\(\int \frac {1}{(d+e x)^3 (c d^2+2 c d e x+c e^2 x^2)} \, dx\) [1006]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4 c e (d+e x)^4} \]

[Out]

-1/4/c/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4 c e (d+e x)^4} \]

[In]

Int[1/((d + e*x)^3*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)),x]

[Out]

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

Mathematica [A] (verified)

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

[In]

Integrate[1/((d + e*x)^3*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)),x]

[Out]

-1/4*1/(c*e*(d + e*x)^4)

Maple [A] (verified)

Time = 2.42 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.00 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4 \, {\left (c e^{5} x^{4} + 4 \, c d e^{4} x^{3} + 6 \, c d^{2} e^{3} x^{2} + 4 \, c d^{3} e^{2} x + c d^{4} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (14) = 28\).

Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.41 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=- \frac {1}{4 c d^{4} e + 16 c d^{3} e^{2} x + 24 c d^{2} e^{3} x^{2} + 16 c d e^{4} x^{3} + 4 c e^{5} x^{4}} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (15) = 30\).

Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.00 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4 \, {\left (c e^{5} x^{4} + 4 \, c d e^{4} x^{3} + 6 \, c d^{2} e^{3} x^{2} + 4 \, c d^{3} e^{2} x + c d^{4} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4 \, {\left (e x + d\right )}^{4} c e} \]

[In]

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

[Out]

-1/4/((e*x + d)^4*c*e)

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.12 \[ \int \frac {1}{(d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )} \, dx=-\frac {1}{4\,c\,d^4\,e+16\,c\,d^3\,e^2\,x+24\,c\,d^2\,e^3\,x^2+16\,c\,d\,e^4\,x^3+4\,c\,e^5\,x^4} \]

[In]

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

[Out]

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