Integrand size = 20, antiderivative size = 140 \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {c \operatorname {Hypergeometric2F1}\left (1,-\frac {2}{n},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) x^2}+\frac {c \operatorname {Hypergeometric2F1}\left (1,-\frac {2}{n},-\frac {2-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) x^2} \] Output:
c*hypergeom([1, -2/n],[-(2-n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a *c-b*(-4*a*c+b^2)^(1/2))/x^2+c*hypergeom([1, -2/n],[-(2-n)/n],-2*c*x^n/(b+ (-4*a*c+b^2)^(1/2)))/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/x^2
Time = 0.76 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.84 \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {2^{\frac {2+n}{n}} c \left (\frac {\left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{2/n} \operatorname {Hypergeometric2F1}\left (\frac {2+n}{n},\frac {2+n}{n},2+\frac {2}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{-b+\sqrt {b^2-4 a c}-2 c x^n}+\frac {x^{-n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {2+n}{n}} \operatorname {Hypergeometric2F1}\left (\frac {2+n}{n},\frac {2+n}{n},2+\frac {2}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{c}\right )}{\sqrt {b^2-4 a c} (2+n) x^2} \] Input:
Integrate[1/(x^3*(a + b*x^n + c*x^(2*n))),x]
Output:
(2^((2 + n)/n)*c*((((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(2/n)*Hyper geometric2F1[(2 + n)/n, (2 + n)/n, 2 + 2/n, (b - Sqrt[b^2 - 4*a*c])/(b - S qrt[b^2 - 4*a*c] + 2*c*x^n)])/(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^n) + (((c*x^ n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((2 + n)/n)*Hypergeometric2F1[(2 + n )/n, (2 + n)/n, 2 + 2/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(c*x^n)))/(Sqrt[b^2 - 4*a*c]*(2 + n)*x^2)
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1719, 888}
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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx\) |
\(\Big \downarrow \) 1719 |
\(\displaystyle \frac {2 c \int \frac {1}{x^3 \left (2 c x^n+b-\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {1}{x^3 \left (2 c x^n+b+\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {c \operatorname {Hypergeometric2F1}\left (1,-\frac {2}{n},-\frac {2-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{x^2 \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {c \operatorname {Hypergeometric2F1}\left (1,-\frac {2}{n},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{x^2 \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}\) |
Input:
Int[1/(x^3*(a + b*x^n + c*x^(2*n))),x]
Output:
-((c*Hypergeometric2F1[1, -2/n, -((2 - n)/n), (-2*c*x^n)/(b - Sqrt[b^2 - 4 *a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*x^2)) + (c*Hypergeomet ric2F1[1, -2/n, -((2 - n)/n), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b ^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*x^2)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symb ol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int[(d*x)^m/(b - q + 2 *c*x^n), x], x] - Simp[2*(c/q) Int[(d*x)^m/(b + q + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
\[\int \frac {1}{x^{3} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]
Input:
int(1/x^3/(a+b*x^n+c*x^(2*n)),x)
Output:
int(1/x^3/(a+b*x^n+c*x^(2*n)),x)
\[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
Output:
integral(1/(c*x^3*x^(2*n) + b*x^3*x^n + a*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**3/(a+b*x**n+c*x**(2*n)),x)
Output:
Timed out
\[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")
Output:
integrate(1/((c*x^(2*n) + b*x^n + a)*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")
Output:
integrate(1/((c*x^(2*n) + b*x^n + a)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {1}{x^3\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \] Input:
int(1/(x^3*(a + b*x^n + c*x^(2*n))),x)
Output:
int(1/(x^3*(a + b*x^n + c*x^(2*n))), x)
\[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {1}{x^{2 n} c \,x^{3}+x^{n} b \,x^{3}+a \,x^{3}}d x \] Input:
int(1/x^3/(a+b*x^n+c*x^(2*n)),x)
Output:
int(1/(x**(2*n)*c*x**3 + x**n*b*x**3 + a*x**3),x)