### 3.1846 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^5} \, dx$$

Optimal. Leaf size=71 $\frac{2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}-\frac{\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac{c^2 d^2 \log (d+e x)}{e^3}$

[Out]

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

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Rubi [A]  time = 0.0519226, antiderivative size = 71, 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, 43} $\frac{2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}-\frac{\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac{c^2 d^2 \log (d+e x)}{e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(c*d^2 - a*e^2)^2/(2*e^3*(d + e*x)^2) + (2*c*d*(c*d^2 - a*e^2))/(e^3*(d + e*x)) + (c^2*d^2*Log[d + e*x])/e^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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx &=\int \frac{(a e+c d x)^2}{(d+e x)^3} \, dx\\ &=\int \left (\frac{\left (-c d^2+a e^2\right )^2}{e^2 (d+e x)^3}-\frac{2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^2}+\frac{c^2 d^2}{e^2 (d+e x)}\right ) \, dx\\ &=-\frac{\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac{2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}+\frac{c^2 d^2 \log (d+e x)}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.0314704, size = 59, normalized size = 0.83 $\frac{\frac{\left (c d^2-a e^2\right ) \left (a e^2+c d (3 d+4 e x)\right )}{(d+e x)^2}+2 c^2 d^2 \log (d+e x)}{2 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(((c*d^2 - a*e^2)*(a*e^2 + c*d*(3*d + 4*e*x)))/(d + e*x)^2 + 2*c^2*d^2*Log[d + e*x])/(2*e^3)

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Maple [A]  time = 0.046, size = 98, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}e}{2\, \left ( ex+d \right ) ^{2}}}+{\frac{ac{d}^{2}}{e \left ( ex+d \right ) ^{2}}}-{\frac{{c}^{2}{d}^{4}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{c}^{2}{d}^{2}\ln \left ( ex+d \right ) }{{e}^{3}}}-2\,{\frac{acd}{e \left ( ex+d \right ) }}+2\,{\frac{{c}^{2}{d}^{3}}{{e}^{3} \left ( ex+d \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07287, size = 122, normalized size = 1.72 \begin{align*} \frac{c^{2} d^{2} \log \left (e x + d\right )}{e^{3}} + \frac{3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4} + 4 \,{\left (c^{2} d^{3} e - a c d e^{3}\right )} x}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65813, size = 224, normalized size = 3.15 \begin{align*} \frac{3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4} + 4 \,{\left (c^{2} d^{3} e - a c d e^{3}\right )} x + 2 \,{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.966675, size = 90, normalized size = 1.27 \begin{align*} \frac{c^{2} d^{2} \log{\left (d + e x \right )}}{e^{3}} - \frac{a^{2} e^{4} + 2 a c d^{2} e^{2} - 3 c^{2} d^{4} + x \left (4 a c d e^{3} - 4 c^{2} d^{3} e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22589, size = 159, normalized size = 2.24 \begin{align*} -c^{2} d^{2} e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + \frac{1}{2} \,{\left (\frac{4 \, c^{2} d^{3} e^{9}}{x e + d} - \frac{c^{2} d^{4} e^{9}}{{\left (x e + d\right )}^{2}} - \frac{4 \, a c d e^{11}}{x e + d} + \frac{2 \, a c d^{2} e^{11}}{{\left (x e + d\right )}^{2}} - \frac{a^{2} e^{13}}{{\left (x e + d\right )}^{2}}\right )} e^{\left (-12\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2*d^2*e^(-3)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) + 1/2*(4*c^2*d^3*e^9/(x*e + d) - c^2*d^4*e^9/(x*e + d)^2
- 4*a*c*d*e^11/(x*e + d) + 2*a*c*d^2*e^11/(x*e + d)^2 - a^2*e^13/(x*e + d)^2)*e^(-12)