\(\int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx\) [2689]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {x^{n/2} \sqrt {a+b x^n}}{b n}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{b^{3/2} n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {362, 294, 212} \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {x^{n/2} \sqrt {a+b x^n}}{b n}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{b^{3/2} n} \]

[In]

Int[x^(-1 + (3*n)/2)/Sqrt[a + b*x^n],x]

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 294

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

Rule 362

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[p]}, Dist[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}, x] && IntegerQ[p + Simplify[(m + 1)/n]] && LtQ[-1, p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(2 a) \text {Subst}\left (\int \frac {x^2}{\left (1-b x^2\right )^2} \, dx,x,\frac {x^{n/2}}{\sqrt {a+b x^n}}\right )}{n} \\ & = \frac {x^{n/2} \sqrt {a+b x^n}}{b n}-\frac {a \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{n/2}}{\sqrt {a+b x^n}}\right )}{b n} \\ & = \frac {x^{n/2} \sqrt {a+b x^n}}{b n}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{b^{3/2} n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.31 \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {\sqrt {b} x^{n/2} \left (a+b x^n\right )-a^{3/2} \sqrt {1+\frac {b x^n}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )}{b^{3/2} n \sqrt {a+b x^n}} \]

[In]

Integrate[x^(-1 + (3*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(Sqrt[b]*x^(n/2)*(a + b*x^n) - a^(3/2)*Sqrt[1 + (b*x^n)/a]*ArcSinh[(Sqrt[b]*x^(n/2))/Sqrt[a]])/(b^(3/2)*n*Sqrt
[a + b*x^n])

Maple [A] (verified)

Time = 3.75 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {n \ln \left (x \right )}{2}} \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{b n}-\frac {a \ln \left (\sqrt {b}\, {\mathrm e}^{\frac {n \ln \left (x \right )}{2}}+\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}\right )}{b^{\frac {3}{2}} n}\)	\(64\)

[In]

int(x^(-1+3/2*n)/(a+b*x^n)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/b/n*exp(1/2*n*ln(x))*(a+b*exp(1/2*n*ln(x))^2)^(1/2)-a/b^(3/2)/n*ln(b^(1/2)*exp(1/2*n*ln(x))+(a+b*exp(1/2*n*l
n(x))^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.84 \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\left [\frac {2 \, \sqrt {b x^{n} + a} b x^{\frac {1}{2} \, n} + a \sqrt {b} \log \left (2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right )}{2 \, b^{2} n}, \frac {\sqrt {b x^{n} + a} b x^{\frac {1}{2} \, n} + a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{\frac {1}{2} \, n}}{\sqrt {b x^{n} + a}}\right )}{b^{2} n}\right ] \]

[In]

integrate(x^(-1+3/2*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(2*sqrt(b*x^n + a)*b*x^(1/2*n) + a*sqrt(b)*log(2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) - 2*b*x^n - a))/(b^2*n
), (sqrt(b*x^n + a)*b*x^(1/2*n) + a*sqrt(-b)*arctan(sqrt(-b)*x^(1/2*n)/sqrt(b*x^n + a)))/(b^2*n)]

Sympy [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79 \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {\sqrt {a} x^{\frac {n}{2}} \sqrt {1 + \frac {b x^{n}}{a}}}{b n} - \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}} n} \]

[In]

integrate(x**(-1+3/2*n)/(a+b*x**n)**(1/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {x^{\frac {3}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \]

[In]

integrate(x^(-1+3/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(3/2*n - 1)/sqrt(b*x^n + a), x)

Giac [F]

\[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {x^{\frac {3}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \]

[In]

integrate(x^(-1+3/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(3/2*n - 1)/sqrt(b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int \frac {x^{\frac {3\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \]

[In]

int(x^((3*n)/2 - 1)/(a + b*x^n)^(1/2),x)

[Out]

int(x^((3*n)/2 - 1)/(a + b*x^n)^(1/2), x)