\(\int \frac {1}{(b x+c x^2)^3} \, dx\) [282]

Optimal result

Integrand size = 11, antiderivative size = 72 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=-\frac {b+2 c x}{2 b^2 \left (b x+c x^2\right )^2}+\frac {3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}+\frac {6 c^2 \log (x)}{b^5}-\frac {6 c^2 \log (b+c x)}{b^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {628, 629} \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {6 c^2 \log (x)}{b^5}-\frac {6 c^2 \log (b+c x)}{b^5}+\frac {3 c (b+2 c x)}{b^4 \left (b x+c x^2\right )}-\frac {b+2 c x}{2 b^2 \left (b x+c x^2\right )^2} \]

[In]

Int[(b*x + c*x^2)^(-3),x]

[Out]

-1/2*(b + 2*c*x)/(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*L
og[b + c*x])/b^5

Rule 628

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 629

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*} \text {integral}& = -\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] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {b \left (-b^3+4 b^2 c x+18 b c^2 x^2+12 c^3 x^3\right )}{x^2 (b+c x)^2}+12 c^2 \log (x)-12 c^2 \log (b+c x)}{2 b^5} \]

[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)

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {-\frac {9 c^{4} x^{4}}{b^{5}}-\frac {1}{2 b}+\frac {2 c x}{b^{2}}-\frac {12 c^{3} x^{3}}{b^{4}}}{x^{2} \left (c x +b \right )^{2}}+\frac {6 c^{2} \ln \left (x \right )}{b^{5}}-\frac {6 c^{2} \ln \left (c x +b \right )}{b^{5}}\)	\(72\)
default	\(-\frac {1}{2 b^{3} x^{2}}+\frac {6 c^{2} \ln \left (x \right )}{b^{5}}+\frac {3 c}{b^{4} x}-\frac {6 c^{2} \ln \left (c x +b \right )}{b^{5}}+\frac {3 c^{2}}{b^{4} \left (c x +b \right )}+\frac {c^{2}}{2 b^{3} \left (c x +b \right )^{2}}\)	\(73\)
risch	\(\frac {\frac {6 c^{3} x^{3}}{b^{4}}+\frac {9 c^{2} x^{2}}{b^{3}}+\frac {2 c x}{b^{2}}-\frac {1}{2 b}}{x^{2} \left (c x +b \right )^{2}}-\frac {6 c^{2} \ln \left (c x +b \right )}{b^{5}}+\frac {6 c^{2} \ln \left (-x \right )}{b^{5}}\)	\(74\)
parallelrisch	\(\frac {12 \ln \left (x \right ) x^{4} c^{6}-12 \ln \left (c x +b \right ) x^{4} c^{6}+24 \ln \left (x \right ) x^{3} b \,c^{5}-24 \ln \left (c x +b \right ) x^{3} b \,c^{5}+12 \ln \left (x \right ) x^{2} b^{2} c^{4}-12 \ln \left (c x +b \right ) x^{2} b^{2} c^{4}+12 x^{3} b \,c^{5}+18 x^{2} b^{2} c^{4}+4 b^{3} c^{3} x -b^{4} c^{2}}{2 b^{5} c^{2} x^{2} \left (c x +b \right )^{2}}\)	\(137\)

[In]

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

[Out]

(-9*c^4/b^5*x^4-1/2/b+2*c/b^2*x-12*c^3/b^4*x^3)/x^2/(c*x+b)^2+6*c^2*ln(x)/b^5-6*c^2*ln(c*x+b)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {12 \, b c^{3} x^{3} + 18 \, b^{2} c^{2} x^{2} + 4 \, b^{3} c x - b^{4} - 12 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + b^{2} c^{2} x^{2}\right )} \log \left (c x + b\right ) + 12 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + b^{2} c^{2} x^{2}\right )} \log \left (x\right )}{2 \, {\left (b^{5} c^{2} x^{4} + 2 \, b^{6} c x^{3} + b^{7} x^{2}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {- b^{3} + 4 b^{2} c x + 18 b c^{2} x^{2} + 12 c^{3} x^{3}}{2 b^{6} x^{2} + 4 b^{5} c x^{3} + 2 b^{4} c^{2} x^{4}} + \frac {6 c^{2} \left (\log {\left (x \right )} - \log {\left (\frac {b}{c} + x \right )}\right )}{b^{5}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x - b^{3}}{2 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} - \frac {6 \, c^{2} \log \left (c x + b\right )}{b^{5}} + \frac {6 \, c^{2} \log \left (x\right )}{b^{5}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=-\frac {6 \, c^{2} \log \left ({\left | c x + b \right |}\right )}{b^{5}} + \frac {6 \, c^{2} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x - b^{3}}{2 \, {\left (c x^{2} + b x\right )}^{2} b^{4}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {9\,c^2\,x^2}{b^3}-\frac {1}{2\,b}+\frac {6\,c^3\,x^3}{b^4}+\frac {2\,c\,x}{b^2}}{b^2\,x^2+2\,b\,c\,x^3+c^2\,x^4}-\frac {12\,c^2\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )}{b^5} \]

[In]

int(1/(b*x + c*x^2)^3,x)

[Out]

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