Integrand size = 20, antiderivative size = 332 \[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=-\frac {e (e+f x)^{1+n}}{b f^2 (1+n)}+\frac {(e+f x)^{2+n}}{b f^2 (2+n)}-\frac {a^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{-1} a^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {(-1)^{2/3} a^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)} \]
-e*(f*x+e)^(1+n)/b/f^2/(1+n)+(f*x+e)^(2+n)/b/f^2/(2+n)-1/3*a^(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))/b^( 4/3)/(b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*(-1)^(1/3)*a^(2/3)*(f*x+e)^(1+n)*hype rgeom([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))/b^(4/3)/((-1)^(2/3)*b^(1/3)*e-a^(1/3)*f)/(1+n)+1/3*(-1)^(2/3)*a^(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))/b^(4/3)/((-1)^(1/3)*b^(1/3)*e+a^(1/3)*f)/(1+ n)
Time = 0.60 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {3 \sqrt [3]{b} e}{f^2 (1+n)}+\frac {3 \sqrt [3]{b} (e+f x)}{f^2 (2+n)}-\frac {a^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{-1} a^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {(-1)^{2/3} a^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{\left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}\right )}{3 b^{4/3}} \]
((e + f*x)^(1 + n)*((-3*b^(1/3)*e)/(f^2*(1 + n)) + (3*b^(1/3)*(e + f*x))/( f^2*(2 + n)) - (a^(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 + n)) + ((-1)^( 1/3)*a^(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 + n)) + ((-1)^(2/3)*a^(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)*(1 + n))))/(3*b^(4/3))
Time = 0.91 (sec) , antiderivative size = 332, 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.
\(\displaystyle \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {a x (e+f x)^n}{b \left (a+b x^3\right )}-\frac {e (e+f x)^n}{b f}+\frac {(e+f x)^{n+1}}{b f}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {\sqrt [3]{-1} a^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {(-1)^{2/3} a^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 b^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}-\frac {e (e+f x)^{n+1}}{b f^2 (n+1)}+\frac {(e+f x)^{n+2}}{b f^2 (n+2)}\) |
-((e*(e + f*x)^(1 + n))/(b*f^2*(1 + n))) + (e + f*x)^(2 + n)/(b*f^2*(2 + n )) - (a^(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)])/(3*b^(4/3)*(b^(1/3)*e - a^(1/3)*f)* (1 + n)) + ((-1)^(1/3)*a^(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*b^(4/3)*((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*a ^(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*b^(4/3)*((-1)^(1/ 3)*b^(1/3)*e + a^(1/3)*f)*(1 + n))
3.2.87.3.1 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]
\[\int \frac {x^{4} \left (f x +e \right )^{n}}{b \,x^{3}+a}d x\]
\[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a} \,d x } \]
Timed out. \[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a} \,d x } \]
\[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{b x^{3} + a} \,d x } \]
Timed out. \[ \int \frac {x^4 (e+f x)^n}{a+b x^3} \, dx=\int \frac {x^4\,{\left (e+f\,x\right )}^n}{b\,x^3+a} \,d x \]