3.2693 \(\int \frac {x^{-1-\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx\)

Optimal. Leaf size=89 \[ -\frac {16 b^2 x^{-n/2} \sqrt {a+b x^n}}{15 a^3 n}+\frac {8 b x^{-3 n/2} \sqrt {a+b x^n}}{15 a^2 n}-\frac {2 x^{-5 n/2} \sqrt {a+b x^n}}{5 a n} \]

[Out]

-2/5*(a+b*x^n)^(1/2)/a/n/(x^(5/2*n))+8/15*b*(a+b*x^n)^(1/2)/a^2/n/(x^(3/2*n))-16/15*b^2*(a+b*x^n)^(1/2)/a^3/n/
(x^(1/2*n))

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Rubi [A]  time = 0.03, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {271, 264} \[ -\frac {16 b^2 x^{-n/2} \sqrt {a+b x^n}}{15 a^3 n}+\frac {8 b x^{-3 n/2} \sqrt {a+b x^n}}{15 a^2 n}-\frac {2 x^{-5 n/2} \sqrt {a+b x^n}}{5 a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x^n])/(5*a*n*x^((5*n)/2)) + (8*b*Sqrt[a + b*x^n])/(15*a^2*n*x^((3*n)/2)) - (16*b^2*Sqrt[a + b*x
^n])/(15*a^3*n*x^(n/2))

Rule 264

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

Rule 271

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

Rubi steps

\begin {align*} \int \frac {x^{-1-\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx &=-\frac {2 x^{-5 n/2} \sqrt {a+b x^n}}{5 a n}-\frac {(4 b) \int \frac {x^{-1-\frac {3 n}{2}}}{\sqrt {a+b x^n}} \, dx}{5 a}\\ &=-\frac {2 x^{-5 n/2} \sqrt {a+b x^n}}{5 a n}+\frac {8 b x^{-3 n/2} \sqrt {a+b x^n}}{15 a^2 n}+\frac {\left (8 b^2\right ) \int \frac {x^{-1-\frac {n}{2}}}{\sqrt {a+b x^n}} \, dx}{15 a^2}\\ &=-\frac {2 x^{-5 n/2} \sqrt {a+b x^n}}{5 a n}+\frac {8 b x^{-3 n/2} \sqrt {a+b x^n}}{15 a^2 n}-\frac {16 b^2 x^{-n/2} \sqrt {a+b x^n}}{15 a^3 n}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 0.57 \[ -\frac {2 x^{-5 n/2} \sqrt {a+b x^n} \left (3 a^2-4 a b x^n+8 b^2 x^{2 n}\right )}{15 a^3 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x^n]*(3*a^2 - 4*a*b*x^n + 8*b^2*x^(2*n)))/(15*a^3*n*x^((5*n)/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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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 [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \frac {x^{-\frac {5 n}{2}-1}}{\sqrt {b \,x^{n}+a}}\, dx \]

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]

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

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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 \[ \int \frac {1}{x^{\frac {5\,n}{2}+1}\,\sqrt {a+b\,x^n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 6.25, size = 354, normalized size = 3.98 \[ - \frac {6 a^{4} b^{\frac {9}{2}} \sqrt {\frac {a x^{- n}}{b} + 1}}{15 a^{5} b^{4} n x^{2 n} + 30 a^{4} b^{5} n x^{3 n} + 15 a^{3} b^{6} n x^{4 n}} - \frac {4 a^{3} b^{\frac {11}{2}} x^{n} \sqrt {\frac {a x^{- n}}{b} + 1}}{15 a^{5} b^{4} n x^{2 n} + 30 a^{4} b^{5} n x^{3 n} + 15 a^{3} b^{6} n x^{4 n}} - \frac {6 a^{2} b^{\frac {13}{2}} x^{2 n} \sqrt {\frac {a x^{- n}}{b} + 1}}{15 a^{5} b^{4} n x^{2 n} + 30 a^{4} b^{5} n x^{3 n} + 15 a^{3} b^{6} n x^{4 n}} - \frac {24 a b^{\frac {15}{2}} x^{3 n} \sqrt {\frac {a x^{- n}}{b} + 1}}{15 a^{5} b^{4} n x^{2 n} + 30 a^{4} b^{5} n x^{3 n} + 15 a^{3} b^{6} n x^{4 n}} - \frac {16 b^{\frac {17}{2}} x^{4 n} \sqrt {\frac {a x^{- n}}{b} + 1}}{15 a^{5} b^{4} n x^{2 n} + 30 a^{4} b^{5} n x^{3 n} + 15 a^{3} b^{6} n x^{4 n}} \]

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]

-6*a**4*b**(9/2)*sqrt(a*x**(-n)/b + 1)/(15*a**5*b**4*n*x**(2*n) + 30*a**4*b**5*n*x**(3*n) + 15*a**3*b**6*n*x**
(4*n)) - 4*a**3*b**(11/2)*x**n*sqrt(a*x**(-n)/b + 1)/(15*a**5*b**4*n*x**(2*n) + 30*a**4*b**5*n*x**(3*n) + 15*a
**3*b**6*n*x**(4*n)) - 6*a**2*b**(13/2)*x**(2*n)*sqrt(a*x**(-n)/b + 1)/(15*a**5*b**4*n*x**(2*n) + 30*a**4*b**5
*n*x**(3*n) + 15*a**3*b**6*n*x**(4*n)) - 24*a*b**(15/2)*x**(3*n)*sqrt(a*x**(-n)/b + 1)/(15*a**5*b**4*n*x**(2*n
) + 30*a**4*b**5*n*x**(3*n) + 15*a**3*b**6*n*x**(4*n)) - 16*b**(17/2)*x**(4*n)*sqrt(a*x**(-n)/b + 1)/(15*a**5*
b**4*n*x**(2*n) + 30*a**4*b**5*n*x**(3*n) + 15*a**3*b**6*n*x**(4*n))

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