### 3.1892 $$\int \frac{(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=107 $\frac{e}{\left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{e^2 \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}-\frac{e^2 \log (d+e x)}{\left (c d^2-a e^2\right )^3}$

[Out]

-1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2) + e/((c*d^2 - a*e^2)^2*(a*e + c*d*x)) + (e^2*Log[a*e + c*d*x])/(c*d^2 -
a*e^2)^3 - (e^2*Log[d + e*x])/(c*d^2 - a*e^2)^3

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Rubi [A]  time = 0.0710984, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.057, Rules used = {626, 44} $\frac{e}{\left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{e^2 \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}-\frac{e^2 \log (d+e x)}{\left (c d^2-a e^2\right )^3}$

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)^3,x]

[Out]

-1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2) + e/((c*d^2 - a*e^2)^2*(a*e + c*d*x)) + (e^2*Log[a*e + c*d*x])/(c*d^2 -
a*e^2)^3 - (e^2*Log[d + e*x])/(c*d^2 - a*e^2)^3

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]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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

Mathematica [A]  time = 0.0770763, size = 83, normalized size = 0.78 $-\frac{\frac{\left (c d^2-a e^2\right ) \left (c d (d-2 e x)-3 a e^2\right )}{(a e+c d x)^2}-2 e^2 \log (a e+c d x)+2 e^2 \log (d+e x)}{2 \left (c d^2-a e^2\right )^3}$

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)^3,x]

[Out]

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

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Maple [A]  time = 0.05, size = 106, normalized size = 1. \begin{align*}{\frac{{e}^{2}\ln \left ( ex+d \right ) }{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}}+{\frac{1}{ \left ( 2\,a{e}^{2}-2\,c{d}^{2} \right ) \left ( cdx+ae \right ) ^{2}}}+{\frac{e}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2} \left ( cdx+ae \right ) }}-{\frac{{e}^{2}\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((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

e^2/(a*e^2-c*d^2)^3*ln(e*x+d)+1/2/(a*e^2-c*d^2)/(c*d*x+a*e)^2+e/(a*e^2-c*d^2)^2/(c*d*x+a*e)-e^2/(a*e^2-c*d^2)^
3*ln(c*d*x+a*e)

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Maxima [B]  time = 1.09754, size = 321, normalized size = 3. \begin{align*} \frac{e^{2} \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{e^{2} \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} + 3 \, a e^{2}}{2 \,{\left (a^{2} c^{2} d^{4} e^{2} - 2 \, a^{3} c d^{2} e^{4} + a^{4} e^{6} +{\left (c^{4} d^{6} - 2 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (a c^{3} d^{5} e - 2 \, a^{2} c^{2} d^{3} e^{3} + a^{3} c d e^{5}\right )} x\right )}} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

e^2*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) - e^2*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) + 1/2*(2*c*d*e*x - c*d^2 + 3*a*e^2)/(a^2*c^2*d^4*e^2 - 2*a^3*c*d^2*e
^4 + a^4*e^6 + (c^4*d^6 - 2*a*c^3*d^4*e^2 + a^2*c^2*d^2*e^4)*x^2 + 2*(a*c^3*d^5*e - 2*a^2*c^2*d^3*e^3 + a^3*c*
d*e^5)*x)

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Fricas [B]  time = 2.00002, size = 556, normalized size = 5.2 \begin{align*} -\frac{c^{2} d^{4} - 4 \, a c d^{2} e^{2} + 3 \, 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} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \log \left (c d x + a e\right ) + 2 \,{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \log \left (e x + d\right )}{2 \,{\left (a^{2} c^{3} d^{6} e^{2} - 3 \, a^{3} c^{2} d^{4} e^{4} + 3 \, a^{4} c d^{2} e^{6} - a^{5} e^{8} +{\left (c^{5} d^{8} - 3 \, a c^{4} d^{6} e^{2} + 3 \, a^{2} c^{3} d^{4} e^{4} - a^{3} c^{2} d^{2} e^{6}\right )} x^{2} + 2 \,{\left (a c^{4} d^{7} e - 3 \, a^{2} c^{3} d^{5} e^{3} + 3 \, a^{3} c^{2} d^{3} e^{5} - a^{4} c d e^{7}\right )} x\right )}} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

-1/2*(c^2*d^4 - 4*a*c*d^2*e^2 + 3*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 + 2*a*c*d*e^3*x +
a^2*e^4)*log(c*d*x + a*e) + 2*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*log(e*x + d))/(a^2*c^3*d^6*e^2 - 3*
a^3*c^2*d^4*e^4 + 3*a^4*c*d^2*e^6 - a^5*e^8 + (c^5*d^8 - 3*a*c^4*d^6*e^2 + 3*a^2*c^3*d^4*e^4 - a^3*c^2*d^2*e^6
)*x^2 + 2*(a*c^4*d^7*e - 3*a^2*c^3*d^5*e^3 + 3*a^3*c^2*d^3*e^5 - a^4*c*d*e^7)*x)

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Sympy [B]  time = 1.61083, size = 457, normalized size = 4.27 \begin{align*} \frac{e^{2} \log{\left (x + \frac{- \frac{a^{4} e^{10}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{4 a^{3} c d^{2} e^{8}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{6 a^{2} c^{2} d^{4} e^{6}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{4 a c^{3} d^{6} e^{4}}{\left (a e^{2} - c d^{2}\right )^{3}} + a e^{4} - \frac{c^{4} d^{8} e^{2}}{\left (a e^{2} - c d^{2}\right )^{3}} + c d^{2} e^{2}}{2 c d e^{3}} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{e^{2} \log{\left (x + \frac{\frac{a^{4} e^{10}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{4 a^{3} c d^{2} e^{8}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{6 a^{2} c^{2} d^{4} e^{6}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac{4 a c^{3} d^{6} e^{4}}{\left (a e^{2} - c d^{2}\right )^{3}} + a e^{4} + \frac{c^{4} d^{8} e^{2}}{\left (a e^{2} - c d^{2}\right )^{3}} + c d^{2} e^{2}}{2 c d e^{3}} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac{3 a e^{2} - c d^{2} + 2 c d e x}{2 a^{4} e^{6} - 4 a^{3} c d^{2} e^{4} + 2 a^{2} c^{2} d^{4} e^{2} + x^{2} \left (2 a^{2} c^{2} d^{2} e^{4} - 4 a c^{3} d^{4} e^{2} + 2 c^{4} d^{6}\right ) + x \left (4 a^{3} c d e^{5} - 8 a^{2} c^{2} d^{3} e^{3} + 4 a c^{3} d^{5} e\right )} \end{align*}

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)**3,x)

[Out]

e**2*log(x + (-a**4*e**10/(a*e**2 - c*d**2)**3 + 4*a**3*c*d**2*e**8/(a*e**2 - c*d**2)**3 - 6*a**2*c**2*d**4*e*
*6/(a*e**2 - c*d**2)**3 + 4*a*c**3*d**6*e**4/(a*e**2 - c*d**2)**3 + a*e**4 - c**4*d**8*e**2/(a*e**2 - c*d**2)*
*3 + c*d**2*e**2)/(2*c*d*e**3))/(a*e**2 - c*d**2)**3 - e**2*log(x + (a**4*e**10/(a*e**2 - c*d**2)**3 - 4*a**3*
c*d**2*e**8/(a*e**2 - c*d**2)**3 + 6*a**2*c**2*d**4*e**6/(a*e**2 - c*d**2)**3 - 4*a*c**3*d**6*e**4/(a*e**2 - c
*d**2)**3 + a*e**4 + c**4*d**8*e**2/(a*e**2 - c*d**2)**3 + c*d**2*e**2)/(2*c*d*e**3))/(a*e**2 - c*d**2)**3 + (
3*a*e**2 - c*d**2 + 2*c*d*e*x)/(2*a**4*e**6 - 4*a**3*c*d**2*e**4 + 2*a**2*c**2*d**4*e**2 + x**2*(2*a**2*c**2*d
**2*e**4 - 4*a*c**3*d**4*e**2 + 2*c**4*d**6) + x*(4*a**3*c*d*e**5 - 8*a**2*c**2*d**3*e**3 + 4*a*c**3*d**5*e))

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Giac [B]  time = 1.24947, size = 535, normalized size = 5. \begin{align*} \frac{2 \,{\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} \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 )}{{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}} + \frac{2 \, c^{3} d^{5} x^{3} e^{3} + 3 \, c^{3} d^{6} x^{2} e^{2} - c^{3} d^{8} - 4 \, a c^{2} d^{3} x^{3} e^{5} - 3 \, a c^{2} d^{4} x^{2} e^{4} + 6 \, a c^{2} d^{5} x e^{3} + 5 \, a c^{2} d^{6} e^{2} + 2 \, a^{2} c d x^{3} e^{7} - 3 \, a^{2} c d^{2} x^{2} e^{6} - 12 \, a^{2} c d^{3} x e^{5} - 7 \, a^{2} c d^{4} e^{4} + 3 \, a^{3} x^{2} e^{8} + 6 \, a^{3} d x e^{7} + 3 \, a^{3} d^{2} e^{6}}{2 \,{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )}{\left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )}^{2}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

2*(c^2*d^4*e^2 - 2*a*c*d^2*e^4 + a^2*e^6)*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))/((c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*sqrt(-c^2*d^4 + 2*a*c*d^
2*e^2 - a^2*e^4)) + 1/2*(2*c^3*d^5*x^3*e^3 + 3*c^3*d^6*x^2*e^2 - c^3*d^8 - 4*a*c^2*d^3*x^3*e^5 - 3*a*c^2*d^4*x
^2*e^4 + 6*a*c^2*d^5*x*e^3 + 5*a*c^2*d^6*e^2 + 2*a^2*c*d*x^3*e^7 - 3*a^2*c*d^2*x^2*e^6 - 12*a^2*c*d^3*x*e^5 -
7*a^2*c*d^4*e^4 + 3*a^3*x^2*e^8 + 6*a^3*d*x*e^7 + 3*a^3*d^2*e^6)/((c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e
^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)^2)