\(\int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx\) [482]

Optimal result

Integrand size = 30, antiderivative size = 89 \[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d n}-\frac {(b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2} n} \] Output:

(a+b*x^n)^(1/2)*(c+d*x^n)^(1/2)/b/d/n-(a*d+b*c)*arctanh(d^(1/2)*(a+b*x^n)^ 
(1/2)/b^(1/2)/(c+d*x^n)^(1/2))/b^(3/2)/d^(3/2)/n
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.38 \[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\frac {b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right )-\sqrt {b c-a d} (b c+a d) \sqrt {\frac {b \left (c+d x^n\right )}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b c-a d}}\right )}{b^2 d^{3/2} n \sqrt {c+d x^n}} \] Input:

Integrate[x^(-1 + 2*n)/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),x]
 

Output:

(b*Sqrt[d]*Sqrt[a + b*x^n]*(c + d*x^n) - Sqrt[b*c - a*d]*(b*c + a*d)*Sqrt[ 
(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^n])/Sqrt[b*c - 
a*d]])/(b^2*d^(3/2)*n*Sqrt[c + d*x^n])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {948, 90, 66, 221}

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.

\(\displaystyle \int \frac {x^{2 n-1}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {\int \frac {x^n}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{n}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}-\frac {(a d+b c) \int \frac {1}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{2 b d}}{n}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}-\frac {(a d+b c) \int \frac {1}{b-d x^{2 n}}d\frac {\sqrt {b x^n+a}}{\sqrt {d x^n+c}}}{b d}}{n}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}-\frac {(a d+b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2}}}{n}\)

Input:

Int[x^(-1 + 2*n)/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),x]
 

Output:

((Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(b*d) - ((b*c + a*d)*ArcTanh[(Sqrt[d]*S 
qrt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d*x^n])])/(b^(3/2)*d^(3/2)))/n
 

Defintions of rubi rules used

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [F]

\[\int \frac {x^{-1+2 n}}{\sqrt {a +b \,x^{n}}\, \sqrt {c +d \,x^{n}}}d x\]

Input:

int(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x)
 

Output:

int(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.16 \[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\left [\frac {4 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} b d + {\left (b c + a d\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2 \, n} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, \sqrt {b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}{4 \, b^{2} d^{2} n}, \frac {2 \, \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} b d + {\left (b c + a d\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, \sqrt {-b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {-b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{2 \, {\left (b^{2} d^{2} x^{2 \, n} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}}\right )}{2 \, b^{2} d^{2} n}\right ] \] Input:

integrate(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x, algorithm="fricas" 
)
 

Output:

[1/4*(4*sqrt(b*x^n + a)*sqrt(d*x^n + c)*b*d + (b*c + a*d)*sqrt(b*d)*log(8* 
b^2*d^2*x^(2*n) + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*sqrt(b*d)*b*d*x^n + 
 (b*c + a*d)*sqrt(b*d))*sqrt(b*x^n + a)*sqrt(d*x^n + c) + 8*(b^2*c*d + a*b 
*d^2)*x^n))/(b^2*d^2*n), 1/2*(2*sqrt(b*x^n + a)*sqrt(d*x^n + c)*b*d + (b*c 
 + a*d)*sqrt(-b*d)*arctan(1/2*(2*sqrt(-b*d)*b*d*x^n + (b*c + a*d)*sqrt(-b* 
d))*sqrt(b*x^n + a)*sqrt(d*x^n + c)/(b^2*d^2*x^(2*n) + a*b*c*d + (b^2*c*d 
+ a*b*d^2)*x^n)))/(b^2*d^2*n)]
 

Sympy [F]

\[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int \frac {x^{2 n - 1}}{\sqrt {a + b x^{n}} \sqrt {c + d x^{n}}}\, dx \] Input:

integrate(x**(-1+2*n)/(a+b*x**n)**(1/2)/(c+d*x**n)**(1/2),x)
 

Output:

Integral(x**(2*n - 1)/(sqrt(a + b*x**n)*sqrt(c + d*x**n)), x)
 

Maxima [F]

\[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int { \frac {x^{2 \, n - 1}}{\sqrt {b x^{n} + a} \sqrt {d x^{n} + c}} \,d x } \] Input:

integrate(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x, algorithm="maxima" 
)
 

Output:

integrate(x^(2*n - 1)/(sqrt(b*x^n + a)*sqrt(d*x^n + c)), x)
 

Giac [F]

\[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int { \frac {x^{2 \, n - 1}}{\sqrt {b x^{n} + a} \sqrt {d x^{n} + c}} \,d x } \] Input:

integrate(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x, algorithm="giac")
 

Output:

integrate(x^(2*n - 1)/(sqrt(b*x^n + a)*sqrt(d*x^n + c)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int \frac {x^{2\,n-1}}{\sqrt {a+b\,x^n}\,\sqrt {c+d\,x^n}} \,d x \] Input:

int(x^(2*n - 1)/((a + b*x^n)^(1/2)*(c + d*x^n)^(1/2)),x)
 

Output:

int(x^(2*n - 1)/((a + b*x^n)^(1/2)*(c + d*x^n)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x^{-1+2 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int \frac {x^{2 n} \sqrt {x^{n} d +c}\, \sqrt {x^{n} b +a}}{x^{2 n} b d x +x^{n} a d x +x^{n} b c x +a c x}d x \] Input:

int(x^(-1+2*n)/(a+b*x^n)^(1/2)/(c+d*x^n)^(1/2),x)
 

Output:

int((x**(2*n)*sqrt(x**n*d + c)*sqrt(x**n*b + a))/(x**(2*n)*b*d*x + x**n*a* 
d*x + x**n*b*c*x + a*c*x),x)