3.13.14 \(\int \frac {1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=81 \[ -\frac {b}{(a+b x) (b d-a e)^2}-\frac {e}{(d+e x) (b d-a e)^2}-\frac {2 b e \log (a+b x)}{(b d-a e)^3}+\frac {2 b e \log (d+e x)}{(b d-a e)^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 81, 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, 44} \begin {gather*} -\frac {b}{(a+b x) (b d-a e)^2}-\frac {e}{(d+e x) (b d-a e)^2}-\frac {2 b e \log (a+b x)}{(b d-a e)^3}+\frac {2 b e \log (d+e x)}{(b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

-(b/((b*d - a*e)^2*(a + b*x))) - e/((b*d - a*e)^2*(d + e*x)) - (2*b*e*Log[a + b*x])/(b*d - a*e)^3 + (2*b*e*Log
[d + e*x])/(b*d - a*e)^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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac {1}{(a+b x)^2 (d+e x)^2} \, dx\\ &=\int \left (\frac {b^2}{(b d-a e)^2 (a+b x)^2}-\frac {2 b^2 e}{(b d-a e)^3 (a+b x)}+\frac {e^2}{(b d-a e)^2 (d+e x)^2}+\frac {2 b e^2}{(b d-a e)^3 (d+e x)}\right ) \, dx\\ &=-\frac {b}{(b d-a e)^2 (a+b x)}-\frac {e}{(b d-a e)^2 (d+e x)}-\frac {2 b e \log (a+b x)}{(b d-a e)^3}+\frac {2 b e \log (d+e x)}{(b d-a e)^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 66, normalized size = 0.81 \begin {gather*} \frac {\frac {b (a e-b d)}{a+b x}+\frac {e (a e-b d)}{d+e x}-2 b e \log (a+b x)+2 b e \log (d+e x)}{(b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

IntegrateAlgebraic[1/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)), x]

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fricas [B]  time = 0.39, size = 241, normalized size = 2.98 \begin {gather*} -\frac {b^{2} d^{2} - a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{4} - 3 \, a^{2} b^{2} d^{3} e + 3 \, a^{3} b d^{2} e^{2} - a^{4} d e^{3} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x^{2} + {\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + 2 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 156, normalized size = 1.93 \begin {gather*} -\frac {2 \, b e^{2} \log \left ({\left | b - \frac {b d}{x e + d} + \frac {a e}{x e + d} \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} - \frac {e^{3}}{{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} {\left (x e + d\right )}} - \frac {b^{2} e}{{\left (b d - a e\right )}^{3} {\left (b - \frac {b d}{x e + d} + \frac {a e}{x e + d}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 82, normalized size = 1.01 \begin {gather*} \frac {2 b e \ln \left (b x +a \right )}{\left (a e -b d \right )^{3}}-\frac {2 b e \ln \left (e x +d \right )}{\left (a e -b d \right )^{3}}-\frac {b}{\left (a e -b d \right )^{2} \left (b x +a \right )}-\frac {e}{\left (a e -b d \right )^{2} \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

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

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maxima [B]  time = 1.48, size = 208, normalized size = 2.57 \begin {gather*} -\frac {2 \, b e \log \left (b x + a\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} + \frac {2 \, b e \log \left (e x + d\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} - \frac {2 \, b e x + b d + a e}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.61, size = 182, normalized size = 2.25 \begin {gather*} \frac {4\,b\,e\,\mathrm {atanh}\left (\frac {a^3\,e^3-a^2\,b\,d\,e^2-a\,b^2\,d^2\,e+b^3\,d^3}{{\left (a\,e-b\,d\right )}^3}+\frac {2\,b\,e\,x\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^3}\right )}{{\left (a\,e-b\,d\right )}^3}-\frac {\frac {a\,e+b\,d}{a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2}+\frac {2\,b\,e\,x}{a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2}}{b\,e\,x^2+\left (a\,e+b\,d\right )\,x+a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)),x)

[Out]

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

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sympy [B]  time = 1.15, size = 406, normalized size = 5.01 \begin {gather*} - \frac {2 b e \log {\left (x + \frac {- \frac {2 a^{4} b e^{5}}{\left (a e - b d\right )^{3}} + \frac {8 a^{3} b^{2} d e^{4}}{\left (a e - b d\right )^{3}} - \frac {12 a^{2} b^{3} d^{2} e^{3}}{\left (a e - b d\right )^{3}} + \frac {8 a b^{4} d^{3} e^{2}}{\left (a e - b d\right )^{3}} + 2 a b e^{2} - \frac {2 b^{5} d^{4} e}{\left (a e - b d\right )^{3}} + 2 b^{2} d e}{4 b^{2} e^{2}} \right )}}{\left (a e - b d\right )^{3}} + \frac {2 b e \log {\left (x + \frac {\frac {2 a^{4} b e^{5}}{\left (a e - b d\right )^{3}} - \frac {8 a^{3} b^{2} d e^{4}}{\left (a e - b d\right )^{3}} + \frac {12 a^{2} b^{3} d^{2} e^{3}}{\left (a e - b d\right )^{3}} - \frac {8 a b^{4} d^{3} e^{2}}{\left (a e - b d\right )^{3}} + 2 a b e^{2} + \frac {2 b^{5} d^{4} e}{\left (a e - b d\right )^{3}} + 2 b^{2} d e}{4 b^{2} e^{2}} \right )}}{\left (a e - b d\right )^{3}} + \frac {- a e - b d - 2 b e x}{a^{3} d e^{2} - 2 a^{2} b d^{2} e + a b^{2} d^{3} + x^{2} \left (a^{2} b e^{3} - 2 a b^{2} d e^{2} + b^{3} d^{2} e\right ) + x \left (a^{3} e^{3} - a^{2} b d e^{2} - a b^{2} d^{2} e + b^{3} d^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

-2*b*e*log(x + (-2*a**4*b*e**5/(a*e - b*d)**3 + 8*a**3*b**2*d*e**4/(a*e - b*d)**3 - 12*a**2*b**3*d**2*e**3/(a*
e - b*d)**3 + 8*a*b**4*d**3*e**2/(a*e - b*d)**3 + 2*a*b*e**2 - 2*b**5*d**4*e/(a*e - b*d)**3 + 2*b**2*d*e)/(4*b
**2*e**2))/(a*e - b*d)**3 + 2*b*e*log(x + (2*a**4*b*e**5/(a*e - b*d)**3 - 8*a**3*b**2*d*e**4/(a*e - b*d)**3 +
12*a**2*b**3*d**2*e**3/(a*e - b*d)**3 - 8*a*b**4*d**3*e**2/(a*e - b*d)**3 + 2*a*b*e**2 + 2*b**5*d**4*e/(a*e -
b*d)**3 + 2*b**2*d*e)/(4*b**2*e**2))/(a*e - b*d)**3 + (-a*e - b*d - 2*b*e*x)/(a**3*d*e**2 - 2*a**2*b*d**2*e +
a*b**2*d**3 + x**2*(a**2*b*e**3 - 2*a*b**2*d*e**2 + b**3*d**2*e) + x*(a**3*e**3 - a**2*b*d*e**2 - a*b**2*d**2*
e + b**3*d**3))

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