Integrand size = 20, antiderivative size = 326 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=-\frac {b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a e^2 (1+n)} \]
-1/3*b^(2/3)*(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^(4/3)/(b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*(-1)^(1/3)*b^(2/3 )*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],(-1)^(2/3)*b^(1/3)*(f*x+e)/((-1)^ (2/3)*b^(1/3)*e-a^(1/3)*f))/a^(4/3)/((-1)^(2/3)*b^(1/3)*e-a^(1/3)*f)/(1+n) +1/3*(-1)^(2/3)*b^(2/3)*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],(-1)^(1/3)* b^(1/3)*(f*x+e)/((-1)^(1/3)*b^(1/3)*e+a^(1/3)*f))/a^(4/3)/((-1)^(1/3)*b^(1 /3)*e+a^(1/3)*f)/(1+n)+f*(f*x+e)^(1+n)*hypergeom([2, 1+n],[2+n],1+f*x/e)/a /e^2/(1+n)
Time = 0.34 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {\sqrt [3]{-1} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {(-1)^{2/3} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}+\frac {3 \sqrt [3]{a} f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{e^2}\right )}{3 a^{4/3} (1+n)} \]
((e + f*x)^(1 + n)*(-((b^(2/3)*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)) + ((-1)^(1/ 3)*b^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e + f*x ))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f) + ((-1)^(2/3)*b^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^(1 /3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f) + (3*a^(1/3)*f*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e ])/e^2))/(3*a^(4/3)*(1 + n))
Time = 0.77 (sec) , antiderivative size = 326, 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.
\(\displaystyle \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {(e+f x)^n}{a x^2}-\frac {b x (e+f x)^n}{a \left (a+b x^3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a e^2 (n+1)}\) |
-1/3*(b^(2/3)*(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)])/(a^(4/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(1/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e + f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)] )/(3*a^(4/3)*((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*b^( 2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^(1 /3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/(3*a^(4/3)*((-1)^(1/3) *b^(1/3)*e + a^(1/3)*f)*(1 + n)) + (f*(e + f*x)^(1 + n)*Hypergeometric2F1[ 2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))
3.2.93.3.1 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]
\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (b \,x^{3}+a \right )}d x\]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (b\,x^3+a\right )} \,d x \]