\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^2}{(d+e x)^6} \, dx\) [995]

Optimal result

Integrand size = 30, antiderivative size = 15 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^2}{e (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, 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} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^2}{e (d+e x)} \]

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^2/(d + e*x)^6,x]

[Out]

-(c^2/(e*(d + e*x)))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^2}{e (d+e x)} \]

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^2/(d + e*x)^6,x]

[Out]

-(c^2/(e*(d + e*x)))

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
gosper	\(-\frac {c^{2}}{e \left (e x +d \right )}\)	\(16\)
default	\(-\frac {c^{2}}{e \left (e x +d \right )}\)	\(16\)
risch	\(-\frac {c^{2}}{e \left (e x +d \right )}\)	\(16\)
parallelrisch	\(-\frac {c^{2}}{e \left (e x +d \right )}\)	\(16\)
norman	\(\frac {-\frac {c^{2} d^{4}}{e}-c^{2} x^{4} e^{3}-4 d \,e^{2} c^{2} x^{3}-6 c^{2} d^{2} e \,x^{2}-4 c^{2} d^{3} x}{\left (e x +d \right )^{5}}\)	\(65\)

[In]

int((c*e^2*x^2+2*c*d*e*x+c*d^2)^2/(e*x+d)^6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^{2}}{e^{2} x + d e} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

-c^2/(e^2*x + d*e)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=- \frac {c^{2}}{d e + e^{2} x} \]

[In]

integrate((c*e**2*x**2+2*c*d*e*x+c*d**2)**2/(e*x+d)**6,x)

[Out]

-c**2/(d*e + e**2*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^{2}}{e^{2} x + d e} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

-c^2/(e^2*x + d*e)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^{2}}{{\left (e x + d\right )} e} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

-c^2/((e*x + d)*e)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^6} \, dx=-\frac {c^2}{e\,\left (d+e\,x\right )} \]

[In]

int((c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^2/(d + e*x)^6,x)

[Out]

-c^2/(e*(d + e*x))