3.5.10 \(\int \frac {1}{\sqrt {x^{2-n} (a+b x^n)}} \, dx\) [410]

Optimal. Leaf size=37 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+a x^{2-n}}}\right )}{\sqrt {b} n} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 37, 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 = {2004, 2033, 212} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x^{2-n}+b x^2}}\right )}{\sqrt {b} n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[x^(2 - n)*(a + b*x^n)],x]

[Out]

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

Rule 212

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

Rule 2004

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

Rule 2033

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(76\) vs. \(2(37)=74\).
time = 0.05, size = 76, normalized size = 2.05 \begin {gather*} \frac {2 \sqrt {a} x^{1-\frac {n}{2}} \sqrt {1+\frac {b x^n}{a}} \sinh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )}{\sqrt {b} n \sqrt {x^{2-n} \left (a+b x^n\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {x^{2-n} \left (a +b \,x^{n}\right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 2.16, size = 102, normalized size = 2.76 \begin {gather*} \left [\frac {\log \left (\frac {2 \, b x x^{n} + a x + 2 \, \sqrt {b} x^{n} \sqrt {\frac {b x^{2} x^{n} + a x^{2}}{x^{n}}}}{x}\right )}{\sqrt {b} n}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b x^{2} x^{n} + a x^{2}}{x^{n}}}}{b x}\right )}{b n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {1}{\sqrt {x^{2-n}\,\left (a+b\,x^n\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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