\(\int \frac {d+e x}{(c d^2+2 c d e x+c e^2 x^2)^3} \, dx\) [1025]

Optimal result

Integrand size = 28, antiderivative size = 17 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 c^3 e (d+e x)^4} \]

[Out]

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

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} \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 c^3 e (d+e x)^4} \]

[In]

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

[Out]

-1/4*1/(c^3*e*(d + e*x)^4)

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)^5} \, dx \\ & = \frac {\int \frac {1}{(d+e x)^5} \, dx}{c^3} \\ & = -\frac {1}{4 c^3 e (d+e x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 c^3 e (d+e x)^4} \]

[In]

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

[Out]

-1/4*1/(c^3*e*(d + e*x)^4)

Maple [A] (verified)

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

method	result	size
default	\(-\frac {1}{4 c^{3} e \left (e x +d \right )^{4}}\)	\(16\)
norman	\(\frac {-\frac {d}{4 e c}-\frac {x}{4 c}}{c^{2} \left (e x +d \right )^{5}}\)	\(28\)
gosper	\(-\frac {1}{4 \left (e x +d \right )^{2} \left (x^{2} e^{2}+2 d e x +d^{2}\right ) e \,c^{3}}\)	\(34\)
risch	\(-\frac {1}{4 \left (e x +d \right )^{2} \left (x^{2} e^{2}+2 d e x +d^{2}\right ) e \,c^{3}}\)	\(34\)
parallelrisch	\(-\frac {1}{4 \left (e x +d \right )^{2} \left (x^{2} e^{2}+2 d e x +d^{2}\right ) e \,c^{3}}\)	\(34\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (15) = 30\).

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.59 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (c^{3} e^{5} x^{4} + 4 \, c^{3} d e^{4} x^{3} + 6 \, c^{3} d^{2} e^{3} x^{2} + 4 \, c^{3} d^{3} e^{2} x + c^{3} d^{4} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (15) = 30\).

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.88 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=- \frac {1}{4 c^{3} d^{4} e + 16 c^{3} d^{3} e^{2} x + 24 c^{3} d^{2} e^{3} x^{2} + 16 c^{3} d e^{4} x^{3} + 4 c^{3} e^{5} x^{4}} \]

[In]

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

[Out]

-1/(4*c**3*d**4*e + 16*c**3*d**3*e**2*x + 24*c**3*d**2*e**3*x**2 + 16*c**3*d*e**4*x**3 + 4*c**3*e**5*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{2} c e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (c d^{2} + {\left (e x^{2} + 2 \, d x\right )} c e\right )}^{2} c e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.71 \[ \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {1}{4\,c^3\,d^4\,e+16\,c^3\,d^3\,e^2\,x+24\,c^3\,d^2\,e^3\,x^2+16\,c^3\,d\,e^4\,x^3+4\,c^3\,e^5\,x^4} \]

[In]

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

[Out]

-1/(4*c^3*d^4*e + 4*c^3*e^5*x^4 + 16*c^3*d^3*e^2*x + 16*c^3*d*e^4*x^3 + 24*c^3*d^2*e^3*x^2)