### 3.984 $$\int \frac{c d^2+2 c d e x+c e^2 x^2}{(d+e x)^4} \, dx$$

Optimal. Leaf size=13 $-\frac{c}{e (d+e x)}$

[Out]

-(c/(e*(d + e*x)))

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Rubi [A]  time = 0.0064377, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.107, Rules used = {24, 21, 32} $-\frac{c}{e (d+e x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)/(d + e*x)^4,x]

[Out]

-(c/(e*(d + e*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]

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 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*} \int \frac{c d^2+2 c d e x+c e^2 x^2}{(d+e x)^4} \, dx &=\frac{\int \frac{c d e^2+c e^3 x}{(d+e x)^3} \, dx}{e^2}\\ &=c \int \frac{1}{(d+e x)^2} \, dx\\ &=-\frac{c}{e (d+e x)}\\ \end{align*}

Mathematica [A]  time = 0.0033505, size = 13, normalized size = 1. $-\frac{c}{e (d+e x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)/(d + e*x)^4,x]

[Out]

-(c/(e*(d + e*x)))

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Maple [A]  time = 0.038, size = 14, normalized size = 1.1 \begin{align*} -{\frac{c}{e \left ( ex+d \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c/e/(e*x+d)

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Maxima [A]  time = 1.16147, size = 19, normalized size = 1.46 \begin{align*} -\frac{c}{e^{2} x + d e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

-c/(e^2*x + d*e)

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Fricas [A]  time = 1.98543, size = 24, normalized size = 1.85 \begin{align*} -\frac{c}{e^{2} x + d e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-c/(e^2*x + d*e)

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Sympy [A]  time = 0.314067, size = 10, normalized size = 0.77 \begin{align*} - \frac{c}{d e + e^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e**2*x**2+2*c*d*e*x+c*d**2)/(e*x+d)**4,x)

[Out]

-c/(d*e + e**2*x)

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Giac [B]  time = 1.22292, size = 46, normalized size = 3.54 \begin{align*} -\frac{{\left (c x^{2} e^{4} + 2 \, c d x e^{3} + c d^{2} e^{2}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)/(e*x+d)^4,x, algorithm="giac")

[Out]

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