### 3.1475 $$\int \frac{(a^2+2 a b x+b^2 x^2)^2}{(d+e x)^6} \, dx$$

Optimal. Leaf size=28 $\frac{(a+b x)^5}{5 (d+e x)^5 (b d-a e)}$

[Out]

(a + b*x)^5/(5*(b*d - a*e)*(d + e*x)^5)

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Rubi [A]  time = 0.004715, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {27, 37} $\frac{(a+b x)^5}{5 (d+e x)^5 (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^2/(d + e*x)^6,x]

[Out]

(a + b*x)^5/(5*(b*d - a*e)*(d + e*x)^5)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^6} \, dx &=\int \frac{(a+b x)^4}{(d+e x)^6} \, dx\\ &=\frac{(a+b x)^5}{5 (b d-a e) (d+e x)^5}\\ \end{align*}

Mathematica [B]  time = 0.0535618, size = 140, normalized size = 5. $-\frac{a^2 b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+a^3 b e^3 (d+5 e x)+a^4 e^4+a b^3 e \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )+b^4 \left (10 d^2 e^2 x^2+5 d^3 e x+d^4+10 d e^3 x^3+5 e^4 x^4\right )}{5 e^5 (d+e x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^2/(d + e*x)^6,x]

[Out]

-(a^4*e^4 + a^3*b*e^3*(d + 5*e*x) + a^2*b^2*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + a*b^3*e*(d^3 + 5*d^2*e*x + 10*d
*e^2*x^2 + 10*e^3*x^3) + b^4*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4))/(5*e^5*(d + e*x)^5
)

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Maple [B]  time = 0.046, size = 186, normalized size = 6.6 \begin{align*} -2\,{\frac{{b}^{2} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) }{{e}^{5} \left ( ex+d \right ) ^{3}}}-{\frac{b \left ({a}^{3}{e}^{3}-3\,{a}^{2}bd{e}^{2}+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) }{{e}^{5} \left ( ex+d \right ) ^{4}}}-2\,{\frac{{b}^{3} \left ( ae-bd \right ) }{{e}^{5} \left ( ex+d \right ) ^{2}}}-{\frac{{b}^{4}}{{e}^{5} \left ( ex+d \right ) }}-{\frac{{a}^{4}{e}^{4}-4\,{a}^{3}bd{e}^{3}+6\,{d}^{2}{e}^{2}{b}^{2}{a}^{2}-4\,{d}^{3}ea{b}^{3}+{b}^{4}{d}^{4}}{5\,{e}^{5} \left ( ex+d \right ) ^{5}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^6,x)

[Out]

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

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Maxima [B]  time = 1.16982, size = 290, normalized size = 10.36 \begin{align*} -\frac{5 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} + a^{3} b d e^{3} + a^{4} e^{4} + 10 \,{\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + 10 \,{\left (b^{4} d^{2} e^{2} + a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 5 \,{\left (b^{4} d^{3} e + a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x}{5 \,{\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

-1/5*(5*b^4*e^4*x^4 + b^4*d^4 + a*b^3*d^3*e + a^2*b^2*d^2*e^2 + a^3*b*d*e^3 + a^4*e^4 + 10*(b^4*d*e^3 + a*b^3*
e^4)*x^3 + 10*(b^4*d^2*e^2 + a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 5*(b^4*d^3*e + a*b^3*d^2*e^2 + a^2*b^2*d*e^3 + a
^3*b*e^4)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 10*d^2*e^8*x^3 + 10*d^3*e^7*x^2 + 5*d^4*e^6*x + d^5*e^5)

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Fricas [B]  time = 1.92392, size = 428, normalized size = 15.29 \begin{align*} -\frac{5 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} + a^{3} b d e^{3} + a^{4} e^{4} + 10 \,{\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + 10 \,{\left (b^{4} d^{2} e^{2} + a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 5 \,{\left (b^{4} d^{3} e + a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x}{5 \,{\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/5*(5*b^4*e^4*x^4 + b^4*d^4 + a*b^3*d^3*e + a^2*b^2*d^2*e^2 + a^3*b*d*e^3 + a^4*e^4 + 10*(b^4*d*e^3 + a*b^3*
e^4)*x^3 + 10*(b^4*d^2*e^2 + a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 5*(b^4*d^3*e + a*b^3*d^2*e^2 + a^2*b^2*d*e^3 + a
^3*b*e^4)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 10*d^2*e^8*x^3 + 10*d^3*e^7*x^2 + 5*d^4*e^6*x + d^5*e^5)

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Sympy [B]  time = 6.00661, size = 233, normalized size = 8.32 \begin{align*} - \frac{a^{4} e^{4} + a^{3} b d e^{3} + a^{2} b^{2} d^{2} e^{2} + a b^{3} d^{3} e + b^{4} d^{4} + 5 b^{4} e^{4} x^{4} + x^{3} \left (10 a b^{3} e^{4} + 10 b^{4} d e^{3}\right ) + x^{2} \left (10 a^{2} b^{2} e^{4} + 10 a b^{3} d e^{3} + 10 b^{4} d^{2} e^{2}\right ) + x \left (5 a^{3} b e^{4} + 5 a^{2} b^{2} d e^{3} + 5 a b^{3} d^{2} e^{2} + 5 b^{4} d^{3} e\right )}{5 d^{5} e^{5} + 25 d^{4} e^{6} x + 50 d^{3} e^{7} x^{2} + 50 d^{2} e^{8} x^{3} + 25 d e^{9} x^{4} + 5 e^{10} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**2/(e*x+d)**6,x)

[Out]

-(a**4*e**4 + a**3*b*d*e**3 + a**2*b**2*d**2*e**2 + a*b**3*d**3*e + b**4*d**4 + 5*b**4*e**4*x**4 + x**3*(10*a*
b**3*e**4 + 10*b**4*d*e**3) + x**2*(10*a**2*b**2*e**4 + 10*a*b**3*d*e**3 + 10*b**4*d**2*e**2) + x*(5*a**3*b*e*
*4 + 5*a**2*b**2*d*e**3 + 5*a*b**3*d**2*e**2 + 5*b**4*d**3*e))/(5*d**5*e**5 + 25*d**4*e**6*x + 50*d**3*e**7*x*
*2 + 50*d**2*e**8*x**3 + 25*d*e**9*x**4 + 5*e**10*x**5)

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Giac [B]  time = 1.171, size = 230, normalized size = 8.21 \begin{align*} -\frac{{\left (5 \, b^{4} x^{4} e^{4} + 10 \, b^{4} d x^{3} e^{3} + 10 \, b^{4} d^{2} x^{2} e^{2} + 5 \, b^{4} d^{3} x e + b^{4} d^{4} + 10 \, a b^{3} x^{3} e^{4} + 10 \, a b^{3} d x^{2} e^{3} + 5 \, a b^{3} d^{2} x e^{2} + a b^{3} d^{3} e + 10 \, a^{2} b^{2} x^{2} e^{4} + 5 \, a^{2} b^{2} d x e^{3} + a^{2} b^{2} d^{2} e^{2} + 5 \, a^{3} b x e^{4} + a^{3} b d e^{3} + a^{4} e^{4}\right )} e^{\left (-5\right )}}{5 \,{\left (x e + d\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/5*(5*b^4*x^4*e^4 + 10*b^4*d*x^3*e^3 + 10*b^4*d^2*x^2*e^2 + 5*b^4*d^3*x*e + b^4*d^4 + 10*a*b^3*x^3*e^4 + 10*
a*b^3*d*x^2*e^3 + 5*a*b^3*d^2*x*e^2 + a*b^3*d^3*e + 10*a^2*b^2*x^2*e^4 + 5*a^2*b^2*d*x*e^3 + a^2*b^2*d^2*e^2 +
5*a^3*b*x*e^4 + a^3*b*d*e^3 + a^4*e^4)*e^(-5)/(x*e + d)^5