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

Optimal. Leaf size=43 \[ -\frac {2 c \log (x)}{b^3}+\frac {2 c \log (b+c x)}{b^3}-\frac {b+2 c x}{b^2 \left (b x+c x^2\right )} \]

[Out]

(-2*c*x-b)/b^2/(c*x^2+b*x)-2*c*ln(x)/b^3+2*c*ln(c*x+b)/b^3

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {614, 615} \[ -\frac {b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac {2 c \log (x)}{b^3}+\frac {2 c \log (b+c x)}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-2),x]

[Out]

-((b + 2*c*x)/(b^2*(b*x + c*x^2))) - (2*c*Log[x])/b^3 + (2*c*Log[b + c*x])/b^3

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 615

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 35, normalized size = 0.81 \[ -\frac {b \left (\frac {c}{b+c x}+\frac {1}{x}\right )-2 c \log (b+c x)+2 c \log (x)}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-2),x]

[Out]

-((b*(x^(-1) + c/(b + c*x)) + 2*c*Log[x] - 2*c*Log[b + c*x])/b^3)

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fricas [A]  time = 0.88, size = 63, normalized size = 1.47 \[ -\frac {2 \, b c x + b^{2} - 2 \, {\left (c^{2} x^{2} + b c x\right )} \log \left (c x + b\right ) + 2 \, {\left (c^{2} x^{2} + b c x\right )} \log \relax (x)}{b^{3} c x^{2} + b^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 45, normalized size = 1.05 \[ \frac {2 \, c \log \left ({\left | c x + b \right |}\right )}{b^{3}} - \frac {2 \, c \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {2 \, c x + b}{{\left (c x^{2} + b x\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*log(abs(c*x + b))/b^3 - 2*c*log(abs(x))/b^3 - (2*c*x + b)/((c*x^2 + b*x)*b^2)

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maple [A]  time = 0.05, size = 43, normalized size = 1.00 \[ -\frac {c}{\left (c x +b \right ) b^{2}}-\frac {2 c \ln \relax (x )}{b^{3}}+\frac {2 c \ln \left (c x +b \right )}{b^{3}}-\frac {1}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^2*c/(c*x+b)+2*c*ln(c*x+b)/b^3-1/b^2/x-2*c*ln(x)/b^3

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maxima [A]  time = 1.28, size = 45, normalized size = 1.05 \[ -\frac {2 \, c x + b}{b^{2} c x^{2} + b^{3} x} + \frac {2 \, c \log \left (c x + b\right )}{b^{3}} - \frac {2 \, c \log \relax (x)}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 41, normalized size = 0.95 \[ \frac {4\,c\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^3}-\frac {\frac {1}{b}+\frac {2\,c\,x}{b^2}}{c\,x^2+b\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^2,x)

[Out]

(4*c*atanh((2*c*x)/b + 1))/b^3 - (1/b + (2*c*x)/b^2)/(b*x + c*x^2)

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sympy [A]  time = 0.27, size = 37, normalized size = 0.86 \[ \frac {- b - 2 c x}{b^{3} x + b^{2} c x^{2}} + \frac {2 c \left (- \log {\relax (x )} + \log {\left (\frac {b}{c} + x \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**2,x)

[Out]

(-b - 2*c*x)/(b**3*x + b**2*c*x**2) + 2*c*(-log(x) + log(b/c + x))/b**3

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