3.5.30 \(\int \frac {1}{(a x^j+b x^n)^{5/2}} \, dx\) [430]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2036, 372, 371} \begin {gather*} \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}} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 2036

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]

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*}

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Mathematica [A]
time = 0.21, size = 185, normalized size = 1.83 \begin {gather*} \frac {2 x^{1-2 j} \left (-\left ((-2+4 j+n) \left (a (-2+j+4 n) x^j+b (-2+2 j+3 n) x^n\right )\right )+\left (4+8 j^2-8 n+3 n^2+2 j (-6+7 n)\right ) \sqrt {1+\frac {a x^{j-n}}{b}} \left (a x^j+b x^n\right ) \, _2F_1\left (\frac {1}{2},-\frac {-2+4 j+n}{2 (j-n)};\frac {2-2 j-3 n}{2 j-2 n};-\frac {a x^{j-n}}{b}\right )\right )}{3 a^2 (2-4 j-n) (j-n)^2 \left (a x^j+b x^n\right )^{3/2}} \end {gather*}

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, -1/2*(-2 + 4*j + n)/(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 = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a \,x^{j}+b \,x^{n}\right )^{\frac {5}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a x^{j} + b x^{n}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 5.60, size = 83, normalized size = 0.82 \begin {gather*} -\frac {x\,{\left (\frac {b\,x^{n-j}}{a}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {\frac {5\,j}{2}-1}{j-n};\ \frac {\frac {5\,j}{2}-1}{j-n}+1;\ -\frac {b\,x^{n-j}}{a}\right )}{\left (\frac {5\,j}{2}-1\right )\,{\left (a\,x^j+b\,x^n\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x^j + b*x^n)^(5/2),x)

[Out]

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

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