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

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

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {698} \[ -\frac {c^2 \log (b+c x)}{b (c d-b e)^2}+\frac {e (2 c d-b e) \log (d+e x)}{d^2 (c d-b e)^2}-\frac {e}{d (d+e x) (c d-b e)}+\frac {\log (x)}{b d^2} \]

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)

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.09, 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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fricas [B]  time = 2.85, size = 207, normalized size = 2.38 \[ -\frac {b c d^{2} e - b^{2} d e^{2} + {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )} \log \left (c x + b\right ) - {\left (2 \, b c d^{2} e - b^{2} d e^{2} + {\left (2 \, b c d e^{2} - b^{2} e^{3}\right )} x\right )} \log \left (e x + d\right ) - {\left (c^{2} d^{3} - 2 \, b c d^{2} e + b^{2} d e^{2} + {\left (c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \log \relax (x)}{b c^{2} d^{5} - 2 \, b^{2} c d^{4} e + b^{3} d^{3} e^{2} + {\left (b c^{2} d^{4} e - 2 \, b^{2} c d^{3} e^{2} + b^{3} d^{2} e^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x),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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giac [B]  time = 0.22, size = 288, normalized size = 3.31 \[ -\frac {{\left (2 \, c^{2} d^{2} e^{2} - 2 \, b c d e^{3} + b^{2} e^{4}\right )} e^{\left (-2\right )} \log \left (\frac {{\left | -2 \, c d e + \frac {2 \, c d^{2} e}{x e + d} + b e^{2} - \frac {2 \, b d e^{2}}{x e + d} - {\left | b \right |} e^{2} \right |}}{{\left | -2 \, c d e + \frac {2 \, c d^{2} e}{x e + d} + b e^{2} - \frac {2 \, b d e^{2}}{x e + d} + {\left | b \right |} e^{2} \right |}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} {\left | b \right |}} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | -c + \frac {2 \, c d}{x e + d} - \frac {c d^{2}}{{\left (x e + d\right )}^{2}} - \frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} \right |}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}} - \frac {e^{3}}{{\left (c d^{2} e^{2} - b d e^{3}\right )} {\left (x e + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x),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)*log(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)*log(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))

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maple [A]  time = 0.06, size = 105, normalized size = 1.21 \[ -\frac {b \,e^{2} \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d^{2}}-\frac {c^{2} \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b}+\frac {2 c e \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d}+\frac {e}{\left (b e -c d \right ) \left (e x +d \right ) d}+\frac {\ln \relax (x )}{b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, size = 128, normalized size = 1.47 \[ -\frac {c^{2} \log \left (c x + b\right )}{b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}} + \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac {e}{c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x} + \frac {\log \relax (x)}{b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x),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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mupad [B]  time = 0.47, size = 116, normalized size = 1.33 \[ \frac {\ln \relax (x)}{b\,d^2}-\frac {c^2\,\ln \left (b+c\,x\right )}{b^3\,e^2-2\,b^2\,c\,d\,e+b\,c^2\,d^2}-\frac {\ln \left (d+e\,x\right )\,\left (b\,e^2-2\,c\,d\,e\right )}{b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4}+\frac {e}{d\,\left (b\,e-c\,d\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x + c*x^2)*(d + e*x)^2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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