\(\int \frac {(e+f x)^n}{x^2 (a+b x^3)} \, dx\) [46]

Optimal result

Integrand size = 20, antiderivative size = 326 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=-\frac {b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a e^2 (1+n)} \] Output:

-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)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {\sqrt [3]{-1} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {(-1)^{2/3} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}+\frac {3 \sqrt [3]{a} f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{e^2}\right )}{3 a^{4/3} (1+n)} \] Input:

Integrate[(e + f*x)^n/(x^2*(a + b*x^3)),x]
 

Output:

((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))
 

Rubi [A] (verified)

Time = 1.31 (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

\(\displaystyle -\frac {b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a e^2 (n+1)}\)

Input:

Int[(e + f*x)^n/(x^2*(a + b*x^3)),x]
 

Output:

-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))
 

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]

\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (b \,x^{3}+a \right )}d x\]

Input:

int((f*x+e)^n/x^2/(b*x^3+a),x)
 

Output:

int((f*x+e)^n/x^2/(b*x^3+a),x)
 

Fricas [F]

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

integrate((f*x+e)^n/x^2/(b*x^3+a),x, algorithm="fricas")
 

Output:

integral((f*x + e)^n/(b*x^5 + a*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:

integrate((f*x+e)**n/x**2/(b*x**3+a),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((f*x+e)^n/x^2/(b*x^3+a),x, algorithm="maxima")
 

Output:

integrate((f*x + e)^n/((b*x^3 + a)*x^2), x)
 

Giac [F]

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

integrate((f*x+e)^n/x^2/(b*x^3+a),x, algorithm="giac")
 

Output:

integrate((f*x + e)^n/((b*x^3 + a)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (b\,x^3+a\right )} \,d x \] Input:

int((e + f*x)^n/(x^2*(a + b*x^3)),x)
 

Output:

int((e + f*x)^n/(x^2*(a + b*x^3)), x)
 

Reduce [F]

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

int((f*x+e)^n/x^2/(b*x^3+a),x)
 

Output:

int((e + f*x)**n/(a*x**2 + b*x**5),x)