3.1460 \(\int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^5} \, dx\)

Optimal. Leaf size=65 \[ \frac {2 b (b d-a e)}{3 e^3 (d+e x)^3}-\frac {(b d-a e)^2}{4 e^3 (d+e x)^4}-\frac {b^2}{2 e^3 (d+e x)^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 43} \[ \frac {2 b (b d-a e)}{3 e^3 (d+e x)^3}-\frac {(b d-a e)^2}{4 e^3 (d+e x)^4}-\frac {b^2}{2 e^3 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 {a^2+2 a b x+b^2 x^2}{(d+e x)^5} \, dx &=\int \frac {(a+b x)^2}{(d+e x)^5} \, dx\\ &=\int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^5}-\frac {2 b (b d-a e)}{e^2 (d+e x)^4}+\frac {b^2}{e^2 (d+e x)^3}\right ) \, dx\\ &=-\frac {(b d-a e)^2}{4 e^3 (d+e x)^4}+\frac {2 b (b d-a e)}{3 e^3 (d+e x)^3}-\frac {b^2}{2 e^3 (d+e x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.85 \[ -\frac {3 a^2 e^2+2 a b e (d+4 e x)+b^2 \left (d^2+4 d e x+6 e^2 x^2\right )}{12 e^3 (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(3*a^2*e^2 + 2*a*b*e*(d + 4*e*x) + b^2*(d^2 + 4*d*e*x + 6*e^2*x^2))/(e^3*(d + e*x)^4)

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fricas [A]  time = 0.97, size = 98, normalized size = 1.51 \[ -\frac {6 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 2 \, a b d e + 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 96, normalized size = 1.48 \[ -\frac {1}{12} \, {\left (\frac {6 \, b^{2} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac {8 \, b^{2} d e^{\left (-2\right )}}{{\left (x e + d\right )}^{3}} + \frac {3 \, b^{2} d^{2} e^{\left (-2\right )}}{{\left (x e + d\right )}^{4}} + \frac {8 \, a b e^{\left (-1\right )}}{{\left (x e + d\right )}^{3}} - \frac {6 \, a b d e^{\left (-1\right )}}{{\left (x e + d\right )}^{4}} + \frac {3 \, a^{2}}{{\left (x e + d\right )}^{4}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*b^2*e^(-2)/(x*e + d)^2 - 8*b^2*d*e^(-2)/(x*e + d)^3 + 3*b^2*d^2*e^(-2)/(x*e + d)^4 + 8*a*b*e^(-1)/(x*
e + d)^3 - 6*a*b*d*e^(-1)/(x*e + d)^4 + 3*a^2/(x*e + d)^4)*e^(-1)

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maple [A]  time = 0.05, size = 71, normalized size = 1.09 \[ -\frac {b^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {2 \left (a e -b d \right ) b}{3 \left (e x +d \right )^{3} e^{3}}-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{4 \left (e x +d \right )^{4} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.34, size = 98, normalized size = 1.51 \[ -\frac {6 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 2 \, a b d e + 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 96, normalized size = 1.48 \[ -\frac {\frac {3\,a^2\,e^2+2\,a\,b\,d\,e+b^2\,d^2}{12\,e^3}+\frac {b^2\,x^2}{2\,e}+\frac {b\,x\,\left (2\,a\,e+b\,d\right )}{3\,e^2}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((3*a^2*e^2 + b^2*d^2 + 2*a*b*d*e)/(12*e^3) + (b^2*x^2)/(2*e) + (b*x*(2*a*e + b*d))/(3*e^2))/(d^4 + e^4*x^4 +
4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)

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sympy [A]  time = 0.77, size = 104, normalized size = 1.60 \[ \frac {- 3 a^{2} e^{2} - 2 a b d e - b^{2} d^{2} - 6 b^{2} e^{2} x^{2} + x \left (- 8 a b e^{2} - 4 b^{2} d e\right )}{12 d^{4} e^{3} + 48 d^{3} e^{4} x + 72 d^{2} e^{5} x^{2} + 48 d e^{6} x^{3} + 12 e^{7} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a**2*e**2 - 2*a*b*d*e - b**2*d**2 - 6*b**2*e**2*x**2 + x*(-8*a*b*e**2 - 4*b**2*d*e))/(12*d**4*e**3 + 48*d*
*3*e**4*x + 72*d**2*e**5*x**2 + 48*d*e**6*x**3 + 12*e**7*x**4)

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