### 3.282 $$\int \frac{1}{(b x+c x^2)^3} \, dx$$

Optimal. Leaf size=72 $\frac{6 c^2 \log (x)}{b^5}-\frac{6 c^2 \log (b+c x)}{b^5}+\frac{3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}-\frac{b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}$

[Out]

-(b + 2*c*x)/(2*b^2*(b*x + c*x^2)^2) + (3*c*(b + 2*c*x))/(b^4*(b*x + c*x^2)) + (6*c^2*Log[x])/b^5 - (6*c^2*Log
[b + c*x])/b^5

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Rubi [A]  time = 0.0192287, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {614, 615} $\frac{6 c^2 \log (x)}{b^5}-\frac{6 c^2 \log (b+c x)}{b^5}+\frac{3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}-\frac{b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(-3),x]

[Out]

-(b + 2*c*x)/(2*b^2*(b*x + c*x^2)^2) + (3*c*(b + 2*c*x))/(b^4*(b*x + c*x^2)) + (6*c^2*Log[x])/b^5 - (6*c^2*Log
[b + c*x])/b^5

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 )^3} \, dx &=-\frac{b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}-\frac{(3 c) \int \frac{1}{\left (b x+c x^2\right )^2} \, dx}{b^2}\\ &=-\frac{b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}+\frac{\left (6 c^2\right ) \int \frac{1}{b x+c x^2} \, dx}{b^4}\\ &=-\frac{b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}+\frac{3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}+\frac{6 c^2 \log (x)}{b^5}-\frac{6 c^2 \log (b+c x)}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0459729, size = 68, normalized size = 0.94 $\frac{\frac{b \left (4 b^2 c x-b^3+18 b c^2 x^2+12 c^3 x^3\right )}{x^2 (b+c x)^2}-12 c^2 \log (b+c x)+12 c^2 \log (x)}{2 b^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(-3),x]

[Out]

((b*(-b^3 + 4*b^2*c*x + 18*b*c^2*x^2 + 12*c^3*x^3))/(x^2*(b + c*x)^2) + 12*c^2*Log[x] - 12*c^2*Log[b + c*x])/(
2*b^5)

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Maple [A]  time = 0.051, size = 73, normalized size = 1. \begin{align*} -{\frac{1}{2\,{b}^{3}{x}^{2}}}+6\,{\frac{{c}^{2}\ln \left ( x \right ) }{{b}^{5}}}+3\,{\frac{c}{{b}^{4}x}}-6\,{\frac{{c}^{2}\ln \left ( cx+b \right ) }{{b}^{5}}}+3\,{\frac{{c}^{2}}{{b}^{4} \left ( cx+b \right ) }}+{\frac{{c}^{2}}{2\,{b}^{3} \left ( cx+b \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^3,x)

[Out]

-1/2/b^3/x^2+6*c^2*ln(x)/b^5+3/b^4*c/x-6*c^2*ln(c*x+b)/b^5+3/b^4*c^2/(c*x+b)+1/2*c^2/b^3/(c*x+b)^2

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Maxima [A]  time = 1.12503, size = 116, normalized size = 1.61 \begin{align*} \frac{12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x - b^{3}}{2 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} - \frac{6 \, c^{2} \log \left (c x + b\right )}{b^{5}} + \frac{6 \, c^{2} \log \left (x\right )}{b^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

1/2*(12*c^3*x^3 + 18*b*c^2*x^2 + 4*b^2*c*x - b^3)/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2) - 6*c^2*log(c*x + b)/b
^5 + 6*c^2*log(x)/b^5

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Fricas [A]  time = 1.69016, size = 269, normalized size = 3.74 \begin{align*} \frac{12 \, b c^{3} x^{3} + 18 \, b^{2} c^{2} x^{2} + 4 \, b^{3} c x - b^{4} - 12 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + b^{2} c^{2} x^{2}\right )} \log \left (c x + b\right ) + 12 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + b^{2} c^{2} x^{2}\right )} \log \left (x\right )}{2 \,{\left (b^{5} c^{2} x^{4} + 2 \, b^{6} c x^{3} + b^{7} x^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

1/2*(12*b*c^3*x^3 + 18*b^2*c^2*x^2 + 4*b^3*c*x - b^4 - 12*(c^4*x^4 + 2*b*c^3*x^3 + b^2*c^2*x^2)*log(c*x + b) +
12*(c^4*x^4 + 2*b*c^3*x^3 + b^2*c^2*x^2)*log(x))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2)

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Sympy [A]  time = 1.50587, size = 78, normalized size = 1.08 \begin{align*} \frac{- b^{3} + 4 b^{2} c x + 18 b c^{2} x^{2} + 12 c^{3} x^{3}}{2 b^{6} x^{2} + 4 b^{5} c x^{3} + 2 b^{4} c^{2} x^{4}} + \frac{6 c^{2} \left (\log{\left (x \right )} - \log{\left (\frac{b}{c} + x \right )}\right )}{b^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**3,x)

[Out]

(-b**3 + 4*b**2*c*x + 18*b*c**2*x**2 + 12*c**3*x**3)/(2*b**6*x**2 + 4*b**5*c*x**3 + 2*b**4*c**2*x**4) + 6*c**2
*(log(x) - log(b/c + x))/b**5

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Giac [A]  time = 1.29595, size = 99, normalized size = 1.38 \begin{align*} -\frac{6 \, c^{2} \log \left ({\left | c x + b \right |}\right )}{b^{5}} + \frac{6 \, c^{2} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac{12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x - b^{3}}{2 \,{\left (c x^{2} + b x\right )}^{2} b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-6*c^2*log(abs(c*x + b))/b^5 + 6*c^2*log(abs(x))/b^5 + 1/2*(12*c^3*x^3 + 18*b*c^2*x^2 + 4*b^2*c*x - b^3)/((c*x
^2 + b*x)^2*b^4)