Integrand size = 26, antiderivative size = 95 \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=-\frac {a^2}{b^2 (b c-a d) n \left (a+b x^n\right )}-\frac {a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2 n}+\frac {c^2 \log \left (c+d x^n\right )}{d (b c-a d)^2 n} \] Output:
-a^2/b^2/(-a*d+b*c)/n/(a+b*x^n)-a*(-a*d+2*b*c)*ln(a+b*x^n)/b^2/(-a*d+b*c)^ 2/n+c^2*ln(c+d*x^n)/d/(-a*d+b*c)^2/n
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=-\frac {a^2}{b^2 (b c-a d) n \left (a+b x^n\right )}+\frac {a (-2 b c+a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2 n}+\frac {c^2 \log \left (c+d x^n\right )}{d (-b c+a d)^2 n} \] Input:
Integrate[x^(-1 + 3*n)/((a + b*x^n)^2*(c + d*x^n)),x]
Output:
-(a^2/(b^2*(b*c - a*d)*n*(a + b*x^n))) + (a*(-2*b*c + a*d)*Log[a + b*x^n]) /(b^2*(b*c - a*d)^2*n) + (c^2*Log[c + d*x^n])/(d*(-(b*c) + a*d)^2*n)
Time = 0.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {948, 99, 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.
\(\displaystyle \int \frac {x^{3 n-1}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\int \frac {x^{2 n}}{\left (b x^n+a\right )^2 \left (d x^n+c\right )}dx^n}{n}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {a^2}{b (b c-a d) \left (b x^n+a\right )^2}+\frac {(a d-2 b c) a}{b (b c-a d)^2 \left (b x^n+a\right )}+\frac {c^2}{(b c-a d)^2 \left (d x^n+c\right )}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2}{b^2 (b c-a d) \left (a+b x^n\right )}-\frac {a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2}+\frac {c^2 \log \left (c+d x^n\right )}{d (b c-a d)^2}}{n}\) |
Input:
Int[x^(-1 + 3*n)/((a + b*x^n)^2*(c + d*x^n)),x]
Output:
(-(a^2/(b^2*(b*c - a*d)*(a + b*x^n))) - (a*(2*b*c - a*d)*Log[a + b*x^n])/( b^2*(b*c - a*d)^2) + (c^2*Log[c + d*x^n])/(d*(b*c - a*d)^2))/n
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.84 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32
method | result | size |
norman | \(\frac {a^{2}}{\left (a d -c b \right ) b^{2} n \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}+\frac {c^{2} \ln \left (c +d \,{\mathrm e}^{n \ln \left (x \right )}\right )}{d n \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a \left (a d -2 c b \right ) \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} n}\) | \(125\) |
risch | \(\frac {\ln \left (x \right )}{b^{2} d}-\frac {\ln \left (x \right ) c^{2}}{d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\ln \left (x \right ) a^{2} d}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2}}+\frac {2 \ln \left (x \right ) a c}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b}+\frac {a^{2}}{\left (a d -c b \right ) b^{2} n \left (a +b \,x^{n}\right )}+\frac {c^{2} \ln \left (x^{n}+\frac {c}{d}\right )}{d n \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {a^{2} \ln \left (x^{n}+\frac {a}{b}\right ) d}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} n}-\frac {2 a \ln \left (x^{n}+\frac {a}{b}\right ) c}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b n}\) | \(269\) |
Input:
int(x^(-1+3*n)/(a+b*x^n)^2/(c+d*x^n),x,method=_RETURNVERBOSE)
Output:
a^2/(a*d-b*c)/b^2/n/(a+b*exp(n*ln(x)))+c^2/d/n/(a^2*d^2-2*a*b*c*d+b^2*c^2) *ln(c+d*exp(n*ln(x)))+a*(a*d-2*b*c)/(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^2/n*ln(a +b*exp(n*ln(x)))
Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.75 \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=-\frac {a^{2} b c d - a^{3} d^{2} + {\left (2 \, a^{2} b c d - a^{3} d^{2} + {\left (2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{n}\right )} \log \left (b x^{n} + a\right ) - {\left (b^{3} c^{2} x^{n} + a b^{2} c^{2}\right )} \log \left (d x^{n} + c\right )}{{\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} n x^{n} + {\left (a b^{4} c^{2} d - 2 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3}\right )} n} \] Input:
integrate(x^(-1+3*n)/(a+b*x^n)^2/(c+d*x^n),x, algorithm="fricas")
Output:
-(a^2*b*c*d - a^3*d^2 + (2*a^2*b*c*d - a^3*d^2 + (2*a*b^2*c*d - a^2*b*d^2) *x^n)*log(b*x^n + a) - (b^3*c^2*x^n + a*b^2*c^2)*log(d*x^n + c))/((b^5*c^2 *d - 2*a*b^4*c*d^2 + a^2*b^3*d^3)*n*x^n + (a*b^4*c^2*d - 2*a^2*b^3*c*d^2 + a^3*b^2*d^3)*n)
Exception generated. \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate(x**(-1+3*n)/(a+b*x**n)**2/(c+d*x**n),x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\frac {c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{b^{2} c^{2} d n - 2 \, a b c d^{2} n + a^{2} d^{3} n} - \frac {a^{2}}{a b^{3} c n - a^{2} b^{2} d n + {\left (b^{4} c n - a b^{3} d n\right )} x^{n}} - \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{4} c^{2} n - 2 \, a b^{3} c d n + a^{2} b^{2} d^{2} n} \] Input:
integrate(x^(-1+3*n)/(a+b*x^n)^2/(c+d*x^n),x, algorithm="maxima")
Output:
c^2*log((d*x^n + c)/d)/(b^2*c^2*d*n - 2*a*b*c*d^2*n + a^2*d^3*n) - a^2/(a* b^3*c*n - a^2*b^2*d*n + (b^4*c*n - a*b^3*d*n)*x^n) - (2*a*b*c - a^2*d)*log ((b*x^n + a)/b)/(b^4*c^2*n - 2*a*b^3*c*d*n + a^2*b^2*d^2*n)
\[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\int { \frac {x^{3 \, n - 1}}{{\left (b x^{n} + a\right )}^{2} {\left (d x^{n} + c\right )}} \,d x } \] Input:
integrate(x^(-1+3*n)/(a+b*x^n)^2/(c+d*x^n),x, algorithm="giac")
Output:
integrate(x^(3*n - 1)/((b*x^n + a)^2*(d*x^n + c)), x)
Timed out. \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\int \frac {x^{3\,n-1}}{{\left (a+b\,x^n\right )}^2\,\left (c+d\,x^n\right )} \,d x \] Input:
int(x^(3*n - 1)/((a + b*x^n)^2*(c + d*x^n)),x)
Output:
int(x^(3*n - 1)/((a + b*x^n)^2*(c + d*x^n)), x)
Time = 0.20 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.04 \[ \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\frac {x^{n} \mathrm {log}\left (x^{n} b +a \right ) a^{2} b \,d^{2}-2 x^{n} \mathrm {log}\left (x^{n} b +a \right ) a \,b^{2} c d +x^{n} \mathrm {log}\left (x^{n} d +c \right ) b^{3} c^{2}-x^{n} a^{2} b \,d^{2}+x^{n} a \,b^{2} c d +\mathrm {log}\left (x^{n} b +a \right ) a^{3} d^{2}-2 \,\mathrm {log}\left (x^{n} b +a \right ) a^{2} b c d +\mathrm {log}\left (x^{n} d +c \right ) a \,b^{2} c^{2}}{b^{2} d n \left (x^{n} a^{2} b \,d^{2}-2 x^{n} a \,b^{2} c d +x^{n} b^{3} c^{2}+a^{3} d^{2}-2 a^{2} b c d +a \,b^{2} c^{2}\right )} \] Input:
int(x^(-1+3*n)/(a+b*x^n)^2/(c+d*x^n),x)
Output:
(x**n*log(x**n*b + a)*a**2*b*d**2 - 2*x**n*log(x**n*b + a)*a*b**2*c*d + x* *n*log(x**n*d + c)*b**3*c**2 - x**n*a**2*b*d**2 + x**n*a*b**2*c*d + log(x* *n*b + a)*a**3*d**2 - 2*log(x**n*b + a)*a**2*b*c*d + log(x**n*d + c)*a*b** 2*c**2)/(b**2*d*n*(x**n*a**2*b*d**2 - 2*x**n*a*b**2*c*d + x**n*b**3*c**2 + a**3*d**2 - 2*a**2*b*c*d + a*b**2*c**2))