Integrand size = 13, antiderivative size = 58 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {1}{2 a n \left (a+b x^n\right )^2}+\frac {1}{a^2 n \left (a+b x^n\right )}+\frac {\log (x)}{a^3}-\frac {\log \left (a+b x^n\right )}{a^3 n} \] Output:
1/2/a/n/(a+b*x^n)^2+1/a^2/n/(a+b*x^n)+ln(x)/a^3-ln(a+b*x^n)/a^3/n
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {\frac {a \left (3 a+2 b x^n\right )}{\left (a+b x^n\right )^2}+2 \log \left (x^n\right )-2 \log \left (a+b x^n\right )}{2 a^3 n} \] Input:
Integrate[1/(x*(a + b*x^n)^3),x]
Output:
((a*(3*a + 2*b*x^n))/(a + b*x^n)^2 + 2*Log[x^n] - 2*Log[a + b*x^n])/(2*a^3 *n)
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {798, 54, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x \left (a+b x^n\right )^3} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {x^{-n}}{\left (b x^n+a\right )^3}dx^n}{n}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {x^{-n}}{a^3}-\frac {b}{a^3 \left (b x^n+a\right )}-\frac {b}{a^2 \left (b x^n+a\right )^2}-\frac {b}{a \left (b x^n+a\right )^3}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\log \left (a+b x^n\right )}{a^3}+\frac {\log \left (x^n\right )}{a^3}+\frac {1}{a^2 \left (a+b x^n\right )}+\frac {1}{2 a \left (a+b x^n\right )^2}}{n}\) |
Input:
Int[1/(x*(a + b*x^n)^3),x]
Output:
(1/(2*a*(a + b*x^n)^2) + 1/(a^2*(a + b*x^n)) + Log[x^n]/a^3 - Log[a + b*x^ n]/a^3)/n
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.55 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\ln \left (x \right )}{a^{3}}+\frac {2 b \,x^{n}+3 a}{2 a^{2} n \left (a +b \,x^{n}\right )^{2}}-\frac {\ln \left (x^{n}+\frac {a}{b}\right )}{a^{3} n}\) | \(53\) |
derivativedivides | \(\frac {-\frac {\ln \left (a +b \,x^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \,x^{n}\right )}+\frac {1}{2 a \left (a +b \,x^{n}\right )^{2}}+\frac {\ln \left (x^{n}\right )}{a^{3}}}{n}\) | \(54\) |
default | \(\frac {-\frac {\ln \left (a +b \,x^{n}\right )}{a^{3}}+\frac {1}{a^{2} \left (a +b \,x^{n}\right )}+\frac {1}{2 a \left (a +b \,x^{n}\right )^{2}}+\frac {\ln \left (x^{n}\right )}{a^{3}}}{n}\) | \(54\) |
norman | \(\frac {\frac {\ln \left (x \right )}{a}+\frac {2 b \ln \left (x \right ) {\mathrm e}^{n \ln \left (x \right )}}{a^{2}}-\frac {2 b \,{\mathrm e}^{n \ln \left (x \right )}}{a^{2} n}+\frac {b^{2} \ln \left (x \right ) {\mathrm e}^{2 n \ln \left (x \right )}}{a^{3}}-\frac {3 b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2 a^{3} n}}{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}}-\frac {\ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{a^{3} n}\) | \(100\) |
parallelrisch | \(\frac {2 x^{2 n} \ln \left (x \right ) b^{2} n +4 x^{n} \ln \left (x \right ) a b n -2 \ln \left (a +b \,x^{n}\right ) x^{2 n} b^{2}+2 a^{2} \ln \left (x \right ) n -4 \ln \left (a +b \,x^{n}\right ) x^{n} a b -3 b^{2} x^{2 n}-2 \ln \left (a +b \,x^{n}\right ) a^{2}-4 a b \,x^{n}}{2 a^{3} n \left (a +b \,x^{n}\right )^{2}}\) | \(113\) |
Input:
int(1/x/(a+b*x^n)^3,x,method=_RETURNVERBOSE)
Output:
ln(x)/a^3+1/2*(2*b*x^n+3*a)/a^2/n/(a+b*x^n)^2-1/a^3/n*ln(x^n+a/b)
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.83 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {2 \, b^{2} n x^{2 \, n} \log \left (x\right ) + 2 \, a^{2} n \log \left (x\right ) + 3 \, a^{2} + 2 \, {\left (2 \, a b n \log \left (x\right ) + a b\right )} x^{n} - 2 \, {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )} \log \left (b x^{n} + a\right )}{2 \, {\left (a^{3} b^{2} n x^{2 \, n} + 2 \, a^{4} b n x^{n} + a^{5} n\right )}} \] Input:
integrate(1/x/(a+b*x^n)^3,x, algorithm="fricas")
Output:
1/2*(2*b^2*n*x^(2*n)*log(x) + 2*a^2*n*log(x) + 3*a^2 + 2*(2*a*b*n*log(x) + a*b)*x^n - 2*(b^2*x^(2*n) + 2*a*b*x^n + a^2)*log(b*x^n + a))/(a^3*b^2*n*x ^(2*n) + 2*a^4*b*n*x^n + a^5*n)
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (48) = 96\).
Time = 1.44 (sec) , antiderivative size = 415, normalized size of antiderivative = 7.16 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\begin {cases} \tilde {\infty } \log {\left (x \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\\frac {\log {\left (x \right )}}{a^{3}} & \text {for}\: b = 0 \\- \frac {x^{- 3 n}}{3 b^{3} n} & \text {for}\: a = 0 \\\tilde {\infty } \log {\left (x \right )} & \text {for}\: b = - a x^{- n} \\\frac {\log {\left (x \right )}}{\left (a + b\right )^{3}} & \text {for}\: n = 0 \\\frac {2 a^{2} n \log {\left (x \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} - \frac {2 a^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} + \frac {3 a^{2}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} + \frac {4 a b n x^{n} \log {\left (x \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} - \frac {4 a b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} + \frac {2 a b x^{n}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} + \frac {2 b^{2} n x^{2 n} \log {\left (x \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} - \frac {2 b^{2} x^{2 n} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{5} n + 4 a^{4} b n x^{n} + 2 a^{3} b^{2} n x^{2 n}} & \text {otherwise} \end {cases} \] Input:
integrate(1/x/(a+b*x**n)**3,x)
Output:
Piecewise((zoo*log(x), Eq(a, 0) & Eq(b, 0) & Eq(n, 0)), (log(x)/a**3, Eq(b , 0)), (-1/(3*b**3*n*x**(3*n)), Eq(a, 0)), (zoo*log(x), Eq(b, -a/x**n)), ( log(x)/(a + b)**3, Eq(n, 0)), (2*a**2*n*log(x)/(2*a**5*n + 4*a**4*b*n*x**n + 2*a**3*b**2*n*x**(2*n)) - 2*a**2*log(a/b + x**n)/(2*a**5*n + 4*a**4*b*n *x**n + 2*a**3*b**2*n*x**(2*n)) + 3*a**2/(2*a**5*n + 4*a**4*b*n*x**n + 2*a **3*b**2*n*x**(2*n)) + 4*a*b*n*x**n*log(x)/(2*a**5*n + 4*a**4*b*n*x**n + 2 *a**3*b**2*n*x**(2*n)) - 4*a*b*x**n*log(a/b + x**n)/(2*a**5*n + 4*a**4*b*n *x**n + 2*a**3*b**2*n*x**(2*n)) + 2*a*b*x**n/(2*a**5*n + 4*a**4*b*n*x**n + 2*a**3*b**2*n*x**(2*n)) + 2*b**2*n*x**(2*n)*log(x)/(2*a**5*n + 4*a**4*b*n *x**n + 2*a**3*b**2*n*x**(2*n)) - 2*b**2*x**(2*n)*log(a/b + x**n)/(2*a**5* n + 4*a**4*b*n*x**n + 2*a**3*b**2*n*x**(2*n)), True))
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {2 \, b x^{n} + 3 \, a}{2 \, {\left (a^{2} b^{2} n x^{2 \, n} + 2 \, a^{3} b n x^{n} + a^{4} n\right )}} - \frac {\log \left (b x^{n} + a\right )}{a^{3} n} + \frac {\log \left (x^{n}\right )}{a^{3} n} \] Input:
integrate(1/x/(a+b*x^n)^3,x, algorithm="maxima")
Output:
1/2*(2*b*x^n + 3*a)/(a^2*b^2*n*x^(2*n) + 2*a^3*b*n*x^n + a^4*n) - log(b*x^ n + a)/(a^3*n) + log(x^n)/(a^3*n)
\[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\int { \frac {1}{{\left (b x^{n} + a\right )}^{3} x} \,d x } \] Input:
integrate(1/x/(a+b*x^n)^3,x, algorithm="giac")
Output:
integrate(1/((b*x^n + a)^3*x), x)
Time = 0.38 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {\ln \left (x\right )}{a^3}+\frac {1}{a^2\,n\,\left (a+b\,x^n\right )}-\frac {\ln \left (a+b\,x^n\right )}{a^3\,n}+\frac {1}{2\,a\,n\,\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )} \] Input:
int(1/(x*(a + b*x^n)^3),x)
Output:
log(x)/a^3 + 1/(a^2*n*(a + b*x^n)) - log(a + b*x^n)/(a^3*n) + 1/(2*a*n*(a^ 2 + b^2*x^(2*n) + 2*a*b*x^n))
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.12 \[ \int \frac {1}{x \left (a+b x^n\right )^3} \, dx=\frac {-2 x^{2 n} \mathrm {log}\left (x^{n} b +a \right ) b^{2}+2 x^{2 n} \mathrm {log}\left (x \right ) b^{2} n -x^{2 n} b^{2}-4 x^{n} \mathrm {log}\left (x^{n} b +a \right ) a b +4 x^{n} \mathrm {log}\left (x \right ) a b n -2 \,\mathrm {log}\left (x^{n} b +a \right ) a^{2}+2 \,\mathrm {log}\left (x \right ) a^{2} n +2 a^{2}}{2 a^{3} n \left (x^{2 n} b^{2}+2 x^{n} a b +a^{2}\right )} \] Input:
int(1/x/(a+b*x^n)^3,x)
Output:
( - 2*x**(2*n)*log(x**n*b + a)*b**2 + 2*x**(2*n)*log(x)*b**2*n - x**(2*n)* b**2 - 4*x**n*log(x**n*b + a)*a*b + 4*x**n*log(x)*a*b*n - 2*log(x**n*b + a )*a**2 + 2*log(x)*a**2*n + 2*a**2)/(2*a**3*n*(x**(2*n)*b**2 + 2*x**n*a*b + a**2))