\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^2}{d+e x} \, dx\) [990]

Optimal result

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

[Out]

1/4*c^2*(e*x+d)^4/e

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} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=\frac {c^2 (d+e x)^4}{4 e} \]

[In]

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

[Out]

(c^2*(d + e*x)^4)/(4*e)

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 c^2 (d+e x)^3 \, dx \\ & = c^2 \int (d+e x)^3 \, dx \\ & = \frac {c^2 (d+e x)^4}{4 e} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(c^2*(d + e*x)^4)/(4*e)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {c^{2} \left (x^{2} e^{2}+2 d e x +d^{2}\right )^{2}}{4 e}\)	\(27\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) c^{2}}{4}\)	\(36\)
norman	\(c^{2} d^{3} x +d \,e^{2} c^{2} x^{3}+\frac {1}{4} c^{2} x^{4} e^{3}+\frac {3}{2} c^{2} d^{2} e \,x^{2}\)	\(44\)
parallelrisch	\(c^{2} d^{3} x +d \,e^{2} c^{2} x^{3}+\frac {1}{4} c^{2} x^{4} e^{3}+\frac {3}{2} c^{2} d^{2} e \,x^{2}\)	\(44\)
risch	\(\frac {c^{2} x^{4} e^{3}}{4}+d \,e^{2} c^{2} x^{3}+\frac {3 c^{2} d^{2} e \,x^{2}}{2}+c^{2} d^{3} x +\frac {c^{2} d^{4}}{4 e}\)	\(55\)

[In]

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

[Out]

1/4*c^2*(e^2*x^2+2*d*e*x+d^2)^2/e

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (15) = 30\).

Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=\frac {1}{4} \, c^{2} e^{3} x^{4} + c^{2} d e^{2} x^{3} + \frac {3}{2} \, c^{2} d^{2} e x^{2} + c^{2} d^{3} x \]

[In]

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

[Out]

1/4*c^2*e^3*x^4 + c^2*d*e^2*x^3 + 3/2*c^2*d^2*e*x^2 + c^2*d^3*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (12) = 24\).

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.71 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=c^{2} d^{3} x + \frac {3 c^{2} d^{2} e x^{2}}{2} + c^{2} d e^{2} x^{3} + \frac {c^{2} e^{3} x^{4}}{4} \]

[In]

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

[Out]

c**2*d**3*x + 3*c**2*d**2*e*x**2/2 + c**2*d*e**2*x**3 + c**2*e**3*x**4/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (15) = 30\).

Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=\frac {1}{4} \, c^{2} e^{3} x^{4} + c^{2} d e^{2} x^{3} + \frac {3}{2} \, c^{2} d^{2} e x^{2} + c^{2} d^{3} x \]

[In]

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

[Out]

1/4*c^2*e^3*x^4 + c^2*d*e^2*x^3 + 3/2*c^2*d^2*e*x^2 + c^2*d^3*x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (15) = 30\).

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=\frac {1}{4} \, c^{2} e^{3} x^{4} + c^{2} d e^{2} x^{3} + \frac {3}{2} \, c^{2} d^{2} e x^{2} + c^{2} d^{3} x \]

[In]

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

[Out]

1/4*c^2*e^3*x^4 + c^2*d*e^2*x^3 + 3/2*c^2*d^2*e*x^2 + c^2*d^3*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{d+e x} \, dx=c^2\,d^3\,x+\frac {3\,c^2\,d^2\,e\,x^2}{2}+c^2\,d\,e^2\,x^3+\frac {c^2\,e^3\,x^4}{4} \]

[In]

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

[Out]

c^2*d^3*x + (c^2*e^3*x^4)/4 + (3*c^2*d^2*e*x^2)/2 + c^2*d*e^2*x^3