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

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 = 0.0565193, 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, Rules used = {2011, 365, 364} \[ \frac{2 x^{1-2 n} \sqrt{\frac{a x^{j-n}}{b}+1} \, _2F_1\left (\frac{5}{2},\frac{1-\frac{5 n}{2}}{j-n};\frac{1-\frac{5 n}{2}}{j-n}+1;-\frac{a x^{j-n}}{b}\right )}{b^2 (2-5 n) \sqrt{a x^j+b x^n}} \]

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])

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a x^j+b x^n\right )^{5/2}} \, dx &=\frac{\left (x^{n/2} \sqrt{b+a x^{j-n}}\right ) \int \frac{x^{-5 n/2}}{\left (b+a x^{j-n}\right )^{5/2}} \, dx}{\sqrt{a x^j+b x^n}}\\ &=\frac{\left (x^{n/2} \sqrt{1+\frac{a x^{j-n}}{b}}\right ) \int \frac{x^{-5 n/2}}{\left (1+\frac{a x^{j-n}}{b}\right )^{5/2}} \, dx}{b^2 \sqrt{a x^j+b x^n}}\\ &=\frac{2 x^{1-2 n} \sqrt{1+\frac{a x^{j-n}}{b}} \, _2F_1\left (\frac{5}{2},\frac{1-\frac{5 n}{2}}{j-n};1+\frac{1-\frac{5 n}{2}}{j-n};-\frac{a x^{j-n}}{b}\right )}{b^2 (2-5 n) \sqrt{a x^j+b x^n}}\\ \end{align*}

Mathematica [A]  time = 0.243232, 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 = 0.348, size = 0, normalized size = 0. \begin{align*} \int \left ( a{x}^{j}+b{x}^{n} \right ) ^{-{\frac{5}{2}}}\, dx \end{align*}

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. \begin{align*} \int \frac{1}{{\left (a x^{j} + b x^{n}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a x^{j} + b x^{n}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**j + b*x**n)**(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^j + b*x^n)^(-5/2), x)