\(\int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx\) [678]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=-\frac {3 a x^{-2 n} (c x)^{5 n/2} \sqrt {a+b x^n}}{4 b^2 c n}+\frac {x^{-n} (c x)^{5 n/2} \sqrt {a+b x^n}}{2 b c n}+\frac {3 a^2 x^{-5 n/2} (c x)^{5 n/2} \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} c n} \] Output:

-3/4*a*(c*x)^(5/2*n)*(a+b*x^n)^(1/2)/b^2/c/n/(x^(2*n))+1/2*(c*x)^(5/2*n)*( 
a+b*x^n)^(1/2)/b/c/n/(x^n)+3/4*a^2*(c*x)^(5/2*n)*arctanh(b^(1/2)*x^(1/2*n) 
/(a+b*x^n)^(1/2))/b^(5/2)/c/n/(x^(5/2*n))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {a x^{-5 n/2} (c x)^{5 n/2} \sqrt {1+\frac {b x^n}{a}} \left (\sqrt {b} x^{n/2} \left (-3 a+2 b x^n\right ) \sqrt {1+\frac {b x^n}{a}}+3 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )\right )}{4 b^{5/2} c n \sqrt {a+b x^n}} \] Input:

Integrate[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]
 

Output:

(a*(c*x)^((5*n)/2)*Sqrt[1 + (b*x^n)/a]*(Sqrt[b]*x^(n/2)*(-3*a + 2*b*x^n)*S 
qrt[1 + (b*x^n)/a] + 3*a^(3/2)*ArcSinh[(Sqrt[b]*x^(n/2))/Sqrt[a]]))/(4*b^( 
5/2)*c*n*x^((5*n)/2)*Sqrt[a + b*x^n])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {883, 880, 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.

\(\displaystyle \int \frac {(c x)^{\frac {5 n}{2}-1}}{\sqrt {a+b x^n}} \, dx\)

\(\Big \downarrow \) 883

\(\displaystyle \frac {x^{-5 n/2} (c x)^{5 n/2} \int \frac {x^{\frac {5 n}{2}-1}}{\sqrt {b x^n+a}}dx}{c}\)

\(\Big \downarrow \) 880

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 219

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

Input:

Int[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]
 

Output:

(2*a^2*(c*x)^((5*n)/2)*(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)))/(c*n* 
x^((5*n)/2))
 

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]
 

rule 883

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^Int 
Part[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a + b*x^n)^p, x], x] / 
; FreeQ[{a, b, c, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n + p]]
 

Maple [F]

\[\int \frac {\left (c x \right )^{-1+\frac {5 n}{2}}}{\sqrt {a +b \,x^{n}}}d x\]

Input:

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)
 

Output:

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.44 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} c^{\frac {5}{2} \, n - 1} \log \left (-2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \, {\left (2 \, b^{2} c^{\frac {5}{2} \, n - 1} x^{\frac {3}{2} \, n} - 3 \, a b c^{\frac {5}{2} \, n - 1} x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{8 \, b^{3} n}, -\frac {3 \, a^{2} \sqrt {-b} c^{\frac {5}{2} \, n - 1} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-b}}{b x^{\frac {1}{2} \, n}}\right ) - {\left (2 \, b^{2} c^{\frac {5}{2} \, n - 1} x^{\frac {3}{2} \, n} - 3 \, a b c^{\frac {5}{2} \, n - 1} x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{4 \, b^{3} n}\right ] \] Input:

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")
 

Output:

[1/8*(3*a^2*sqrt(b)*c^(5/2*n - 1)*log(-2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) 
 - 2*b*x^n - a) + 2*(2*b^2*c^(5/2*n - 1)*x^(3/2*n) - 3*a*b*c^(5/2*n - 1)*x 
^(1/2*n))*sqrt(b*x^n + a))/(b^3*n), -1/4*(3*a^2*sqrt(-b)*c^(5/2*n - 1)*arc 
tan(sqrt(b*x^n + a)*sqrt(-b)/(b*x^(1/2*n))) - (2*b^2*c^(5/2*n - 1)*x^(3/2* 
n) - 3*a*b*c^(5/2*n - 1)*x^(1/2*n))*sqrt(b*x^n + a))/(b^3*n)]
 

Sympy [A] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.09 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=- \frac {3 a^{\frac {3}{2}} c^{\frac {5 n}{2} - 1} x^{\frac {n}{2}}}{4 b^{2} n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {\sqrt {a} c^{\frac {5 n}{2} - 1} x^{\frac {3 n}{2}}}{4 b n \sqrt {1 + \frac {b x^{n}}{a}}} + \frac {3 a^{2} c^{\frac {5 n}{2} - 1} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}} n} + \frac {c^{\frac {5 n}{2} - 1} x^{\frac {5 n}{2}}}{2 \sqrt {a} n \sqrt {1 + \frac {b x^{n}}{a}}} \] Input:

integrate((c*x)**(-1+5/2*n)/(a+b*x**n)**(1/2),x)
 

Output:

-3*a**(3/2)*c**(5*n/2 - 1)*x**(n/2)/(4*b**2*n*sqrt(1 + b*x**n/a)) - sqrt(a 
)*c**(5*n/2 - 1)*x**(3*n/2)/(4*b*n*sqrt(1 + b*x**n/a)) + 3*a**2*c**(5*n/2 
- 1)*asinh(sqrt(b)*x**(n/2)/sqrt(a))/(4*b**(5/2)*n) + c**(5*n/2 - 1)*x**(5 
*n/2)/(2*sqrt(a)*n*sqrt(1 + b*x**n/a))
 

Maxima [F]

\[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \] Input:

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)
 

Giac [F]

\[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \] Input:

integrate((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int \frac {{\left (c\,x\right )}^{\frac {5\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \] Input:

int((c*x)^((5*n)/2 - 1)/(a + b*x^n)^(1/2),x)
 

Output:

int((c*x)^((5*n)/2 - 1)/(a + b*x^n)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {c^{\frac {5 n}{2}} \left (\int \frac {x^{\frac {5 n}{2}} \sqrt {x^{n} b +a}}{x^{n} b x +a x}d x \right )}{c} \] Input:

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)
 

Output:

(c**((5*n)/2)*int((x**((5*n)/2)*sqrt(x**n*b + a))/(x**n*b*x + a*x),x))/c