\(\int \frac {1}{x (a+b x)^5} \, dx\) [57]

Optimal result

Integrand size = 11, antiderivative size = 64 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {\frac {25}{12 a}+\frac {13 b x}{3 a^2}+\frac {7 b^2 x^2}{2 a^3}+\frac {b^3 x^3}{a^4}}{(a+b x)^4}-\frac {\log \left (\frac {a+b x}{x}\right )}{a^5} \]

[Out]

(25/12/a+13/3*b*x/a^2+7/2*b^2*x^2/a^3+b^3*x^3/a^4)/(b*x+a)^4-1/a^5*ln((b*x+a)/x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x (a+b x)^5} \, dx=-\frac {\log (a+b x)}{a^5}+\frac {\log (x)}{a^5}+\frac {1}{a^4 (a+b x)}+\frac {1}{2 a^3 (a+b x)^2}+\frac {1}{3 a^2 (a+b x)^3}+\frac {1}{4 a (a+b x)^4} \]

[In]

Int[1/(x*(a + b*x)^5),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{a^5 x}-\frac {b}{a (a+b x)^5}-\frac {b}{a^2 (a+b x)^4}-\frac {b}{a^3 (a+b x)^3}-\frac {b}{a^4 (a+b x)^2}-\frac {b}{a^5 (a+b x)}\right ) \, dx \\ & = \frac {1}{4 a (a+b x)^4}+\frac {1}{3 a^2 (a+b x)^3}+\frac {1}{2 a^3 (a+b x)^2}+\frac {1}{a^4 (a+b x)}+\frac {\log (x)}{a^5}-\frac {\log (a+b x)}{a^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {\frac {a \left (25 a^3+52 a^2 b x+42 a b^2 x^2+12 b^3 x^3\right )}{(a+b x)^4}+12 \log (x)-12 \log (a+b x)}{12 a^5} \]

[In]

Integrate[1/(x*(a + b*x)^5),x]

[Out]

((a*(25*a^3 + 52*a^2*b*x + 42*a*b^2*x^2 + 12*b^3*x^3))/(a + b*x)^4 + 12*Log[x] - 12*Log[a + b*x])/(12*a^5)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {\frac {25}{12 a}+\frac {13 b x}{3 a^{2}}+\frac {7 b^{2} x^{2}}{2 a^{3}}+\frac {b^{3} x^{3}}{a^{4}}}{\left (b x +a \right )^{4}}+\frac {\ln \left (-x \right )}{a^{5}}-\frac {\ln \left (b x +a \right )}{a^{5}}\)	\(63\)
default	\(-\frac {\ln \left (b x +a \right )}{a^{5}}+\frac {1}{a^{4} \left (b x +a \right )}+\frac {1}{2 a^{3} \left (b x +a \right )^{2}}+\frac {1}{3 a^{2} \left (b x +a \right )^{3}}+\frac {1}{4 a \left (b x +a \right )^{4}}+\frac {\ln \left (x \right )}{a^{5}}\)	\(66\)
norman	\(\frac {-\frac {4 b x}{a^{2}}-\frac {9 b^{2} x^{2}}{a^{3}}-\frac {22 b^{3} x^{3}}{3 a^{4}}-\frac {25 b^{4} x^{4}}{12 a^{5}}}{\left (b x +a \right )^{4}}+\frac {\ln \left (x \right )}{a^{5}}-\frac {\ln \left (b x +a \right )}{a^{5}}\)	\(68\)
parallelrisch	\(\frac {12 \ln \left (x \right ) x^{4} b^{4}-12 \ln \left (b x +a \right ) x^{4} b^{4}+48 \ln \left (x \right ) x^{3} a \,b^{3}-48 \ln \left (b x +a \right ) x^{3} a \,b^{3}-25 b^{4} x^{4}+72 \ln \left (x \right ) x^{2} a^{2} b^{2}-72 \ln \left (b x +a \right ) x^{2} a^{2} b^{2}-88 a \,b^{3} x^{3}+48 \ln \left (x \right ) x \,a^{3} b -48 \ln \left (b x +a \right ) x \,a^{3} b -108 b^{2} a^{2} x^{2}+12 \ln \left (x \right ) a^{4}-12 \ln \left (b x +a \right ) a^{4}-48 b \,a^{3} x}{12 a^{5} \left (b x +a \right )^{4}}\)	\(169\)

[In]

int(1/x/(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

(25/12/a+13/3*b*x/a^2+7/2*b^2*x^2/a^3+b^3*x^3/a^4)/(b*x+a)^4+1/a^5*ln(-x)-1/a^5*ln(b*x+a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (60) = 120\).

Time = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.62 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {12 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} + 52 \, a^{3} b x + 25 \, a^{4} - 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{5} b^{4} x^{4} + 4 \, a^{6} b^{3} x^{3} + 6 \, a^{7} b^{2} x^{2} + 4 \, a^{8} b x + a^{9}\right )}} \]

[In]

integrate(1/x/(b*x+a)^5,x, algorithm="fricas")

[Out]

1/12*(12*a*b^3*x^3 + 42*a^2*b^2*x^2 + 52*a^3*b*x + 25*a^4 - 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*
b*x + a^4)*log(b*x + a) + 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*log(x))/(a^5*b^4*x^4 +
4*a^6*b^3*x^3 + 6*a^7*b^2*x^2 + 4*a^8*b*x + a^9)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {25 a^{3} + 52 a^{2} b x + 42 a b^{2} x^{2} + 12 b^{3} x^{3}}{12 a^{8} + 48 a^{7} b x + 72 a^{6} b^{2} x^{2} + 48 a^{5} b^{3} x^{3} + 12 a^{4} b^{4} x^{4}} + \frac {\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}}{a^{5}} \]

[In]

integrate(1/x/(b*x+a)**5,x)

[Out]

(25*a**3 + 52*a**2*b*x + 42*a*b**2*x**2 + 12*b**3*x**3)/(12*a**8 + 48*a**7*b*x + 72*a**6*b**2*x**2 + 48*a**5*b
**3*x**3 + 12*a**4*b**4*x**4) + (log(x) - log(a/b + x))/a**5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {12 \, b^{3} x^{3} + 42 \, a b^{2} x^{2} + 52 \, a^{2} b x + 25 \, a^{3}}{12 \, {\left (a^{4} b^{4} x^{4} + 4 \, a^{5} b^{3} x^{3} + 6 \, a^{6} b^{2} x^{2} + 4 \, a^{7} b x + a^{8}\right )}} - \frac {\log \left (b x + a\right )}{a^{5}} + \frac {\log \left (x\right )}{a^{5}} \]

[In]

integrate(1/x/(b*x+a)^5,x, algorithm="maxima")

[Out]

1/12*(12*b^3*x^3 + 42*a*b^2*x^2 + 52*a^2*b*x + 25*a^3)/(a^4*b^4*x^4 + 4*a^5*b^3*x^3 + 6*a^6*b^2*x^2 + 4*a^7*b*
x + a^8) - log(b*x + a)/a^5 + log(x)/a^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {1}{12} \, b {\left (\frac {12 \, \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{5} b} + \frac {\frac {12 \, b^{3}}{b x + a} + \frac {6 \, a b^{3}}{{\left (b x + a\right )}^{2}} + \frac {4 \, a^{2} b^{3}}{{\left (b x + a\right )}^{3}} + \frac {3 \, a^{3} b^{3}}{{\left (b x + a\right )}^{4}}}{a^{4} b^{4}}\right )} \]

[In]

integrate(1/x/(b*x+a)^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.87 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x (a+b x)^5} \, dx=\frac {\frac {\frac {\frac {1}{a^2+b\,x\,a}-\frac {\ln \left (\frac {a+b\,x}{x}\right )}{a^2}}{a}+\frac {1}{2\,a\,{\left (a+b\,x\right )}^2}}{a}+\frac {1}{3\,a\,{\left (a+b\,x\right )}^3}}{a}+\frac {1}{4\,a\,{\left (a+b\,x\right )}^4} \]

[In]

int(1/(x*(a + b*x)^5),x)

[Out]

(((1/(a^2 + a*b*x) - log((a + b*x)/x)/a^2)/a + 1/(2*a*(a + b*x)^2))/a + 1/(3*a*(a + b*x)^3))/a + 1/(4*a*(a + b
*x)^4)