Integrand size = 20, antiderivative size = 300 \[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\frac {\sqrt [3]{b} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{b} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 a \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{b} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 a \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}-\frac {(e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{a e (1+n)} \]
1/3*b^(1/3)*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3 )*e-a^(1/3)*f))/a/(b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*b^(1/3)*(f*x+e)^(1+n)*hy pergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e+(-1)^(1/3)*a^(1/3)*f))/a /(b^(1/3)*e+(-1)^(1/3)*a^(1/3)*f)/(1+n)+1/3*b^(1/3)*(f*x+e)^(1+n)*hypergeo m([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e-(-1)^(2/3)*a^(1/3)*f))/a/(b^(1 /3)*e-(-1)^(2/3)*a^(1/3)*f)/(1+n)-(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],1 +f*x/e)/a/e/(1+n)
Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.81 \[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\frac {(e+f x)^{1+n} \left (\frac {\sqrt [3]{b} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {\sqrt [3]{b} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}+\frac {\sqrt [3]{b} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}-\frac {3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{e}\right )}{3 a (1+n)} \]
((e + f*x)^(1 + n)*((b^(1/3)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*( e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(b^(1/3)*e - a^(1/3)*f) + (b^(1/3)*Hyp ergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e + (-1)^(1/3 )*a^(1/3)*f)])/(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f) + (b^(1/3)*Hypergeometri c2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)* f)])/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f) - (3*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e])/e))/(3*a*(1 + n))
Time = 0.77 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {7276, 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.
\(\displaystyle \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {(e+f x)^n}{a x}-\frac {b x^2 (e+f x)^n}{a \left (a+b x^3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [3]{b} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {\sqrt [3]{b} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {\sqrt [3]{b} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}-\frac {(e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {f x}{e}+1\right )}{a e (n+1)}\) |
(b^(1/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(3*a*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + (b^(1/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)])/(3*a*(b^(1/3)*e + (-1)^(1/3)* a^(1/3)*f)*(1 + n)) + (b^(1/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)])/(3*a*(b ^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)*(1 + n)) - ((e + f*x)^(1 + n)*Hypergeomet ric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e*(1 + n))
3.2.92.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
\[\int \frac {\left (f x +e \right )^{n}}{x \left (b \,x^{3}+a \right )}d x\]
\[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x} \,d x } \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\int \frac {\left (e + f x\right )^{n}}{x \left (a + b x^{3}\right )}\, dx \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x} \,d x } \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x \left (a+b x^3\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x\,\left (b\,x^3+a\right )} \,d x \]