3.1882 \(\int \frac{1}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=114 \[ -\frac{a e^2+c d^2+2 c d e x}{\left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )}-\frac{2 c d e \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}+\frac{2 c d e \log (d+e x)}{\left (c d^2-a e^2\right )^3} \]

[Out]

-((c*d^2 + a*e^2 + 2*c*d*e*x)/((c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2))) - (2*c*d*e*Log[a*e
+ c*d*x])/(c*d^2 - a*e^2)^3 + (2*c*d*e*Log[d + e*x])/(c*d^2 - a*e^2)^3

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Rubi [A]  time = 0.0331131, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {614, 616, 31} \[ -\frac{a e^2+c d^2+2 c d e x}{\left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )}-\frac{2 c d e \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}+\frac{2 c d e \log (d+e x)}{\left (c d^2-a e^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*d^2 + a*e^2 + 2*c*d*e*x)/((c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2))) - (2*c*d*e*Log[a*e
+ c*d*x])/(c*d^2 - a*e^2)^3 + (2*c*d*e*Log[d + e*x])/(c*d^2 - a*e^2)^3

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.094808, size = 86, normalized size = 0.75 \[ \frac{\frac{\left (c d^2-a e^2\right ) \left (a e^2+c d (d+2 e x)\right )}{(d+e x) (a e+c d x)}+2 c d e \log (a e+c d x)-2 c d e \log (d+e x)}{\left (a e^2-c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((c*d^2 - a*e^2)*(a*e^2 + c*d*(d + 2*e*x)))/((a*e + c*d*x)*(d + e*x)) + 2*c*d*e*Log[a*e + c*d*x] - 2*c*d*e*Lo
g[d + e*x])/(-(c*d^2) + a*e^2)^3

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Maple [A]  time = 0.052, size = 107, normalized size = 0.9 \begin{align*} -{\frac{e}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}-2\,{\frac{dec\ln \left ( ex+d \right ) }{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}}-{\frac{cd}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2} \left ( cdx+ae \right ) }}+2\,{\frac{dec\ln \left ( cdx+ae \right ) }{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06506, size = 319, normalized size = 2.8 \begin{align*} -\frac{2 \, c d e \log \left (c d x + a e\right )}{c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}} + \frac{2 \, c d e \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}} - \frac{2 \, c d e x + c d^{2} + a e^{2}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} +{\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} +{\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="maxima")

[Out]

-2*c*d*e*log(c*d*x + a*e)/(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6) + 2*c*d*e*log(e*x + d)/(c^3*
d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6) - (2*c*d*e*x + c*d^2 + a*e^2)/(a*c^2*d^5*e - 2*a^2*c*d^3*e^
3 + a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a
^3*e^6)*x)

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Fricas [B]  time = 1.92847, size = 545, normalized size = 4.78 \begin{align*} -\frac{c^{2} d^{4} - a^{2} e^{4} + 2 \,{\left (c^{2} d^{3} e - a c d e^{3}\right )} x + 2 \,{\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} +{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \log \left (c d x + a e\right ) - 2 \,{\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} +{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \log \left (e x + d\right )}{a c^{3} d^{7} e - 3 \, a^{2} c^{2} d^{5} e^{3} + 3 \, a^{3} c d^{3} e^{5} - a^{4} d e^{7} +{\left (c^{4} d^{7} e - 3 \, a c^{3} d^{5} e^{3} + 3 \, a^{2} c^{2} d^{3} e^{5} - a^{3} c d e^{7}\right )} x^{2} +{\left (c^{4} d^{8} - 2 \, a c^{3} d^{6} e^{2} + 2 \, a^{3} c d^{2} e^{6} - a^{4} e^{8}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

-(c^2*d^4 - a^2*e^4 + 2*(c^2*d^3*e - a*c*d*e^3)*x + 2*(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)
*x)*log(c*d*x + a*e) - 2*(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)*log(e*x + d))/(a*c^3*d^7*
e - 3*a^2*c^2*d^5*e^3 + 3*a^3*c*d^3*e^5 - a^4*d*e^7 + (c^4*d^7*e - 3*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5 - a^3*c
*d*e^7)*x^2 + (c^4*d^8 - 2*a*c^3*d^6*e^2 + 2*a^3*c*d^2*e^6 - a^4*e^8)*x)

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Sympy [B]  time = 1.53199, size = 484, normalized size = 4.25 \begin{align*} - \frac{2 c d e \log{\left (x + \frac{- \frac{2 a^{4} c d e^{9}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{8 a^{3} c^{2} d^{3} e^{7}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{12 a^{2} c^{3} d^{5} e^{5}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{8 a c^{4} d^{7} e^{3}}{\left (a e^{2} - c d^{2}\right )^{3}} + 2 a c d e^{3} - \frac{2 c^{5} d^{9} e}{\left (a e^{2} - c d^{2}\right )^{3}} + 2 c^{2} d^{3} e}{4 c^{2} d^{2} e^{2}} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{2 c d e \log{\left (x + \frac{\frac{2 a^{4} c d e^{9}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{8 a^{3} c^{2} d^{3} e^{7}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{12 a^{2} c^{3} d^{5} e^{5}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{8 a c^{4} d^{7} e^{3}}{\left (a e^{2} - c d^{2}\right )^{3}} + 2 a c d e^{3} + \frac{2 c^{5} d^{9} e}{\left (a e^{2} - c d^{2}\right )^{3}} + 2 c^{2} d^{3} e}{4 c^{2} d^{2} e^{2}} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{a e^{2} + c d^{2} + 2 c d e x}{a^{3} d e^{5} - 2 a^{2} c d^{3} e^{3} + a c^{2} d^{5} e + x^{2} \left (a^{2} c d e^{5} - 2 a c^{2} d^{3} e^{3} + c^{3} d^{5} e\right ) + x \left (a^{3} e^{6} - a^{2} c d^{2} e^{4} - a c^{2} d^{4} e^{2} + c^{3} d^{6}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

-2*c*d*e*log(x + (-2*a**4*c*d*e**9/(a*e**2 - c*d**2)**3 + 8*a**3*c**2*d**3*e**7/(a*e**2 - c*d**2)**3 - 12*a**2
*c**3*d**5*e**5/(a*e**2 - c*d**2)**3 + 8*a*c**4*d**7*e**3/(a*e**2 - c*d**2)**3 + 2*a*c*d*e**3 - 2*c**5*d**9*e/
(a*e**2 - c*d**2)**3 + 2*c**2*d**3*e)/(4*c**2*d**2*e**2))/(a*e**2 - c*d**2)**3 + 2*c*d*e*log(x + (2*a**4*c*d*e
**9/(a*e**2 - c*d**2)**3 - 8*a**3*c**2*d**3*e**7/(a*e**2 - c*d**2)**3 + 12*a**2*c**3*d**5*e**5/(a*e**2 - c*d**
2)**3 - 8*a*c**4*d**7*e**3/(a*e**2 - c*d**2)**3 + 2*a*c*d*e**3 + 2*c**5*d**9*e/(a*e**2 - c*d**2)**3 + 2*c**2*d
**3*e)/(4*c**2*d**2*e**2))/(a*e**2 - c*d**2)**3 - (a*e**2 + c*d**2 + 2*c*d*e*x)/(a**3*d*e**5 - 2*a**2*c*d**3*e
**3 + a*c**2*d**5*e + x**2*(a**2*c*d*e**5 - 2*a*c**2*d**3*e**3 + c**3*d**5*e) + x*(a**3*e**6 - a**2*c*d**2*e**
4 - a*c**2*d**4*e**2 + c**3*d**6))

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Giac [A]  time = 1.31586, size = 238, normalized size = 2.09 \begin{align*} -\frac{4 \, c d \arctan \left (\frac{2 \, c d x e + c d^{2} + a e^{2}}{\sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}}\right ) e}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}} - \frac{2 \, c d x e + c d^{2} + a e^{2}}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}{\left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

-4*c*d*arctan((2*c*d*x*e + c*d^2 + a*e^2)/sqrt(-c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4))*e/((c^2*d^4 - 2*a*c*d^2*e^
2 + a^2*e^4)*sqrt(-c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4)) - (2*c*d*x*e + c*d^2 + a*e^2)/((c^2*d^4 - 2*a*c*d^2*e^2
+ a^2*e^4)*(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))