Integrand size = 20, antiderivative size = 324 \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)} \]
e^2*(f*x+e)^(1+n)/b/f^3/(1+n)-2*e*(f*x+e)^(2+n)/b/f^3/(2+n)+(f*x+e)^(3+n)/ b/f^3/(3+n)+1/3*a*(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))/b^(5/3)/(b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*a*(f*x+e)^(1 +n)*hypergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e+(-1)^(1/3)*a^(1/3) *f))/b^(5/3)/(b^(1/3)*e+(-1)^(1/3)*a^(1/3)*f)/(1+n)+1/3*a*(f*x+e)^(1+n)*hy pergeom([1, 1+n],[2+n],b^(1/3)*(f*x+e)/(b^(1/3)*e-(-1)^(2/3)*a^(1/3)*f))/b ^(5/3)/(b^(1/3)*e-(-1)^(2/3)*a^(1/3)*f)/(1+n)
Time = 0.56 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.88 \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\frac {(e+f x)^{1+n} \left (\frac {3 b^{2/3} e^2}{f^3 (1+n)}-\frac {6 b^{2/3} e (e+f x)}{f^3 (2+n)}+\frac {3 b^{2/3} (e+f x)^2}{f^3 (3+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}\right )}{3 b^{5/3}} \]
((e + f*x)^(1 + n)*((3*b^(2/3)*e^2)/(f^3*(1 + n)) - (6*b^(2/3)*e*(e + f*x) )/(f^3*(2 + n)) + (3*b^(2/3)*(e + f*x)^2)/(f^3*(3 + n)) + (a*Hypergeometri c2F1[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 + n)) + (a*Hypergeometric2F1[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)*(1 + n)) + (a*Hypergeometric2F1[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)*(1 + n))))/(3*b^(5/3))
Time = 0.97 (sec) , antiderivative size = 324, 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.
\(\displaystyle \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {a x^2 (e+f x)^n}{b \left (a+b x^3\right )}+\frac {e^2 (e+f x)^n}{b f^2}-\frac {2 e (e+f x)^{n+1}}{b f^2}+\frac {(e+f x)^{n+2}}{b f^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}+\frac {e^2 (e+f x)^{n+1}}{b f^3 (n+1)}-\frac {2 e (e+f x)^{n+2}}{b f^3 (n+2)}+\frac {(e+f x)^{n+3}}{b f^3 (n+3)}\) |
(e^2*(e + f*x)^(1 + n))/(b*f^3*(1 + n)) - (2*e*(e + f*x)^(2 + n))/(b*f^3*( 2 + n)) + (e + f*x)^(3 + n)/(b*f^3*(3 + n)) + (a*(e + f*x)^(1 + n)*Hyperge ometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/ (3*b^(5/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + (a*(e + f*x)^(1 + n)*Hyperge ometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e + (-1)^(1/3)*a^ (1/3)*f)])/(3*b^(5/3)*(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)*(1 + n)) + (a*(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*b^(5/3)*(b^(1/3)*e - (-1)^(2/3)*a^(1/3 )*f)*(1 + n))
3.2.86.3.1 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]
\[\int \frac {x^{5} \left (f x +e \right )^{n}}{b \,x^{3}+a}d x\]
\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \]
Timed out. \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \]
\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \]
Timed out. \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int \frac {x^5\,{\left (e+f\,x\right )}^n}{b\,x^3+a} \,d x \]