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

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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 = {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 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])

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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {b x^{2 \, m + 2} + a}}\,{d x} \]

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)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {b \,x^{2 m +2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {b x^{2 \, m + 2} + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^m}{\sqrt {a+b\,x^{2\,m+2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(2*m + 2))^(1/2),x)

[Out]

int(x^m/(a + b*x^(2*m + 2))^(1/2), x)

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sympy [C]  time = 3.42, size = 117, normalized size = 3.08 \[ \frac {\sqrt {\pi } x x^{m} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2} x^{2 m} e^{i \pi }}{a}} \right )}}{2 a^{\frac {m}{2 m + 2}} a^{\frac {1}{2 m + 2}} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 a^{\frac {m}{2 m + 2}} a^{\frac {1}{2 m + 2}} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \]

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

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