\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^2}{(d+e x)^3} \, dx\) [992]

Optimal result

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

[Out]

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

[In]

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

[Out]

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

Mathematica [A] (verified)

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

[In]

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

[Out]

c^2*(d*x + (e*x^2)/2)

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {x \left (e x +2 d \right ) c^{2}}{2}\)	\(14\)
default	\(c^{2} \left (\frac {1}{2} e \,x^{2}+d x \right )\)	\(15\)
risch	\(\frac {1}{2} c^{2} e \,x^{2}+c^{2} d x\)	\(17\)
parallelrisch	\(\frac {1}{2} c^{2} e \,x^{2}+c^{2} d x\)	\(17\)
norman	\(\frac {\frac {c^{2} x^{4} e^{3}}{2}-\frac {5 c^{2} d^{4}}{2 e}+2 d \,e^{2} c^{2} x^{3}-4 c^{2} d^{3} x}{\left (e x +d \right )^{2}}\)	\(53\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {1}{2} \, c^{2} e x^{2} + c^{2} d x \]

[In]

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

[Out]

1/2*c^2*e*x^2 + c^2*d*x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^3} \, dx=c^{2} d x + \frac {c^{2} e x^{2}}{2} \]

[In]

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

[Out]

c**2*d*x + c**2*e*x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {1}{2} \, c^{2} e x^{2} + c^{2} d x \]

[In]

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

[Out]

1/2*c^2*e*x^2 + c^2*d*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {1}{2} \, c^{2} e x^{2} + c^{2} d x \]

[In]

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

[Out]

1/2*c^2*e*x^2 + c^2*d*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^2\,x\,\left (2\,d+e\,x\right )}{2} \]

[In]

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

[Out]

(c^2*x*(2*d + e*x))/2