\right )}} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="maxima")

[Out]

-1/5/(c^3*e^6*x^5 + 5*c^3*d*e^5*x^4 + 10*c^3*d^2*e^4*x^3 + 10*c^3*d^3*e^3*x^2 + 5*c^3*d^4*e^2*x + c^3*d^5*e)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5 \, {\left (e x + d\right )}^{5} c^{3} e} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="giac")

[Out]

-1/5/((e*x + d)^5*c^3*e)

Mupad [B] (verification not implemented)

Time = 9.67 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.53 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{5\,c^3\,d^5\,e+25\,c^3\,d^4\,e^2\,x+50\,c^3\,d^3\,e^3\,x^2+50\,c^3\,d^2\,e^4\,x^3+25\,c^3\,d\,e^5\,x^4+5\,c^3\,e^6\,x^5} \]

[In]

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

[Out]

-1/(5*c^3*d^5*e + 5*c^3*e^6*x^5 + 25*c^3*d^4*e^2*x + 25*c^3*d*e^5*x^4 + 50*c^3*d^3*e^3*x^2 + 50*c^3*d^2*e^4*x^
3)