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

Optimal. Leaf size=57 \[ -\frac {1}{(a+b x) (b d-a e)}-\frac {e \log (a+b x)}{(b d-a e)^2}+\frac {e \log (d+e x)}{(b d-a e)^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 57, 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 {1}{(a+b x) (b d-a e)}-\frac {e \log (a+b x)}{(b d-a e)^2}+\frac {e \log (d+e x)}{(b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.93 \begin {gather*} \frac {e (a+b x) \log (d+e x)-e (a+b x) \log (a+b x)+a e-b d}{(a+b x) (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [A]  time = 0.40, size = 93, normalized size = 1.63 \begin {gather*} -\frac {b d - a e + {\left (b e x + a e\right )} \log \left (b x + a\right ) - {\left (b e x + a e\right )} \log \left (e x + d\right )}{a b^{2} d^{2} - 2 \, a^{2} b d e + a^{3} e^{2} + {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 95, normalized size = 1.67 \begin {gather*} -\frac {b e \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} + \frac {e^{2} \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} - \frac {1}{{\left (b d - a e\right )} {\left (b x + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 57, normalized size = 1.00 \begin {gather*} -\frac {e \ln \left (b x +a \right )}{\left (a e -b d \right )^{2}}+\frac {e \ln \left (e x +d \right )}{\left (a e -b d \right )^{2}}+\frac {1}{\left (a e -b d \right ) \left (b x +a \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 92, normalized size = 1.61 \begin {gather*} -\frac {e \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac {e \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac {1}{a b d - a^{2} e + {\left (b^{2} d - a b e\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.58, size = 76, normalized size = 1.33 \begin {gather*} \frac {1}{\left (a\,e-b\,d\right )\,\left (a+b\,x\right )}-\frac {2\,e\,\mathrm {atanh}\left (\frac {a^2\,e^2-b^2\,d^2}{{\left (a\,e-b\,d\right )}^2}+\frac {2\,b\,e\,x}{a\,e-b\,d}\right )}{{\left (a\,e-b\,d\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.70, size = 233, normalized size = 4.09 \begin {gather*} \frac {e \log {\left (x + \frac {- \frac {a^{3} e^{4}}{\left (a e - b d\right )^{2}} + \frac {3 a^{2} b d e^{3}}{\left (a e - b d\right )^{2}} - \frac {3 a b^{2} d^{2} e^{2}}{\left (a e - b d\right )^{2}} + a e^{2} + \frac {b^{3} d^{3} e}{\left (a e - b d\right )^{2}} + b d e}{2 b e^{2}} \right )}}{\left (a e - b d\right )^{2}} - \frac {e \log {\left (x + \frac {\frac {a^{3} e^{4}}{\left (a e - b d\right )^{2}} - \frac {3 a^{2} b d e^{3}}{\left (a e - b d\right )^{2}} + \frac {3 a b^{2} d^{2} e^{2}}{\left (a e - b d\right )^{2}} + a e^{2} - \frac {b^{3} d^{3} e}{\left (a e - b d\right )^{2}} + b d e}{2 b e^{2}} \right )}}{\left (a e - b d\right )^{2}} + \frac {1}{a^{2} e - a b d + x \left (a b e - b^{2} d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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