3.1880 \(\int \frac {(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=18 \[ -\frac {1}{c d (a e+c d x)} \]

[Out]

-1/c/d/(c*d*x+a*e)

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {626, 32} \[ -\frac {1}{c d (a e+c d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(c*d*(a*e + c*d*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]

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac {1}{(a e+c d x)^2} \, dx\\ &=-\frac {1}{c d (a e+c d x)}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.00 \[ -\frac {1}{c d (a e+c d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(c*d*(a*e + c*d*x)))

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fricas [A]  time = 0.80, size = 18, normalized size = 1.00 \[ -\frac {1}{c^{2} d^{2} x + a c d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(c^2*d^2*x + a*c*d*e)

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giac [B]  time = 0.19, size = 109, normalized size = 6.06 \[ -\frac {c^{2} d^{4} x e + c^{2} d^{5} - 2 \, a c d^{2} x e^{3} - 2 \, a c d^{3} e^{2} + a^{2} x e^{5} + a^{2} d e^{4}}{{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} {\left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^2*d^4*x*e + c^2*d^5 - 2*a*c*d^2*x*e^3 - 2*a*c*d^3*e^2 + a^2*x*e^5 + a^2*d*e^4)/((c^3*d^5 - 2*a*c^2*d^3*e^2
+ a^2*c*d*e^4)*(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))

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maple [A]  time = 0.04, size = 19, normalized size = 1.06 \[ -\frac {1}{\left (c d x +a e \right ) c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c/d/(c*d*x+a*e)

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maxima [A]  time = 1.00, size = 18, normalized size = 1.00 \[ -\frac {1}{c^{2} d^{2} x + a c d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(c^2*d^2*x + a*c*d*e)

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mupad [B]  time = 0.55, size = 18, normalized size = 1.00 \[ -\frac {1}{c\,d\,\left (a\,e+c\,d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(c*d*(a*e + c*d*x))

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sympy [A]  time = 0.19, size = 17, normalized size = 0.94 \[ - \frac {1}{a c d e + c^{2} d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*c*d*e + c**2*d**2*x)

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