\(\int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx\) [998]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^4}{4 c e} \]

[Out]

1/4*(e*x+d)^4/c/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} \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^4}{4 c e} \]

[In]

Int[(d + e*x)^5/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^4}{4 c e} \]

[In]

Integrate[(d + e*x)^5/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

(d + e*x)^4/(4*c*e)

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (e x +d \right )^{4}}{4 c e}\)	\(16\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right )}{4 c}\)	\(36\)
parallelrisch	\(\frac {e^{3} x^{4}+4 d \,e^{2} x^{3}+6 d^{2} e \,x^{2}+4 d^{3} x}{4 c}\)	\(38\)
risch	\(\frac {e^{3} x^{4}}{4 c}+\frac {e^{2} d \,x^{3}}{c}+\frac {3 e \,d^{2} x^{2}}{2 c}+\frac {d^{3} x}{c}+\frac {d^{4}}{4 c e}\)	\(55\)
norman	\(\frac {-\frac {d^{5}}{c e}+\frac {e^{4} x^{5}}{4 c}+\frac {5 d \,e^{3} x^{4}}{4 c}+\frac {5 d^{2} e^{2} x^{3}}{2 c}+\frac {5 d^{3} e \,x^{2}}{2 c}}{e x +d}\)	\(70\)

[In]

int((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x}{4 \, c} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="fricas")

[Out]

1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)/c

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (10) = 20\).

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.29 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {d^{3} x}{c} + \frac {3 d^{2} e x^{2}}{2 c} + \frac {d e^{2} x^{3}}{c} + \frac {e^{3} x^{4}}{4 c} \]

[In]

integrate((e*x+d)**5/(c*e**2*x**2+2*c*d*e*x+c*d**2),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x}{4 \, c} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="maxima")

[Out]

1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)/c

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x}{4 \, c} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="giac")

[Out]

1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)/c

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {(d+e x)^5}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {d^3\,x}{c}+\frac {e^3\,x^4}{4\,c}+\frac {3\,d^2\,e\,x^2}{2\,c}+\frac {d\,e^2\,x^3}{c} \]

[In]

int((d + e*x)^5/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x),x)

[Out]

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