3.2689 \(\int \frac{x^{-1+\frac{3 n}{2}}}{\sqrt{a+b x^n}} \, dx\)

Optimal. Leaf size=62 \[ \frac{x^{n/2} \sqrt{a+b x^n}}{b n}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{b^{3/2} n} \]

[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)

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Rubi [A]  time = 0.0265925, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {355, 288, 206} \[ \frac{x^{n/2} \sqrt{a+b x^n}}{b n}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{b^{3/2} n} \]

Antiderivative was successfully verified.

[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 355

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]

Rule 288

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 206

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

Rubi steps

\begin{align*} \int \frac{x^{-1+\frac{3 n}{2}}}{\sqrt{a+b x^n}} \, dx &=\frac{(2 a) \operatorname{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 \operatorname{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]  time = 0.0504219, size = 81, normalized size = 1.31 \[ \frac{\sqrt{b} x^{n/2} \left (a+b x^n\right )-a^{3/2} \sqrt{\frac{b x^n}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a}}\right )}{b^{3/2} n \sqrt{a+b x^n}} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.029, size = 64, normalized size = 1. \begin{align*}{\frac{1}{nb}{{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}}\sqrt{a+b \left ({{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}} \right ) ^{2}}}-{\frac{a}{n}\ln \left ( \sqrt{b}{{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}}+\sqrt{a+b \left ({{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}} \right ) ^{2}} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2} \, n - 1}}{\sqrt{b x^{n} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.46157, size = 285, normalized size = 4.6 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [A]  time = 18.4395, size = 49, normalized size = 0.79 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2} \, n - 1}}{\sqrt{b x^{n} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)