\(\int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} (e+f x^n)} \, dx\) [4]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F(-2)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 186 \[ \int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} \left (e+f x^n\right )} \, dx=\frac {2 C x \sqrt {a+b x^n}}{b f (2+n)}+\frac {\left (C e^2-B e f+A f^2\right ) x \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )}{e f^2 \sqrt {a+b x^n}}-\frac {(2 a C f+b (C e-B f) (2+n)) x \sqrt {1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b f^2 (2+n) \sqrt {a+b x^n}} \] Output:

2*C*x*(a+b*x^n)^(1/2)/b/f/(2+n)+(A*f^2-B*e*f+C*e^2)*x*(1+b*x^n/a)^(1/2)*Ap 
pellF1(1/n,1/2,1,1+1/n,-b*x^n/a,-f*x^n/e)/e/f^2/(a+b*x^n)^(1/2)-(2*a*C*f+b 
*(-B*f+C*e)*(2+n))*x*(1+b*x^n/a)^(1/2)*hypergeom([1/2, 1/n],[1+1/n],-b*x^n 
/a)/b/f^2/(2+n)/(a+b*x^n)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(414\) vs. \(2(186)=372\).

Time = 1.14 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.23 \[ \int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} \left (e+f x^n\right )} \, dx=\frac {x \left (-\frac {(2 a C f+b (C e-B f) (2+n)) x^n \sqrt {1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},1,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )}{e (1+n)}+\frac {4 a C f n x^n \left (a+b x^n\right ) \left (e+f x^n\right ) \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},2,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )+2 b C e n x^n \left (a+b x^n\right ) \left (e+f x^n\right ) \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},1,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )-2 a e (1+n) \left (A b f (2+n)+2 C x^n \left (a f+b \left (e+f x^n\right )\right )\right ) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )}{\left (e+f x^n\right ) \left (2 a f n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},2,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )+b e n x^n \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {3}{2},1,2+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )-2 a e (1+n) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )\right )}\right )}{b f (2+n) \sqrt {a+b x^n}} \] Input:

Integrate[(A + B*x^n + C*x^(2*n))/(Sqrt[a + b*x^n]*(e + f*x^n)),x]
 

Output:

(x*(-(((2*a*C*f + b*(C*e - B*f)*(2 + n))*x^n*Sqrt[1 + (b*x^n)/a]*AppellF1[ 
1 + n^(-1), 1/2, 1, 2 + n^(-1), -((b*x^n)/a), -((f*x^n)/e)])/(e*(1 + n))) 
+ (4*a*C*f*n*x^n*(a + b*x^n)*(e + f*x^n)*AppellF1[1 + n^(-1), 1/2, 2, 2 + 
n^(-1), -((b*x^n)/a), -((f*x^n)/e)] + 2*b*C*e*n*x^n*(a + b*x^n)*(e + f*x^n 
)*AppellF1[1 + n^(-1), 3/2, 1, 2 + n^(-1), -((b*x^n)/a), -((f*x^n)/e)] - 2 
*a*e*(1 + n)*(A*b*f*(2 + n) + 2*C*x^n*(a*f + b*(e + f*x^n)))*AppellF1[n^(- 
1), 1/2, 1, 1 + n^(-1), -((b*x^n)/a), -((f*x^n)/e)])/((e + f*x^n)*(2*a*f*n 
*x^n*AppellF1[1 + n^(-1), 1/2, 2, 2 + n^(-1), -((b*x^n)/a), -((f*x^n)/e)] 
+ b*e*n*x^n*AppellF1[1 + n^(-1), 3/2, 1, 2 + n^(-1), -((b*x^n)/a), -((f*x^ 
n)/e)] - 2*a*e*(1 + n)*AppellF1[n^(-1), 1/2, 1, 1 + n^(-1), -((b*x^n)/a), 
-((f*x^n)/e)]))))/(b*f*(2 + n)*Sqrt[a + b*x^n])
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {7293, 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.

\(\displaystyle \int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} \left (e+f x^n\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {A f^2-B e f+C e^2}{f^2 \sqrt {a+b x^n} \left (e+f x^n\right )}+\frac {B f-C e}{f^2 \sqrt {a+b x^n}}+\frac {C x^n}{f \sqrt {a+b x^n}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \sqrt {\frac {b x^n}{a}+1} \left (A f^2-B e f+C e^2\right ) \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},1,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {f x^n}{e}\right )}{e f^2 \sqrt {a+b x^n}}-\frac {x \sqrt {\frac {b x^n}{a}+1} (C e-B f) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{f^2 \sqrt {a+b x^n}}+\frac {C x^{n+1} \sqrt {\frac {b x^n}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {1}{n},2+\frac {1}{n},-\frac {b x^n}{a}\right )}{f (n+1) \sqrt {a+b x^n}}\)

Input:

Int[(A + B*x^n + C*x^(2*n))/(Sqrt[a + b*x^n]*(e + f*x^n)),x]
 

Output:

((C*e^2 - B*e*f + A*f^2)*x*Sqrt[1 + (b*x^n)/a]*AppellF1[n^(-1), 1/2, 1, 1 
+ n^(-1), -((b*x^n)/a), -((f*x^n)/e)])/(e*f^2*Sqrt[a + b*x^n]) + (C*x^(1 + 
 n)*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, 1 + n^(-1), 2 + n^(-1), -(( 
b*x^n)/a)])/(f*(1 + n)*Sqrt[a + b*x^n]) - ((C*e - B*f)*x*Sqrt[1 + (b*x^n)/ 
a]*Hypergeometric2F1[1/2, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(f^2*Sqrt[a + 
 b*x^n])
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
Maple [F]

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

Input:

int((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x)
 

Output:

int((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} \left (e+f x^n\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x, algorithm="fric 
as")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

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

integrate((A+B*x**n+C*x**(2*n))/(a+b*x**n)**(1/2)/(e+f*x**n),x)
 

Output:

Integral((A + B*x**n + C*x**(2*n))/(sqrt(a + b*x**n)*(e + f*x**n)), x)
 

Maxima [F]

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

integrate((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x, algorithm="maxi 
ma")
 

Output:

integrate((C*x^(2*n) + B*x^n + A)/(sqrt(b*x^n + a)*(f*x^n + e)), x)
 

Giac [F]

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

integrate((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x, algorithm="giac 
")
 

Output:

integrate((C*x^(2*n) + B*x^n + A)/(sqrt(b*x^n + a)*(f*x^n + e)), x)
 

Mupad [F(-1)]

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

int((A + C*x^(2*n) + B*x^n)/((a + b*x^n)^(1/2)*(e + f*x^n)),x)
 

Output:

int((A + C*x^(2*n) + B*x^n)/((a + b*x^n)^(1/2)*(e + f*x^n)), x)
 

Reduce [F]

\[ \int \frac {A+B x^n+C x^{2 n}}{\sqrt {a+b x^n} \left (e+f x^n\right )} \, dx=\text {too large to display} \] Input:

int((A+B*x^n+C*x^(2*n))/(a+b*x^n)^(1/2)/(e+f*x^n),x)
 

Output:

(2*sqrt(x**n*b + a)*b*x + 4*int(sqrt(x**n*b + a)/(2*x**(2*n)*a*b*f**2 + x* 
*(2*n)*b**2*e*f*n + 2*x**(2*n)*b**2*e*f + 2*x**n*a**2*f**2 + x**n*a*b*e*f* 
n + 4*x**n*a*b*e*f + x**n*b**2*e**2*n + 2*x**n*b**2*e**2 + 2*a**2*e*f + a* 
b*e**2*n + 2*a*b*e**2),x)*a**3*f**2 + 4*int(sqrt(x**n*b + a)/(2*x**(2*n)*a 
*b*f**2 + x**(2*n)*b**2*e*f*n + 2*x**(2*n)*b**2*e*f + 2*x**n*a**2*f**2 + x 
**n*a*b*e*f*n + 4*x**n*a*b*e*f + x**n*b**2*e**2*n + 2*x**n*b**2*e**2 + 2*a 
**2*e*f + a*b*e**2*n + 2*a*b*e**2),x)*a**2*b*e*f*n + 4*int(sqrt(x**n*b + a 
)/(2*x**(2*n)*a*b*f**2 + x**(2*n)*b**2*e*f*n + 2*x**(2*n)*b**2*e*f + 2*x** 
n*a**2*f**2 + x**n*a*b*e*f*n + 4*x**n*a*b*e*f + x**n*b**2*e**2*n + 2*x**n* 
b**2*e**2 + 2*a**2*e*f + a*b*e**2*n + 2*a*b*e**2),x)*a**2*b*e*f + int(sqrt 
(x**n*b + a)/(2*x**(2*n)*a*b*f**2 + x**(2*n)*b**2*e*f*n + 2*x**(2*n)*b**2* 
e*f + 2*x**n*a**2*f**2 + x**n*a*b*e*f*n + 4*x**n*a*b*e*f + x**n*b**2*e**2* 
n + 2*x**n*b**2*e**2 + 2*a**2*e*f + a*b*e**2*n + 2*a*b*e**2),x)*a*b**2*e** 
2*n**2 + 2*int(sqrt(x**n*b + a)/(2*x**(2*n)*a*b*f**2 + x**(2*n)*b**2*e*f*n 
 + 2*x**(2*n)*b**2*e*f + 2*x**n*a**2*f**2 + x**n*a*b*e*f*n + 4*x**n*a*b*e* 
f + x**n*b**2*e**2*n + 2*x**n*b**2*e**2 + 2*a**2*e*f + a*b*e**2*n + 2*a*b* 
e**2),x)*a*b**2*e**2*n + 4*int((x**(2*n)*sqrt(x**n*b + a))/(2*x**(2*n)*a*b 
*f**2 + x**(2*n)*b**2*e*f*n + 2*x**(2*n)*b**2*e*f + 2*x**n*a**2*f**2 + x** 
n*a*b*e*f*n + 4*x**n*a*b*e*f + x**n*b**2*e**2*n + 2*x**n*b**2*e**2 + 2*a** 
2*e*f + a*b*e**2*n + 2*a*b*e**2),x)*a**2*c*f**2 - 2*int((x**(2*n)*sqrt(...