3.430 \(\int \frac{1}{\left (a x^j+b x^n\right )^{5/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{2 x^{1-2 n} \sqrt{\frac{a x^{j-n}}{b}+1} \, _2F_1\left (\frac{5}{2},\frac{1-\frac{5 n}{2}}{j-n};\frac{1-\frac{5 n}{2}}{j-n}+1;-\frac{a x^{j-n}}{b}\right )}{b^2 (2-5 n) \sqrt{a x^j+b x^n}} \]

[Out]

(2*x^(1 - 2*n)*Sqrt[1 + (a*x^(j - n))/b]*Hypergeometric2F1[5/2, (1 - (5*n)/2)/(j
- n), 1 + (1 - (5*n)/2)/(j - n), -((a*x^(j - n))/b)])/(b^2*(2 - 5*n)*Sqrt[a*x^j
+ b*x^n])

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Rubi [A]  time = 0.141901, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ \frac{2 x^{1-2 n} \sqrt{\frac{a x^{j-n}}{b}+1} \, _2F_1\left (\frac{5}{2},\frac{1-\frac{5 n}{2}}{j-n};\frac{1-\frac{5 n}{2}}{j-n}+1;-\frac{a x^{j-n}}{b}\right )}{b^2 (2-5 n) \sqrt{a x^j+b x^n}} \]

Antiderivative was successfully verified.

[In]  Int[(a*x^j + b*x^n)^(-5/2),x]

[Out]

(2*x^(1 - 2*n)*Sqrt[1 + (a*x^(j - n))/b]*Hypergeometric2F1[5/2, (1 - (5*n)/2)/(j
- n), 1 + (1 - (5*n)/2)/(j - n), -((a*x^(j - n))/b)])/(b^2*(2 - 5*n)*Sqrt[a*x^j
+ b*x^n])

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Rubi in Sympy [A]  time = 13.6373, size = 82, normalized size = 0.81 \[ \frac{2 x^{- \frac{n}{2}} x^{- \frac{5 n}{2} + 1} \sqrt{a x^{j} + b x^{n}}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, \frac{- \frac{5 n}{2} + 1}{j - n} \\ \frac{j - \frac{7 n}{2} + 1}{j - n} \end{matrix}\middle |{- \frac{a x^{j - n}}{b}} \right )}}{b^{3} \left (- 5 n + 2\right ) \sqrt{\frac{a x^{j - n}}{b} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/(a*x**j+b*x**n)**(5/2),x)

[Out]

2*x**(-n/2)*x**(-5*n/2 + 1)*sqrt(a*x**j + b*x**n)*hyper((5/2, (-5*n/2 + 1)/(j -
n)), ((j - 7*n/2 + 1)/(j - n),), -a*x**(j - n)/b)/(b**3*(-5*n + 2)*sqrt(a*x**(j
- n)/b + 1))

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Mathematica [A]  time = 0.404327, size = 185, normalized size = 1.83 \[ \frac{2 x^{1-2 j} \left (\left (8 j^2+2 j (7 n-6)+3 n^2-8 n+4\right ) \sqrt{\frac{a x^{j-n}}{b}+1} \left (a x^j+b x^n\right ) \, _2F_1\left (\frac{1}{2},-\frac{4 j+n-2}{2 (j-n)};\frac{-2 j-3 n+2}{2 j-2 n};-\frac{a x^{j-n}}{b}\right )-(4 j+n-2) \left (a (j+4 n-2) x^j+b (2 j+3 n-2) x^n\right )\right )}{3 a^2 (-4 j-n+2) (j-n)^2 \left (a x^j+b x^n\right )^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(2*x^(1 - 2*j)*(-((-2 + 4*j + n)*(a*(-2 + j + 4*n)*x^j + b*(-2 + 2*j + 3*n)*x^n)
) + (4 + 8*j^2 - 8*n + 3*n^2 + 2*j*(-6 + 7*n))*Sqrt[1 + (a*x^(j - n))/b]*(a*x^j
+ b*x^n)*Hypergeometric2F1[1/2, -(-2 + 4*j + n)/(2*(j - n)), (2 - 2*j - 3*n)/(2*
j - 2*n), -((a*x^(j - n))/b)]))/(3*a^2*(2 - 4*j - n)*(j - n)^2*(a*x^j + b*x^n)^(
3/2))

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \[ \int \left ( a{x}^{j}+b{x}^{n} \right ) ^{-{\frac{5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/(a*x^j+b*x^n)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{{\left (a x^{j} + b x^{n}\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((a*x^j + b*x^n)^(-5/2),x, algorithm="maxima")

[Out]

integrate((a*x^j + b*x^n)^(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((a*x^j + b*x^n)^(-5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(a*x**j+b*x**n)**(5/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{{\left (a x^{j} + b x^{n}\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((a*x^j + b*x^n)^(-5/2),x, algorithm="giac")

[Out]

integrate((a*x^j + b*x^n)^(-5/2), x)