\(\int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx\) [1000]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 9} \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^2}{2 c e} \]

[In]

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

[Out]

(d + e*x)^2/(2*c*e)

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, 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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {d x+\frac {e x^2}{2}}{c} \]

[In]

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

[Out]

(d*x + (e*x^2)/2)/c

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {x \left (e x +2 d \right )}{2 c}\)	\(14\)
default	\(\frac {\frac {1}{2} e \,x^{2}+d x}{c}\)	\(15\)
parallelrisch	\(\frac {e \,x^{2}+2 d x}{2 c}\)	\(16\)
risch	\(\frac {e \,x^{2}}{2 c}+\frac {d x}{c}\)	\(17\)
norman	\(\frac {\frac {d^{2} x}{c}+\frac {e^{2} x^{3}}{2 c}+\frac {3 e d \,x^{2}}{2 c}}{e x +d}\)	\(39\)

[In]

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

[Out]

1/2*x*(e*x+2*d)/c

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c} \]

[In]

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

[Out]

1/2*(e*x^2 + 2*d*x)/c

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {d x}{c} + \frac {e x^{2}}{2 c} \]

[In]

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

[Out]

d*x/c + e*x**2/(2*c)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c} \]

[In]

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

[Out]

1/2*(e*x^2 + 2*d*x)/c

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

1/2*(e*x^2 + 2*d*x)/c

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {(d+e x)^3}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {x\,\left (2\,d+e\,x\right )}{2\,c} \]

[In]

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

[Out]

(x*(2*d + e*x))/(2*c)