\(\int \frac {A+B x}{x^4 (a+b x)^2} \, dx\) [192]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=-\frac {A}{3 a^2 x^3}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {b (3 A b-2 a B)}{a^4 x}-\frac {b^2 (A b-a B)}{a^4 (a+b x)}-\frac {b^2 (4 A b-3 a B) \log (x)}{a^5}+\frac {b^2 (4 A b-3 a B) \log (a+b x)}{a^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=-\frac {b^2 \log (x) (4 A b-3 a B)}{a^5}+\frac {b^2 (4 A b-3 a B) \log (a+b x)}{a^5}-\frac {b^2 (A b-a B)}{a^4 (a+b x)}-\frac {b (3 A b-2 a B)}{a^4 x}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {A}{3 a^2 x^3} \]

[In]

Int[(A + B*x)/(x^4*(a + b*x)^2),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=\frac {-\frac {2 a^3 A}{x^3}-\frac {3 a^2 (-2 A b+a B)}{x^2}+\frac {6 a b (-3 A b+2 a B)}{x}+\frac {6 a b^2 (-A b+a B)}{a+b x}+6 b^2 (-4 A b+3 a B) \log (x)+6 b^2 (4 A b-3 a B) \log (a+b x)}{6 a^5} \]

[In]

Integrate[(A + B*x)/(x^4*(a + b*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {A}{3 a^{2} x^{3}}-\frac {-2 A b +B a}{2 x^{2} a^{3}}-\frac {b \left (3 A b -2 B a \right )}{a^{4} x}-\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (x \right )}{a^{5}}+\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{5}}-\frac {b^{2} \left (A b -B a \right )}{a^{4} \left (b x +a \right )}\)	\(109\)
norman	\(\frac {\frac {b \left (4 b^{3} A -3 a \,b^{2} B \right ) x^{4}}{a^{5}}-\frac {A}{3 a}+\frac {\left (4 A b -3 B a \right ) x}{6 a^{2}}-\frac {b \left (4 A b -3 B a \right ) x^{2}}{2 a^{3}}}{x^{3} \left (b x +a \right )}+\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{5}}-\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (x \right )}{a^{5}}\)	\(116\)
risch	\(\frac {-\frac {b^{2} \left (4 A b -3 B a \right ) x^{3}}{a^{4}}-\frac {b \left (4 A b -3 B a \right ) x^{2}}{2 a^{3}}+\frac {\left (4 A b -3 B a \right ) x}{6 a^{2}}-\frac {A}{3 a}}{x^{3} \left (b x +a \right )}-\frac {4 b^{3} \ln \left (x \right ) A}{a^{5}}+\frac {3 b^{2} \ln \left (x \right ) B}{a^{4}}+\frac {4 b^{3} \ln \left (-b x -a \right ) A}{a^{5}}-\frac {3 b^{2} \ln \left (-b x -a \right ) B}{a^{4}}\)	\(131\)
parallelrisch	\(-\frac {24 A \ln \left (x \right ) x^{4} b^{4}-24 A \ln \left (b x +a \right ) x^{4} b^{4}-18 B \ln \left (x \right ) x^{4} a \,b^{3}+18 B \ln \left (b x +a \right ) x^{4} a \,b^{3}+24 A \ln \left (x \right ) x^{3} a \,b^{3}-24 A \ln \left (b x +a \right ) x^{3} a \,b^{3}-24 A \,b^{4} x^{4}-18 B \ln \left (x \right ) x^{3} a^{2} b^{2}+18 B \ln \left (b x +a \right ) x^{3} a^{2} b^{2}+18 B a \,b^{3} x^{4}+12 A \,a^{2} b^{2} x^{2}-9 B \,a^{3} b \,x^{2}-4 A \,a^{3} b x +3 B \,a^{4} x +2 A \,a^{4}}{6 a^{5} x^{3} \left (b x +a \right )}\)	\(193\)

[In]

int((B*x+A)/x^4/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=-\frac {2 \, A a^{4} - 6 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} + {\left (3 \, B a^{4} - 4 \, A a^{3} b\right )} x + 6 \, {\left ({\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} + {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3}\right )} \log \left (b x + a\right ) - 6 \, {\left ({\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} + {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}} \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

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

Time = 0.34 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.94 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=\frac {- 2 A a^{3} + x^{3} \left (- 24 A b^{3} + 18 B a b^{2}\right ) + x^{2} \left (- 12 A a b^{2} + 9 B a^{2} b\right ) + x \left (4 A a^{2} b - 3 B a^{3}\right )}{6 a^{5} x^{3} + 6 a^{4} b x^{4}} + \frac {b^{2} \left (- 4 A b + 3 B a\right ) \log {\left (x + \frac {- 4 A a b^{3} + 3 B a^{2} b^{2} - a b^{2} \left (- 4 A b + 3 B a\right )}{- 8 A b^{4} + 6 B a b^{3}} \right )}}{a^{5}} - \frac {b^{2} \left (- 4 A b + 3 B a\right ) \log {\left (x + \frac {- 4 A a b^{3} + 3 B a^{2} b^{2} + a b^{2} \left (- 4 A b + 3 B a\right )}{- 8 A b^{4} + 6 B a b^{3}} \right )}}{a^{5}} \]

[In]

integrate((B*x+A)/x**4/(b*x+a)**2,x)

[Out]

(-2*A*a**3 + x**3*(-24*A*b**3 + 18*B*a*b**2) + x**2*(-12*A*a*b**2 + 9*B*a**2*b) + x*(4*A*a**2*b - 3*B*a**3))/(
6*a**5*x**3 + 6*a**4*b*x**4) + b**2*(-4*A*b + 3*B*a)*log(x + (-4*A*a*b**3 + 3*B*a**2*b**2 - a*b**2*(-4*A*b + 3
*B*a))/(-8*A*b**4 + 6*B*a*b**3))/a**5 - b**2*(-4*A*b + 3*B*a)*log(x + (-4*A*a*b**3 + 3*B*a**2*b**2 + a*b**2*(-
4*A*b + 3*B*a))/(-8*A*b**4 + 6*B*a*b**3))/a**5

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=-\frac {2 \, A a^{3} - 6 \, {\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} x^{2} + {\left (3 \, B a^{3} - 4 \, A a^{2} b\right )} x}{6 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} - \frac {{\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} \log \left (b x + a\right )}{a^{5}} + \frac {{\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} \log \left (x\right )}{a^{5}} \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.42 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=\frac {{\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{5} b} + \frac {\frac {B a b^{6}}{b x + a} - \frac {A b^{7}}{b x + a}}{a^{4} b^{4}} - \frac {15 \, B a b^{2} - 26 \, A b^{3} - \frac {3 \, {\left (11 \, B a^{2} b^{3} - 20 \, A a b^{4}\right )}}{{\left (b x + a\right )} b} + \frac {18 \, {\left (B a^{3} b^{4} - 2 \, A a^{2} b^{5}\right )}}{{\left (b x + a\right )}^{2} b^{2}}}{6 \, a^{5} {\left (\frac {a}{b x + a} - 1\right )}^{3}} \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^2,x, algorithm="giac")

[Out]

(3*B*a*b^3 - 4*A*b^4)*log(abs(-a/(b*x + a) + 1))/(a^5*b) + (B*a*b^6/(b*x + a) - A*b^7/(b*x + a))/(a^4*b^4) - 1
/6*(15*B*a*b^2 - 26*A*b^3 - 3*(11*B*a^2*b^3 - 20*A*a*b^4)/((b*x + a)*b) + 18*(B*a^3*b^4 - 2*A*a^2*b^5)/((b*x +
a)^2*b^2))/(a^5*(a/(b*x + a) - 1)^3)

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{x^4 (a+b x)^2} \, dx=\frac {2\,b^2\,\mathrm {atanh}\left (\frac {b^2\,\left (4\,A\,b-3\,B\,a\right )\,\left (a+2\,b\,x\right )}{a\,\left (4\,A\,b^3-3\,B\,a\,b^2\right )}\right )\,\left (4\,A\,b-3\,B\,a\right )}{a^5}-\frac {\frac {A}{3\,a}-\frac {x\,\left (4\,A\,b-3\,B\,a\right )}{6\,a^2}+\frac {b^2\,x^3\,\left (4\,A\,b-3\,B\,a\right )}{a^4}+\frac {b\,x^2\,\left (4\,A\,b-3\,B\,a\right )}{2\,a^3}}{b\,x^4+a\,x^3} \]

[In]

int((A + B*x)/(x^4*(a + b*x)^2),x)

[Out]

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