\(\int \frac {A+B x}{(b x+c x^2)^3} \, dx\) [1159]

Optimal result

Integrand size = 17, antiderivative size = 109 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=-\frac {A}{2 b^3 x^2}-\frac {b B-3 A c}{b^4 x}-\frac {c (b B-A c)}{2 b^3 (b+c x)^2}-\frac {c (2 b B-3 A c)}{b^4 (b+c x)}-\frac {3 c (b B-2 A c) \log (x)}{b^5}+\frac {3 c (b B-2 A c) \log (b+c x)}{b^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {645} \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=-\frac {3 c \log (x) (b B-2 A c)}{b^5}+\frac {3 c (b B-2 A c) \log (b+c x)}{b^5}-\frac {b B-3 A c}{b^4 x}-\frac {c (2 b B-3 A c)}{b^4 (b+c x)}-\frac {c (b B-A c)}{2 b^3 (b+c x)^2}-\frac {A}{2 b^3 x^2} \]

[In]

Int[(A + B*x)/(b*x + c*x^2)^3,x]

[Out]

-1/2*A/(b^3*x^2) - (b*B - 3*A*c)/(b^4*x) - (c*(b*B - A*c))/(2*b^3*(b + c*x)^2) - (c*(2*b*B - 3*A*c))/(b^4*(b +
c*x)) - (3*c*(b*B - 2*A*c)*Log[x])/b^5 + (3*c*(b*B - 2*A*c)*Log[b + c*x])/b^5

Rule 645

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A}{b^3 x^3}+\frac {b B-3 A c}{b^4 x^2}-\frac {3 c (b B-2 A c)}{b^5 x}+\frac {c^2 (b B-A c)}{b^3 (b+c x)^3}+\frac {c^2 (2 b B-3 A c)}{b^4 (b+c x)^2}+\frac {3 c^2 (b B-2 A c)}{b^5 (b+c x)}\right ) \, dx \\ & = -\frac {A}{2 b^3 x^2}-\frac {b B-3 A c}{b^4 x}-\frac {c (b B-A c)}{2 b^3 (b+c x)^2}-\frac {c (2 b B-3 A c)}{b^4 (b+c x)}-\frac {3 c (b B-2 A c) \log (x)}{b^5}+\frac {3 c (b B-2 A c) \log (b+c x)}{b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=\frac {-\frac {b \left (b B x \left (2 b^2+9 b c x+6 c^2 x^2\right )+A \left (b^3-4 b^2 c x-18 b c^2 x^2-12 c^3 x^3\right )\right )}{x^2 (b+c x)^2}+6 c (-b B+2 A c) \log (x)+6 c (b B-2 A c) \log (b+c x)}{2 b^5} \]

[In]

Integrate[(A + B*x)/(b*x + c*x^2)^3,x]

[Out]

(-((b*(b*B*x*(2*b^2 + 9*b*c*x + 6*c^2*x^2) + A*(b^3 - 4*b^2*c*x - 18*b*c^2*x^2 - 12*c^3*x^3)))/(x^2*(b + c*x)^
2)) + 6*c*(-(b*B) + 2*A*c)*Log[x] + 6*c*(b*B - 2*A*c)*Log[b + c*x])/(2*b^5)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {A}{2 b^{3} x^{2}}-\frac {-3 A c +B b}{b^{4} x}+\frac {3 c \left (2 A c -B b \right ) \ln \left (x \right )}{b^{5}}-\frac {3 c \left (2 A c -B b \right ) \ln \left (c x +b \right )}{b^{5}}+\frac {c \left (3 A c -2 B b \right )}{b^{4} \left (c x +b \right )}+\frac {\left (A c -B b \right ) c}{2 b^{3} \left (c x +b \right )^{2}}\)	\(107\)
norman	\(\frac {\frac {\left (2 A c -B b \right ) x}{b^{2}}-\frac {A}{2 b}-\frac {2 c \left (6 A \,c^{2}-3 B b c \right ) x^{3}}{b^{4}}-\frac {c^{2} \left (18 A \,c^{2}-9 B b c \right ) x^{4}}{2 b^{5}}}{x^{2} \left (c x +b \right )^{2}}+\frac {3 c \left (2 A c -B b \right ) \ln \left (x \right )}{b^{5}}-\frac {3 c \left (2 A c -B b \right ) \ln \left (c x +b \right )}{b^{5}}\)	\(116\)
risch	\(\frac {\frac {3 c^{2} \left (2 A c -B b \right ) x^{3}}{b^{4}}+\frac {9 c \left (2 A c -B b \right ) x^{2}}{2 b^{3}}+\frac {\left (2 A c -B b \right ) x}{b^{2}}-\frac {A}{2 b}}{x^{2} \left (c x +b \right )^{2}}-\frac {6 c^{2} \ln \left (c x +b \right ) A}{b^{5}}+\frac {3 c \ln \left (c x +b \right ) B}{b^{4}}+\frac {6 c^{2} \ln \left (-x \right ) A}{b^{5}}-\frac {3 c \ln \left (-x \right ) B}{b^{4}}\)	\(124\)
parallelrisch	\(\frac {12 A \ln \left (x \right ) x^{4} c^{4}-12 A \ln \left (c x +b \right ) x^{4} c^{4}-6 B \ln \left (x \right ) x^{4} b \,c^{3}+6 B \ln \left (c x +b \right ) x^{4} b \,c^{3}+24 A \ln \left (x \right ) x^{3} b \,c^{3}-24 A \ln \left (c x +b \right ) x^{3} b \,c^{3}-18 A \,c^{4} x^{4}-12 B \ln \left (x \right ) x^{3} b^{2} c^{2}+12 B \ln \left (c x +b \right ) x^{3} b^{2} c^{2}+9 B b \,c^{3} x^{4}+12 A \ln \left (x \right ) x^{2} b^{2} c^{2}-12 A \ln \left (c x +b \right ) x^{2} b^{2} c^{2}-24 A b \,c^{3} x^{3}-6 B \ln \left (x \right ) x^{2} b^{3} c +6 B \ln \left (c x +b \right ) x^{2} b^{3} c +12 B \,b^{2} c^{2} x^{3}+4 A \,b^{3} c x -2 b^{4} B x -A \,b^{4}}{2 b^{5} x^{2} \left (c x +b \right )^{2}}\)	\(253\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (105) = 210\).

Time = 0.29 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.06 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=-\frac {A b^{4} + 6 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{3} + 9 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} + 2 \, {\left (B b^{4} - 2 \, A b^{3} c\right )} x - 6 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{4} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{3} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2}\right )} \log \left (c x + b\right ) + 6 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{4} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{3} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (b^{5} c^{2} x^{4} + 2 \, b^{6} c x^{3} + b^{7} x^{2}\right )}} \]

[In]

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

[Out]

-1/2*(A*b^4 + 6*(B*b^2*c^2 - 2*A*b*c^3)*x^3 + 9*(B*b^3*c - 2*A*b^2*c^2)*x^2 + 2*(B*b^4 - 2*A*b^3*c)*x - 6*((B*
b*c^3 - 2*A*c^4)*x^4 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^3 + (B*b^3*c - 2*A*b^2*c^2)*x^2)*log(c*x + b) + 6*((B*b*c^3
- 2*A*c^4)*x^4 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^3 + (B*b^3*c - 2*A*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (104) = 208\).

Time = 0.35 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.01 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=\frac {- A b^{3} + x^{3} \cdot \left (12 A c^{3} - 6 B b c^{2}\right ) + x^{2} \cdot \left (18 A b c^{2} - 9 B b^{2} c\right ) + x \left (4 A b^{2} c - 2 B b^{3}\right )}{2 b^{6} x^{2} + 4 b^{5} c x^{3} + 2 b^{4} c^{2} x^{4}} - \frac {3 c \left (- 2 A c + B b\right ) \log {\left (x + \frac {- 6 A b c^{2} + 3 B b^{2} c - 3 b c \left (- 2 A c + B b\right )}{- 12 A c^{3} + 6 B b c^{2}} \right )}}{b^{5}} + \frac {3 c \left (- 2 A c + B b\right ) \log {\left (x + \frac {- 6 A b c^{2} + 3 B b^{2} c + 3 b c \left (- 2 A c + B b\right )}{- 12 A c^{3} + 6 B b c^{2}} \right )}}{b^{5}} \]

[In]

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

[Out]

(-A*b**3 + x**3*(12*A*c**3 - 6*B*b*c**2) + x**2*(18*A*b*c**2 - 9*B*b**2*c) + x*(4*A*b**2*c - 2*B*b**3))/(2*b**
6*x**2 + 4*b**5*c*x**3 + 2*b**4*c**2*x**4) - 3*c*(-2*A*c + B*b)*log(x + (-6*A*b*c**2 + 3*B*b**2*c - 3*b*c*(-2*
A*c + B*b))/(-12*A*c**3 + 6*B*b*c**2))/b**5 + 3*c*(-2*A*c + B*b)*log(x + (-6*A*b*c**2 + 3*B*b**2*c + 3*b*c*(-2
*A*c + B*b))/(-12*A*c**3 + 6*B*b*c**2))/b**5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=-\frac {A b^{3} + 6 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{3} + 9 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{2} + 2 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} x}{2 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} + \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (c x + b\right )}{b^{5}} - \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (x\right )}{b^{5}} \]

[In]

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

[Out]

-1/2*(A*b^3 + 6*(B*b*c^2 - 2*A*c^3)*x^3 + 9*(B*b^2*c - 2*A*b*c^2)*x^2 + 2*(B*b^3 - 2*A*b^2*c)*x)/(b^4*c^2*x^4
+ 2*b^5*c*x^3 + b^6*x^2) + 3*(B*b*c - 2*A*c^2)*log(c*x + b)/b^5 - 3*(B*b*c - 2*A*c^2)*log(x)/b^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=-\frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {3 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{5} c} - \frac {6 \, B b c^{2} x^{3} - 12 \, A c^{3} x^{3} + 9 \, B b^{2} c x^{2} - 18 \, A b c^{2} x^{2} + 2 \, B b^{3} x - 4 \, A b^{2} c x + A b^{3}}{2 \, {\left (c x^{2} + b x\right )}^{2} b^{4}} \]

[In]

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

[Out]

-3*(B*b*c - 2*A*c^2)*log(abs(x))/b^5 + 3*(B*b*c^2 - 2*A*c^3)*log(abs(c*x + b))/(b^5*c) - 1/2*(6*B*b*c^2*x^3 -
12*A*c^3*x^3 + 9*B*b^2*c*x^2 - 18*A*b*c^2*x^2 + 2*B*b^3*x - 4*A*b^2*c*x + A*b^3)/((c*x^2 + b*x)^2*b^4)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {x\,\left (2\,A\,c-B\,b\right )}{b^2}-\frac {A}{2\,b}+\frac {3\,c^2\,x^3\,\left (2\,A\,c-B\,b\right )}{b^4}+\frac {9\,c\,x^2\,\left (2\,A\,c-B\,b\right )}{2\,b^3}}{b^2\,x^2+2\,b\,c\,x^3+c^2\,x^4}-\frac {6\,c\,\mathrm {atanh}\left (\frac {3\,c\,\left (2\,A\,c-B\,b\right )\,\left (b+2\,c\,x\right )}{b\,\left (6\,A\,c^2-3\,B\,b\,c\right )}\right )\,\left (2\,A\,c-B\,b\right )}{b^5} \]

[In]

int((A + B*x)/(b*x + c*x^2)^3,x)

[Out]

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