3.265 \(\int \frac{1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx\)

Optimal. Leaf size=87 \[ -\frac{c^2 \log (b+c x)}{b (c d-b e)^2}+\frac{e (2 c d-b e) \log (d+e x)}{d^2 (c d-b e)^2}-\frac{e}{d (d+e x) (c d-b e)}+\frac{\log (x)}{b d^2} \]

[Out]

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

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Rubi [A]  time = 0.192499, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105 \[ -\frac{c^2 \log (b+c x)}{b (c d-b e)^2}+\frac{e (2 c d-b e) \log (d+e x)}{d^2 (c d-b e)^2}-\frac{e}{d (d+e x) (c d-b e)}+\frac{\log (x)}{b d^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 28.7541, size = 71, normalized size = 0.82 \[ \frac{e}{d \left (d + e x\right ) \left (b e - c d\right )} - \frac{e \left (b e - 2 c d\right ) \log{\left (d + e x \right )}}{d^{2} \left (b e - c d\right )^{2}} - \frac{c^{2} \log{\left (b + c x \right )}}{b \left (b e - c d\right )^{2}} + \frac{\log{\left (x \right )}}{b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/(e*x+d)**2/(c*x**2+b*x),x)

[Out]

e/(d*(d + e*x)*(b*e - c*d)) - e*(b*e - 2*c*d)*log(d + e*x)/(d**2*(b*e - c*d)**2)
- c**2*log(b + c*x)/(b*(b*e - c*d)**2) + log(x)/(b*d**2)

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Mathematica [A]  time = 0.169706, size = 83, normalized size = 0.95 \[ \frac{\frac{b e ((d+e x) (2 c d-b e) \log (d+e x)+d (b e-c d))-c^2 d^2 (d+e x) \log (b+c x)}{(d+e x) (c d-b e)^2}+\log (x)}{b d^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [A]  time = 0.027, size = 105, normalized size = 1.2 \[{\frac{\ln \left ( x \right ) }{{d}^{2}b}}-{\frac{{c}^{2}\ln \left ( cx+b \right ) }{ \left ( be-cd \right ) ^{2}b}}+{\frac{e}{d \left ( be-cd \right ) \left ( ex+d \right ) }}-{\frac{{e}^{2}\ln \left ( ex+d \right ) b}{{d}^{2} \left ( be-cd \right ) ^{2}}}+2\,{\frac{e\ln \left ( ex+d \right ) c}{d \left ( be-cd \right ) ^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/(e*x+d)^2/(c*x^2+b*x),x)

[Out]

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

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Maxima [A]  time = 0.701009, size = 173, normalized size = 1.99 \[ -\frac{c^{2} \log \left (c x + b\right )}{b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}} + \frac{{\left (2 \, c d e - b e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac{e}{c d^{3} - b d^{2} e +{\left (c d^{2} e - b d e^{2}\right )} x} + \frac{\log \left (x\right )}{b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((c*x^2 + b*x)*(e*x + d)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.03931, size = 279, normalized size = 3.21 \[ -\frac{b c d^{2} e - b^{2} d e^{2} +{\left (c^{2} d^{2} e x + c^{2} d^{3}\right )} \log \left (c x + b\right ) -{\left (2 \, b c d^{2} e - b^{2} d e^{2} +{\left (2 \, b c d e^{2} - b^{2} e^{3}\right )} x\right )} \log \left (e x + d\right ) -{\left (c^{2} d^{3} - 2 \, b c d^{2} e + b^{2} d e^{2} +{\left (c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \log \left (x\right )}{b c^{2} d^{5} - 2 \, b^{2} c d^{4} e + b^{3} d^{3} e^{2} +{\left (b c^{2} d^{4} e - 2 \, b^{2} c d^{3} e^{2} + b^{3} d^{2} e^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((c*x^2 + b*x)*(e*x + d)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(e*x+d)**2/(c*x**2+b*x),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.228496, size = 389, normalized size = 4.47 \[ -\frac{{\left (2 \, c^{2} d^{2} e^{2} - 2 \, b c d e^{3} + b^{2} e^{4}\right )} e^{\left (-2\right )}{\rm ln}\left (\frac{{\left | -2 \, c d e + \frac{2 \, c d^{2} e}{x e + d} + b e^{2} - \frac{2 \, b d e^{2}}{x e + d} -{\left | b \right |} e^{2} \right |}}{{\left | -2 \, c d e + \frac{2 \, c d^{2} e}{x e + d} + b e^{2} - \frac{2 \, b d e^{2}}{x e + d} +{\left | b \right |} e^{2} \right |}}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}{\left | b \right |}} - \frac{{\left (2 \, c d e - b e^{2}\right )}{\rm ln}\left ({\left | -c + \frac{2 \, c d}{x e + d} - \frac{c d^{2}}{{\left (x e + d\right )}^{2}} - \frac{b e}{x e + d} + \frac{b d e}{{\left (x e + d\right )}^{2}} \right |}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}} - \frac{e^{3}}{{\left (c d^{2} e^{2} - b d e^{3}\right )}{\left (x e + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((c*x^2 + b*x)*(e*x + d)^2),x, algorithm="giac")

[Out]

-1/2*(2*c^2*d^2*e^2 - 2*b*c*d*e^3 + b^2*e^4)*e^(-2)*ln(abs(-2*c*d*e + 2*c*d^2*e/
(x*e + d) + b*e^2 - 2*b*d*e^2/(x*e + d) - abs(b)*e^2)/abs(-2*c*d*e + 2*c*d^2*e/(
x*e + d) + b*e^2 - 2*b*d*e^2/(x*e + d) + abs(b)*e^2))/((c^2*d^4 - 2*b*c*d^3*e +
b^2*d^2*e^2)*abs(b)) - 1/2*(2*c*d*e - b*e^2)*ln(abs(-c + 2*c*d/(x*e + d) - c*d^2
/(x*e + d)^2 - b*e/(x*e + d) + b*d*e/(x*e + d)^2))/(c^2*d^4 - 2*b*c*d^3*e + b^2*
d^2*e^2) - e^3/((c*d^2*e^2 - b*d*e^3)*(x*e + d))