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

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

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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 = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 43} \begin {gather*} -\frac {2 e (b d-a e)}{3 b^3 (a+b x)^3}-\frac {(b d-a e)^2}{4 b^3 (a+b x)^4}-\frac {e^2}{2 b^3 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 56, normalized size = 0.86 \begin {gather*} -\frac {a^2 e^2+2 a b e (d+2 e x)+b^2 \left (3 d^2+8 d e x+6 e^2 x^2\right )}{12 b^3 (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [A]  time = 0.40, size = 98, normalized size = 1.51 \begin {gather*} -\frac {6 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 2 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 60, normalized size = 0.92 \begin {gather*} -\frac {6 \, b^{2} x^{2} e^{2} + 8 \, b^{2} d x e + 3 \, b^{2} d^{2} + 4 \, a b x e^{2} + 2 \, a b d e + a^{2} e^{2}}{12 \, {\left (b x + a\right )}^{4} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 71, normalized size = 1.09 \begin {gather*} -\frac {e^{2}}{2 \left (b x +a \right )^{2} b^{3}}+\frac {2 \left (a e -b d \right ) e}{3 \left (b x +a \right )^{3} b^{3}}-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{4 \left (b x +a \right )^{4} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.67, size = 98, normalized size = 1.51 \begin {gather*} -\frac {6 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 2 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 96, normalized size = 1.48 \begin {gather*} -\frac {\frac {a^2\,e^2+2\,a\,b\,d\,e+3\,b^2\,d^2}{12\,b^3}+\frac {e^2\,x^2}{2\,b}+\frac {e\,x\,\left (a\,e+2\,b\,d\right )}{3\,b^2}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.80, size = 104, normalized size = 1.60 \begin {gather*} \frac {- a^{2} e^{2} - 2 a b d e - 3 b^{2} d^{2} - 6 b^{2} e^{2} x^{2} + x \left (- 4 a b e^{2} - 8 b^{2} d e\right )}{12 a^{4} b^{3} + 48 a^{3} b^{4} x + 72 a^{2} b^{5} x^{2} + 48 a b^{6} x^{3} + 12 b^{7} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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