3.2688 \(\int \frac {x^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx\)

Optimal. Leaf size=98 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} n}-\frac {3 a x^{n/2} \sqrt {a+b x^n}}{4 b^2 n}+\frac {x^{3 n/2} \sqrt {a+b x^n}}{2 b n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {355, 288, 206} \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} n}-\frac {3 a x^{n/2} \sqrt {a+b x^n}}{4 b^2 n}+\frac {x^{3 n/2} \sqrt {a+b x^n}}{2 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

\begin {align*} \int \frac {x^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx &=\frac {\left (2 a^2\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 )}{n}\\ &=\frac {x^{3 n/2} \sqrt {a+b x^n}}{2 b n}-\frac {\left (3 a^2\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 )}{2 b n}\\ &=-\frac {3 a x^{n/2} \sqrt {a+b x^n}}{4 b^2 n}+\frac {x^{3 n/2} \sqrt {a+b x^n}}{2 b n}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^2 n}\\ &=-\frac {3 a x^{n/2} \sqrt {a+b x^n}}{4 b^2 n}+\frac {x^{3 n/2} \sqrt {a+b x^n}}{2 b n}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} n}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 100, normalized size = 1.02 \[ \frac {3 a^{5/2} \sqrt {\frac {b x^n}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )+\sqrt {b} x^{n/2} \left (-3 a^2-a b x^n+2 b^2 x^{2 n}\right )}{4 b^{5/2} n \sqrt {a+b x^n}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 150, normalized size = 1.53 \[ \left [\frac {3 \, a^{2} \sqrt {b} \log \left (-2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \, {\left (2 \, b^{2} x^{\frac {3}{2} \, n} - 3 \, a b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{8 \, b^{3} n}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{\frac {1}{2} \, n}}{\sqrt {b x^{n} + a}}\right ) - {\left (2 \, b^{2} x^{\frac {3}{2} \, n} - 3 \, a b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{4 \, b^{3} n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 82, normalized size = 0.84 \[ \frac {3 a^{2} \ln \left (\sqrt {b}\, {\mathrm e}^{\frac {n \ln \relax (x )}{2}}+\sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}\right )}{4 b^{\frac {5}{2}} n}-\frac {\left (-2 b \,{\mathrm e}^{n \ln \relax (x )}+3 a \right ) \sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}\, {\mathrm e}^{\frac {n \ln \relax (x )}{2}}}{4 b^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{\frac {5\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 12.74, size = 116, normalized size = 1.18 \[ - \frac {3 a^{\frac {3}{2}} x^{\frac {n}{2}}}{4 b^{2} n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {\sqrt {a} x^{\frac {3 n}{2}}}{4 b n \sqrt {1 + \frac {b x^{n}}{a}}} + \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}} n} + \frac {x^{\frac {5 n}{2}}}{2 \sqrt {a} n \sqrt {1 + \frac {b x^{n}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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