\(\int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx\) [982]

Optimal result

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

[Out]

c*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 3, 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 = {24, 21, 8} \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c x \]

[In]

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

[Out]

c*x

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {c d e^2+c e^3 x}{d+e x} \, dx}{e^2} \\ & = c \int 1 \, dx \\ & = c x \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

c*x

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.33

method	result	size
default	\(c x\)	\(4\)
risch	\(c x\)	\(4\)
norman	\(\frac {c e \,x^{2}+c d x}{e x +d}\)	\(20\)

[In]

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

[Out]

c*x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c x \]

[In]

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

[Out]

c*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.67 \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c x \]

[In]

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

[Out]

c*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c x \]

[In]

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

[Out]

c*x

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 37.67 \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c e^{2} {\left (\frac {2 \, d \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {e x + d}{e^{3}} - \frac {d^{2}}{{\left (e x + d\right )} e^{3}}\right )} - 2 \, c d {\left (\frac {\log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e} - \frac {d}{{\left (e x + d\right )} e}\right )} - \frac {c d^{2}}{{\left (e x + d\right )} e} \]

[In]

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

[Out]

c*e^2*(2*d*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^3 + (e*x + d)/e^3 - d^2/((e*x + d)*e^3)) - 2*c*d*(log(abs(
e*x + d)/((e*x + d)^2*abs(e)))/e - d/((e*x + d)*e)) - c*d^2/((e*x + d)*e)

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 3, normalized size of antiderivative = 1.00 \[ \int \frac {c d^2+2 c d e x+c e^2 x^2}{(d+e x)^2} \, dx=c\,x \]

[In]

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

[Out]

c*x