\(\int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx\) [2749]

Optimal result

Integrand size = 19, antiderivative size = 38 \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a+b x^{2 (1+m)}}}\right )}{\sqrt {b} (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {352, 223, 212} \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a+b x^{2 (m+1)}}}\right )}{\sqrt {b} (m+1)} \]

[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 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 223

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 352

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^{1+m}\right )}{1+m} \\ & = \frac {\text {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] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.74 \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\frac {\sqrt {a} \sqrt {1+\frac {b x^{2+2 m}}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a}}\right )}{\sqrt {b} (1+m) \sqrt {a+b x^{2+2 m}}} \]

[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)])

Maple [F]

\[\int \frac {x^{m}}{\sqrt {a +b \,x^{2+2 m}}}d x\]

[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)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.85 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.82 \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\frac {\sqrt {\pi } \sqrt {a} a^{- \frac {m}{2 m + 2} - \frac {1}{2} - \frac {1}{2 m + 2}} x^{m + 1} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2 m + 2} e^{i \pi }}{a}} \right )}}{2 m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\int { \frac {x^{m}}{\sqrt {b x^{2 \, m + 2} + a}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\int { \frac {x^{m}}{\sqrt {b x^{2 \, m + 2} + a}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{\sqrt {a+b x^{2+2 m}}} \, dx=\int \frac {x^m}{\sqrt {a+b\,x^{2\,m+2}}} \,d x \]

[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)