3.2750 \(\int \frac {x^m}{(a+b x^{2+2 m})^{3/2}} \, dx\)

Optimal. Leaf size=29 \[ \frac {x^{m+1}}{a (m+1) \sqrt {a+b x^{2 (m+1)}}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {264} \[ \frac {x^{m+1}}{a (m+1) \sqrt {a+b x^{2 (m+1)}}} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m))^(3/2),x]

[Out]

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{3/2}} \, dx &=\frac {x^{1+m}}{a (1+m) \sqrt {a+b x^{2 (1+m)}}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \[ \frac {x^{m+1}}{a (m+1) \sqrt {a+b x^{2 m+2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^(2 + 2*m))^(3/2),x]

[Out]

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

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fricas [A]  time = 0.58, size = 47, normalized size = 1.62 \[ \frac {\sqrt {b x^{2} x^{2 \, m} + a} x x^{m}}{{\left (a b m + a b\right )} x^{2} x^{2 \, m} + a^{2} m + a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(2+2*m))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(2+2*m))^(3/2),x, algorithm="giac")

[Out]

integrate(x^m/(b*x^(2*m + 2) + a)^(3/2), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\left (b \,x^{2 m +2}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x^(2*m+2)+a)^(3/2),x)

[Out]

int(x^m/(b*x^(2*m+2)+a)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(2+2*m))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^m/(b*x^(2*m + 2) + a)^(3/2), x)

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mupad [B]  time = 1.22, size = 27, normalized size = 0.93 \[ \frac {x^{m+1}}{a\,\sqrt {a+b\,x^{2\,m+2}}\,\left (m+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(2*m + 2))^(3/2),x)

[Out]

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

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sympy [C]  time = 21.63, size = 121, normalized size = 4.17 \[ \frac {\sqrt {\pi } x x^{m} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2} x^{2 m} e^{i \pi }}{a}} \right )}}{2 a a^{\frac {m}{2 m + 2}} a^{\frac {1}{2 m + 2}} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 a a^{\frac {m}{2 m + 2}} a^{\frac {1}{2 m + 2}} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(a+b*x**(2+2*m))**(3/2),x)

[Out]

sqrt(pi)*x*x**m*hyper((3/2, 1/2), (m/(2*m + 2) + 1 + 1/(2*m + 2),), b*x**2*x**(2*m)*exp_polar(I*pi)/a)/(2*a*a*
*(m/(2*m + 2))*a**(1/(2*m + 2))*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) + 2*a*a**(m/(2*m + 2))*a**(1/(2*m + 2))
*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)))

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