\(\int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx\) [605]

Optimal result

Integrand size = 21, antiderivative size = 129 \[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {5 a^2 x^{n/2} \sqrt {a+b x^n}}{8 b^3 n}-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{8 b^{7/2} n} \] Output:

5/8*a^2*x^(1/2*n)*(a+b*x^n)^(1/2)/b^3/n-5/12*a*x^(3/2*n)*(a+b*x^n)^(1/2)/b 
^2/n+1/3*x^(5/2*n)*(a+b*x^n)^(1/2)/b/n-5/8*a^3*arctanh(b^(1/2)*x^(1/2*n)/( 
a+b*x^n)^(1/2))/b^(7/2)/n
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78 \[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {\sqrt {a+b x^n} \left (\sqrt {b} x^{n/2} \left (15 a^2-10 a b x^n+8 b^2 x^{2 n}\right )-\frac {15 a^{5/2} \text {arcsinh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )}{\sqrt {1+\frac {b x^n}{a}}}\right )}{24 b^{7/2} n} \] Input:

Integrate[x^(-1 + (7*n)/2)/Sqrt[a + b*x^n],x]
 

Output:

(Sqrt[a + b*x^n]*(Sqrt[b]*x^(n/2)*(15*a^2 - 10*a*b*x^n + 8*b^2*x^(2*n)) - 
(15*a^(5/2)*ArcSinh[(Sqrt[b]*x^(n/2))/Sqrt[a]])/Sqrt[1 + (b*x^n)/a]))/(24* 
b^(7/2)*n)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {880, 252, 252, 252, 219}

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.

\(\displaystyle \int \frac {x^{\frac {7 n}{2}-1}}{\sqrt {a+b x^n}} \, dx\)

\(\Big \downarrow \) 880

\(\displaystyle \frac {2 a^3 \int \frac {x^{3 n}}{\left (b x^n+a\right )^3 \left (1-\frac {b x^n}{b x^n+a}\right )^4}d\frac {x^{n/2}}{\sqrt {b x^n+a}}}{n}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {2 a^3 \left (\frac {x^{5 n/2}}{6 b \left (a+b x^n\right )^{5/2} \left (1-\frac {b x^n}{a+b x^n}\right )^3}-\frac {5 \int \frac {x^{2 n}}{\left (b x^n+a\right )^2 \left (1-\frac {b x^n}{b x^n+a}\right )^3}d\frac {x^{n/2}}{\sqrt {b x^n+a}}}{6 b}\right )}{n}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {2 a^3 \left (\frac {x^{5 n/2}}{6 b \left (a+b x^n\right )^{5/2} \left (1-\frac {b x^n}{a+b x^n}\right )^3}-\frac {5 \left (\frac {x^{3 n/2}}{4 b \left (a+b x^n\right )^{3/2} \left (1-\frac {b x^n}{a+b x^n}\right )^2}-\frac {3 \int \frac {x^n}{\left (b x^n+a\right ) \left (1-\frac {b x^n}{b x^n+a}\right )^2}d\frac {x^{n/2}}{\sqrt {b x^n+a}}}{4 b}\right )}{6 b}\right )}{n}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {2 a^3 \left (\frac {x^{5 n/2}}{6 b \left (a+b x^n\right )^{5/2} \left (1-\frac {b x^n}{a+b x^n}\right )^3}-\frac {5 \left (\frac {x^{3 n/2}}{4 b \left (a+b x^n\right )^{3/2} \left (1-\frac {b x^n}{a+b x^n}\right )^2}-\frac {3 \left (\frac {x^{n/2}}{2 b \sqrt {a+b x^n} \left (1-\frac {b x^n}{a+b x^n}\right )}-\frac {\int \frac {1}{1-\frac {b x^n}{b x^n+a}}d\frac {x^{n/2}}{\sqrt {b x^n+a}}}{2 b}\right )}{4 b}\right )}{6 b}\right )}{n}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 a^3 \left (\frac {x^{5 n/2}}{6 b \left (a+b x^n\right )^{5/2} \left (1-\frac {b x^n}{a+b x^n}\right )^3}-\frac {5 \left (\frac {x^{3 n/2}}{4 b \left (a+b x^n\right )^{3/2} \left (1-\frac {b x^n}{a+b x^n}\right )^2}-\frac {3 \left (\frac {x^{n/2}}{2 b \sqrt {a+b x^n} \left (1-\frac {b x^n}{a+b x^n}\right )}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{6 b}\right )}{n}\)

Input:

Int[x^(-1 + (7*n)/2)/Sqrt[a + b*x^n],x]
 

Output:

(2*a^3*(x^((5*n)/2)/(6*b*(a + b*x^n)^(5/2)*(1 - (b*x^n)/(a + b*x^n))^3) - 
(5*(x^((3*n)/2)/(4*b*(a + b*x^n)^(3/2)*(1 - (b*x^n)/(a + b*x^n))^2) - (3*( 
x^(n/2)/(2*b*Sqrt[a + b*x^n]*(1 - (b*x^n)/(a + b*x^n))) - ArcTanh[(Sqrt[b] 
*x^(n/2))/Sqrt[a + b*x^n]]/(2*b^(3/2))))/(4*b)))/(6*b)))/n
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 880

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[p]}, Simp[k*(a^(p + Simplify[(m + 1)/n])/n)   Subst[Int[x^(k*Simplify 
[(m + 1)/n] - 1)/(1 - b*x^k)^(p + Simplify[(m + 1)/n] + 1), x], x, x^(n/k)/ 
(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[p + Simpli 
fy[(m + 1)/n]] && LtQ[-1, p, 0]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {n \ln \left (x \right )}{2}} \left (8 \,{\mathrm e}^{2 n \ln \left (x \right )} b^{2}-10 a b \,{\mathrm e}^{n \ln \left (x \right )}+15 a^{2}\right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{24 b^{3} n}-\frac {5 a^{3} \ln \left (\sqrt {b}\, {\mathrm e}^{\frac {n \ln \left (x \right )}{2}}+\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}\right )}{8 b^{\frac {7}{2}} n}\)	\(98\)

Input:

int(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/24*exp(1/2*n*ln(x))*(8*exp(1/2*n*ln(x))^4*b^2-10*a*exp(1/2*n*ln(x))^2*b+ 
15*a^2)*(a+b*exp(1/2*n*ln(x))^2)^(1/2)/b^3/n-5/8*a^3/b^(7/2)/n*ln(b^(1/2)* 
exp(1/2*n*ln(x))+(a+b*exp(1/2*n*ln(x))^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.40 \[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \, {\left (8 \, b^{3} x^{\frac {5}{2} \, n} - 10 \, a b^{2} x^{\frac {3}{2} \, n} + 15 \, a^{2} b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{48 \, b^{4} n}, \frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-b}}{b x^{\frac {1}{2} \, n}}\right ) + {\left (8 \, b^{3} x^{\frac {5}{2} \, n} - 10 \, a b^{2} x^{\frac {3}{2} \, n} + 15 \, a^{2} b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{24 \, b^{4} n}\right ] \] Input:

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")
 

Output:

[1/48*(15*a^3*sqrt(b)*log(2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) - 2*b*x^n - 
a) + 2*(8*b^3*x^(5/2*n) - 10*a*b^2*x^(3/2*n) + 15*a^2*b*x^(1/2*n))*sqrt(b* 
x^n + a))/(b^4*n), 1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x^n + a)*sqrt(-b)/( 
b*x^(1/2*n))) + (8*b^3*x^(5/2*n) - 10*a*b^2*x^(3/2*n) + 15*a^2*b*x^(1/2*n) 
)*sqrt(b*x^n + a))/(b^4*n)]
 

Sympy [A] (verification not implemented)

Time = 10.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15 \[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {5 a^{\frac {5}{2}} x^{\frac {n}{2}}}{8 b^{3} n \sqrt {1 + \frac {b x^{n}}{a}}} + \frac {5 a^{\frac {3}{2}} x^{\frac {3 n}{2}}}{24 b^{2} n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {\sqrt {a} x^{\frac {5 n}{2}}}{12 b n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}} n} + \frac {x^{\frac {7 n}{2}}}{3 \sqrt {a} n \sqrt {1 + \frac {b x^{n}}{a}}} \] Input:

integrate(x**(-1+7/2*n)/(a+b*x**n)**(1/2),x)
 

Output:

5*a**(5/2)*x**(n/2)/(8*b**3*n*sqrt(1 + b*x**n/a)) + 5*a**(3/2)*x**(3*n/2)/ 
(24*b**2*n*sqrt(1 + b*x**n/a)) - sqrt(a)*x**(5*n/2)/(12*b*n*sqrt(1 + b*x** 
n/a)) - 5*a**3*asinh(sqrt(b)*x**(n/2)/sqrt(a))/(8*b**(7/2)*n) + x**(7*n/2) 
/(3*sqrt(a)*n*sqrt(1 + b*x**n/a))
 

Maxima [F]

\[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {x^{\frac {7}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \] Input:

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^(7/2*n - 1)/sqrt(b*x^n + a), x)
 

Giac [F]

\[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {x^{\frac {7}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \] Input:

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")
 

Output:

integrate(x^(7/2*n - 1)/sqrt(b*x^n + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int \frac {x^{\frac {7\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \] Input:

int(x^((7*n)/2 - 1)/(a + b*x^n)^(1/2),x)
 

Output:

int(x^((7*n)/2 - 1)/(a + b*x^n)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int \frac {x^{\frac {7 n}{2}} \sqrt {x^{n} b +a}}{x^{n} b x +a x}d x \] Input:

int(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((x**((7*n)/2)*sqrt(x**n*b + a))/(x**n*b*x + a*x),x)