3.13.33 \(\int \frac {(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=65 \[ -\frac {e (b d-a e)}{2 b^3 (a+b x)^4}-\frac {(b d-a e)^2}{5 b^3 (a+b x)^5}-\frac {e^2}{3 b^3 (a+b x)^3} \]

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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 = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 43} \begin {gather*} -\frac {e (b d-a e)}{2 b^3 (a+b x)^4}-\frac {(b d-a e)^2}{5 b^3 (a+b x)^5}-\frac {e^2}{3 b^3 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

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

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 {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^2}{(a+b x)^6} \, dx\\ &=\int \left (\frac {(b d-a e)^2}{b^2 (a+b x)^6}+\frac {2 e (b d-a e)}{b^2 (a+b x)^5}+\frac {e^2}{b^2 (a+b x)^4}\right ) \, dx\\ &=-\frac {(b d-a e)^2}{5 b^3 (a+b x)^5}-\frac {e (b d-a e)}{2 b^3 (a+b x)^4}-\frac {e^2}{3 b^3 (a+b x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.88 \begin {gather*} -\frac {a^2 e^2+a b e (3 d+5 e x)+b^2 \left (6 d^2+15 d e x+10 e^2 x^2\right )}{30 b^3 (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

-1/30*(a^2*e^2 + a*b*e*(3*d + 5*e*x) + b^2*(6*d^2 + 15*d*e*x + 10*e^2*x^2))/(b^3*(a + b*x)^5)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

IntegrateAlgebraic[(d + e*x)^2/(a^2 + 2*a*b*x + b^2*x^2)^3, x]

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fricas [A]  time = 0.38, size = 109, normalized size = 1.68 \begin {gather*} -\frac {10 \, b^{2} e^{2} x^{2} + 6 \, b^{2} d^{2} + 3 \, a b d e + a^{2} e^{2} + 5 \, {\left (3 \, b^{2} d e + a b e^{2}\right )} x}{30 \, {\left (b^{8} x^{5} + 5 \, a b^{7} x^{4} + 10 \, a^{2} b^{6} x^{3} + 10 \, a^{3} b^{5} x^{2} + 5 \, a^{4} b^{4} x + a^{5} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

-1/30*(10*b^2*e^2*x^2 + 6*b^2*d^2 + 3*a*b*d*e + a^2*e^2 + 5*(3*b^2*d*e + a*b*e^2)*x)/(b^8*x^5 + 5*a*b^7*x^4 +
10*a^2*b^6*x^3 + 10*a^3*b^5*x^2 + 5*a^4*b^4*x + a^5*b^3)

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giac [A]  time = 0.16, size = 60, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{2} x^{2} e^{2} + 15 \, b^{2} d x e + 6 \, b^{2} d^{2} + 5 \, a b x e^{2} + 3 \, a b d e + a^{2} e^{2}}{30 \, {\left (b x + a\right )}^{5} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

-1/30*(10*b^2*x^2*e^2 + 15*b^2*d*x*e + 6*b^2*d^2 + 5*a*b*x*e^2 + 3*a*b*d*e + a^2*e^2)/((b*x + a)^5*b^3)

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maple [A]  time = 0.05, size = 71, normalized size = 1.09 \begin {gather*} -\frac {e^{2}}{3 \left (b x +a \right )^{3} b^{3}}+\frac {\left (a e -b d \right ) e}{2 \left (b x +a \right )^{4} b^{3}}-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{5 \left (b x +a \right )^{5} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

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

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maxima [A]  time = 1.41, size = 109, normalized size = 1.68 \begin {gather*} -\frac {10 \, b^{2} e^{2} x^{2} + 6 \, b^{2} d^{2} + 3 \, a b d e + a^{2} e^{2} + 5 \, {\left (3 \, b^{2} d e + a b e^{2}\right )} x}{30 \, {\left (b^{8} x^{5} + 5 \, a b^{7} x^{4} + 10 \, a^{2} b^{6} x^{3} + 10 \, a^{3} b^{5} x^{2} + 5 \, a^{4} b^{4} x + a^{5} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

-1/30*(10*b^2*e^2*x^2 + 6*b^2*d^2 + 3*a*b*d*e + a^2*e^2 + 5*(3*b^2*d*e + a*b*e^2)*x)/(b^8*x^5 + 5*a*b^7*x^4 +
10*a^2*b^6*x^3 + 10*a^3*b^5*x^2 + 5*a^4*b^4*x + a^5*b^3)

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mupad [B]  time = 0.06, size = 107, normalized size = 1.65 \begin {gather*} -\frac {\frac {a^2\,e^2+3\,a\,b\,d\,e+6\,b^2\,d^2}{30\,b^3}+\frac {e^2\,x^2}{3\,b}+\frac {e\,x\,\left (a\,e+3\,b\,d\right )}{6\,b^2}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

-((a^2*e^2 + 6*b^2*d^2 + 3*a*b*d*e)/(30*b^3) + (e^2*x^2)/(3*b) + (e*x*(a*e + 3*b*d))/(6*b^2))/(a^5 + b^5*x^5 +
5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x)

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sympy [B]  time = 0.97, size = 116, normalized size = 1.78 \begin {gather*} \frac {- a^{2} e^{2} - 3 a b d e - 6 b^{2} d^{2} - 10 b^{2} e^{2} x^{2} + x \left (- 5 a b e^{2} - 15 b^{2} d e\right )}{30 a^{5} b^{3} + 150 a^{4} b^{4} x + 300 a^{3} b^{5} x^{2} + 300 a^{2} b^{6} x^{3} + 150 a b^{7} x^{4} + 30 b^{8} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

(-a**2*e**2 - 3*a*b*d*e - 6*b**2*d**2 - 10*b**2*e**2*x**2 + x*(-5*a*b*e**2 - 15*b**2*d*e))/(30*a**5*b**3 + 150
*a**4*b**4*x + 300*a**3*b**5*x**2 + 300*a**2*b**6*x**3 + 150*a*b**7*x**4 + 30*b**8*x**5)

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