\(\int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx\) [39]

Optimal result

Integrand size = 20, antiderivative size = 324 \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\frac {e^2 (e+f x)^{1+n}}{b f^3 (1+n)}-\frac {2 e (e+f x)^{2+n}}{b f^3 (2+n)}+\frac {(e+f x)^{3+n}}{b f^3 (3+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)} \] Output:

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)
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.88 \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\frac {(e+f x)^{1+n} \left (\frac {3 b^{2/3} e^2}{f^3 (1+n)}-\frac {6 b^{2/3} e (e+f x)}{f^3 (2+n)}+\frac {3 b^{2/3} (e+f x)^2}{f^3 (3+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{\left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}\right )}{3 b^{5/3}} \] Input:

Integrate[(x^5*(e + f*x)^n)/(a + b*x^3),x]
 

Output:

((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))
 

Rubi [A] (verified)

Time = 1.58 (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

\(\displaystyle \frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac {a (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 b^{5/3} (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}+\frac {e^2 (e+f x)^{n+1}}{b f^3 (n+1)}-\frac {2 e (e+f x)^{n+2}}{b f^3 (n+2)}+\frac {(e+f x)^{n+3}}{b f^3 (n+3)}\)

Input:

Int[(x^5*(e + f*x)^n)/(a + b*x^3),x]
 

Output:

(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))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7276

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]
 

Maple [F]

\[\int \frac {x^{5} \left (f x +e \right )^{n}}{b \,x^{3}+a}d x\]

Input:

int(x^5*(f*x+e)^n/(b*x^3+a),x)
 

Output:

int(x^5*(f*x+e)^n/(b*x^3+a),x)
 

Fricas [F]

\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \] Input:

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="fricas")
 

Output:

integral((f*x + e)^n*x^5/(b*x^3 + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\text {Timed out} \] Input:

integrate(x**5*(f*x+e)**n/(b*x**3+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \] Input:

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="maxima")
 

Output:

integrate((f*x + e)^n*x^5/(b*x^3 + a), x)
 

Giac [F]

\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{5}}{b x^{3} + a} \,d x } \] Input:

integrate(x^5*(f*x+e)^n/(b*x^3+a),x, algorithm="giac")
 

Output:

integrate((f*x + e)^n*x^5/(b*x^3 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\int \frac {x^5\,{\left (e+f\,x\right )}^n}{b\,x^3+a} \,d x \] Input:

int((x^5*(e + f*x)^n)/(a + b*x^3),x)
 

Output:

int((x^5*(e + f*x)^n)/(a + b*x^3), x)
 

Reduce [F]

\[ \int \frac {x^5 (e+f x)^n}{a+b x^3} \, dx=\frac {-\left (f x +e \right )^{n} a \,f^{3} n^{3}-6 \left (f x +e \right )^{n} a \,f^{3} n^{2}-11 \left (f x +e \right )^{n} a \,f^{3} n -6 \left (f x +e \right )^{n} a \,f^{3}+2 \left (f x +e \right )^{n} b \,e^{3} n -2 \left (f x +e \right )^{n} b \,e^{2} f \,n^{2} x +\left (f x +e \right )^{n} b e \,f^{2} n^{3} x^{2}+\left (f x +e \right )^{n} b e \,f^{2} n^{2} x^{2}+\left (f x +e \right )^{n} b \,f^{3} n^{3} x^{3}+3 \left (f x +e \right )^{n} b \,f^{3} n^{2} x^{3}+2 \left (f x +e \right )^{n} b \,f^{3} n \,x^{3}+\left (\int \frac {\left (f x +e \right )^{n}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a^{2} f^{4} n^{4}+6 \left (\int \frac {\left (f x +e \right )^{n}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a^{2} f^{4} n^{3}+11 \left (\int \frac {\left (f x +e \right )^{n}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a^{2} f^{4} n^{2}+6 \left (\int \frac {\left (f x +e \right )^{n}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a^{2} f^{4} n -\left (\int \frac {\left (f x +e \right )^{n} x^{2}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a b e \,f^{3} n^{4}-6 \left (\int \frac {\left (f x +e \right )^{n} x^{2}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a b e \,f^{3} n^{3}-11 \left (\int \frac {\left (f x +e \right )^{n} x^{2}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a b e \,f^{3} n^{2}-6 \left (\int \frac {\left (f x +e \right )^{n} x^{2}}{b f \,x^{4}+b e \,x^{3}+a f x +a e}d x \right ) a b e \,f^{3} n}{b^{2} f^{3} n \left (n^{3}+6 n^{2}+11 n +6\right )} \] Input:

int(x^5*(f*x+e)^n/(b*x^3+a),x)
 

Output:

( - (e + f*x)**n*a*f**3*n**3 - 6*(e + f*x)**n*a*f**3*n**2 - 11*(e + f*x)** 
n*a*f**3*n - 6*(e + f*x)**n*a*f**3 + 2*(e + f*x)**n*b*e**3*n - 2*(e + f*x) 
**n*b*e**2*f*n**2*x + (e + f*x)**n*b*e*f**2*n**3*x**2 + (e + f*x)**n*b*e*f 
**2*n**2*x**2 + (e + f*x)**n*b*f**3*n**3*x**3 + 3*(e + f*x)**n*b*f**3*n**2 
*x**3 + 2*(e + f*x)**n*b*f**3*n*x**3 + int((e + f*x)**n/(a*e + a*f*x + b*e 
*x**3 + b*f*x**4),x)*a**2*f**4*n**4 + 6*int((e + f*x)**n/(a*e + a*f*x + b* 
e*x**3 + b*f*x**4),x)*a**2*f**4*n**3 + 11*int((e + f*x)**n/(a*e + a*f*x + 
b*e*x**3 + b*f*x**4),x)*a**2*f**4*n**2 + 6*int((e + f*x)**n/(a*e + a*f*x + 
 b*e*x**3 + b*f*x**4),x)*a**2*f**4*n - int(((e + f*x)**n*x**2)/(a*e + a*f* 
x + b*e*x**3 + b*f*x**4),x)*a*b*e*f**3*n**4 - 6*int(((e + f*x)**n*x**2)/(a 
*e + a*f*x + b*e*x**3 + b*f*x**4),x)*a*b*e*f**3*n**3 - 11*int(((e + f*x)** 
n*x**2)/(a*e + a*f*x + b*e*x**3 + b*f*x**4),x)*a*b*e*f**3*n**2 - 6*int(((e 
 + f*x)**n*x**2)/(a*e + a*f*x + b*e*x**3 + b*f*x**4),x)*a*b*e*f**3*n)/(b** 
2*f**3*n*(n**3 + 6*n**2 + 11*n + 6))