3.2749 \(\int \frac{x^m}{\sqrt{a+b x^{2+2 m}}} \, dx\)

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

[Out]

ArcTanh[(Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2*(1 + m))]]/(Sqrt[b]*(1 + m))

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Rubi [A]  time = 0.0199075, antiderivative size = 38, normalized size of antiderivative = 1., 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 = {345, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a+b x^{2 (m+1)}}}\right )}{\sqrt{b} (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/Sqrt[a + b*x^(2 + 2*m)],x]

[Out]

ArcTanh[(Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2*(1 + m))]]/(Sqrt[b]*(1 + m))

Rule 345

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0379224, size = 66, normalized size = 1.74 \[ \frac{\sqrt{a} \sqrt{\frac{b x^{2 m+2}}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{\sqrt{b} (m+1) \sqrt{a+b x^{2 m+2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/Sqrt[a + b*x^(2 + 2*m)],x]

[Out]

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

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m}{\frac{1}{\sqrt{a+b{x}^{2+2\,m}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sqrt{b x^{2 \, m + 2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [C]  time = 3.71795, size = 117, normalized size = 3.08 \begin{align*} \frac{\sqrt{\pi } x x^{m}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{1}{2} \\ \frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )} \end{matrix}\middle |{\frac{b x^{2} x^{2 m} e^{i \pi }}{a}} \right )}}{2 a^{\frac{m}{2 \left (m + 1\right )}} a^{\frac{1}{2 \left (m + 1\right )}} m \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right ) + 2 a^{\frac{m}{2 \left (m + 1\right )}} a^{\frac{1}{2 \left (m + 1\right )}} \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(pi)*x*x**m*hyper((1/2, 1/2), (m/(2*(m + 1)) + 1 + 1/(2*(m + 1)),), b*x**2*x**(2*m)*exp_polar(I*pi)/a)/(2*
a**(m/(2*(m + 1)))*a**(1/(2*(m + 1)))*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 2*a**(m/(2*(m + 1)))*a**(1/
(2*(m + 1)))*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sqrt{b x^{2 \, m + 2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/sqrt(b*x^(2*m + 2) + a), x)