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

Optimal. Leaf size=38 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x^n-b x^2}}\right )}{\sqrt {b} (2-n)} \]

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1979, 2008, 203} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x^n-b x^2}}\right )}{\sqrt {b} (2-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(Sqrt[b]*x)/Sqrt[-(b*x^2) + a*x^n]])/(Sqrt[b]*(2 - n))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rubi steps

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

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Mathematica [B]  time = 0.02, size = 80, normalized size = 2.11 \begin {gather*} -\frac {2 \sqrt {a} x^{n/2} \sqrt {1-\frac {b x^{2-n}}{a}} \sin ^{-1}\left (\frac {\sqrt {b} x^{1-\frac {n}{2}}}{\sqrt {a}}\right )}{\sqrt {b} (n-2) \sqrt {a x^n-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x^2 \left (-b+a x^{-2+n}\right )}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 109, normalized size = 2.87 \begin {gather*} \left [-\frac {\sqrt {-b} \log \left (\frac {a x x^{n - 2} - 2 \, b x - 2 \, \sqrt {a x^{2} x^{n - 2} - b x^{2}} \sqrt {-b}}{x x^{n - 2}}\right )}{b n - 2 \, b}, \frac {2 \, \sqrt {b} \arctan \left (\frac {\sqrt {a x^{2} x^{n - 2} - b x^{2}}}{\sqrt {b} x}\right )}{b n - 2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-sqrt(-b)*log((a*x*x^(n - 2) - 2*b*x - 2*sqrt(a*x^2*x^(n - 2) - b*x^2)*sqrt(-b))/(x*x^(n - 2)))/(b*n - 2*b),
2*sqrt(b)*arctan(sqrt(a*x^2*x^(n - 2) - b*x^2)/(sqrt(b)*x))/(b*n - 2*b)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {{\left (a x^{n - 2} - b\right )} x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.15, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\left (a \,x^{n -2}-b \right ) x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {{\left (a x^{n - 2} - b\right )} x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.10, size = 66, normalized size = 1.74 \begin {gather*} -\frac {\sqrt {a}\,x^{n/2}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x^{1-\frac {n}{2}}}{\sqrt {a}}\right )\,\sqrt {1-\frac {b\,x^{2-n}}{a}}}{\sqrt {b}\,\left (\frac {n}{2}-1\right )\,\sqrt {a\,x^n-b\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^(1/2)*x^(n/2)*asin((b^(1/2)*x^(1 - n/2))/a^(1/2))*(1 - (b*x^(2 - n))/a)^(1/2))/(b^(1/2)*(n/2 - 1)*(a*x^n -
b*x^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(x**2*(a*x**(n - 2) - b)), x)

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