3.2687 \(\int \frac{x^{-1+\frac{7 n}{2}}}{\sqrt{a+b x^n}} \, dx\)

Optimal. Leaf size=129 \[ \frac{5 a^2 x^{n/2} \sqrt{a+b x^n}}{8 b^3 n}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{8 b^{7/2} n}-\frac{5 a x^{3 n/2} \sqrt{a+b x^n}}{12 b^2 n}+\frac{x^{5 n/2} \sqrt{a+b x^n}}{3 b n} \]

[Out]

(5*a^2*x^(n/2)*Sqrt[a + b*x^n])/(8*b^3*n) - (5*a*x^((3*n)/2)*Sqrt[a + b*x^n])/(12*b^2*n) + (x^((5*n)/2)*Sqrt[a
 + b*x^n])/(3*b*n) - (5*a^3*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(8*b^(7/2)*n)

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Rubi [A]  time = 0.0590889, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {355, 288, 206} \[ \frac{5 a^2 x^{n/2} \sqrt{a+b x^n}}{8 b^3 n}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{8 b^{7/2} n}-\frac{5 a x^{3 n/2} \sqrt{a+b x^n}}{12 b^2 n}+\frac{x^{5 n/2} \sqrt{a+b x^n}}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*x^(n/2)*Sqrt[a + b*x^n])/(8*b^3*n) - (5*a*x^((3*n)/2)*Sqrt[a + b*x^n])/(12*b^2*n) + (x^((5*n)/2)*Sqrt[a
 + b*x^n])/(3*b*n) - (5*a^3*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(8*b^(7/2)*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{7 n}{2}}}{\sqrt{a+b x^n}} \, dx &=\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{x^6}{\left (1-b x^2\right )^4} \, dx,x,\frac{x^{n/2}}{\sqrt{a+b x^n}}\right )}{n}\\ &=\frac{x^{5 n/2} \sqrt{a+b x^n}}{3 b n}-\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-b x^2\right )^3} \, dx,x,\frac{x^{n/2}}{\sqrt{a+b x^n}}\right )}{3 b 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{\left (5 a^3\right ) \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 )}{4 b^2 n}\\ &=\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{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{n/2}}{\sqrt{a+b x^n}}\right )}{8 b^3 n}\\ &=\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 \tanh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a+b x^n}}\right )}{8 b^{7/2} n}\\ \end{align*}

Mathematica [A]  time = 0.206233, size = 100, normalized size = 0.78 \[ \frac{\sqrt{a+b x^n} \left (\sqrt{b} x^{n/2} \left (15 a^2-10 a b x^n+8 b^2 x^{2 n}\right )-\frac{15 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a}}\right )}{\sqrt{\frac{b x^n}{a}+1}}\right )}{24 b^{7/2} n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [A]  time = 0.094, size = 98, normalized size = 0.8 \begin{align*}{\frac{1}{24\,{b}^{3}n}{{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}} \left ( 8\, \left ({{\rm e}^{1/2\,n\ln \left ( x \right ) }} \right ) ^{4}{b}^{2}-10\,a \left ({{\rm e}^{1/2\,n\ln \left ( x \right ) }} \right ) ^{2}b+15\,{a}^{2} \right ) \sqrt{a+b \left ({{\rm e}^{{\frac{n\ln \left ( x \right ) }{2}}}} \right ) ^{2}}}-{\frac{5\,{a}^{3}}{8\,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{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{7}{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+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.26573, size = 436, normalized size = 3.38 \begin{align*} \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^{\frac{1}{2} \, n}}{\sqrt{b x^{n} + a}}\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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^(1/2*n)/sqrt(
b*x^n + a)) + (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)]

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Sympy [A]  time = 25.4921, size = 148, normalized size = 1.15 \begin{align*} \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)) - sqr
t(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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{7}{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+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

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