3.2743 \(\int \frac{x^m}{a+b x^{2+2 m}} \, dx\)

Optimal. Leaf size=33 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0172386, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {345, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m)),x]

[Out]

ArcTan[(Sqrt[b]*x^(1 + m))/Sqrt[a]]/(Sqrt[a]*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0118577, size = 33, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^(2 + 2*m)),x]

[Out]

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

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Maple [B]  time = 0.038, size = 61, normalized size = 1.9 \begin{align*} -{\frac{1}{2+2\,m}\ln \left ({x}^{m}-{\frac{a}{x}{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{2+2\,m}\ln \left ({x}^{m}+{\frac{a}{x}{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{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)),x, algorithm="maxima")

[Out]

integrate(x^m/(b*x^(2*m + 2) + a), x)

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Fricas [A]  time = 1.35306, size = 209, normalized size = 6.33 \begin{align*} \left [-\frac{\sqrt{-a b} \log \left (\frac{b x^{2} x^{2 \, m} - 2 \, \sqrt{-a b} x x^{m} - a}{b x^{2} x^{2 \, m} + a}\right )}{2 \,{\left (a b m + a b\right )}}, -\frac{\sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b x x^{m}}\right )}{a b m + a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a*b)*log((b*x^2*x^(2*m) - 2*sqrt(-a*b)*x*x^m - a)/(b*x^2*x^(2*m) + a))/(a*b*m + a*b), -sqrt(a*b)*a
rctan(sqrt(a*b)/(b*x*x^m))/(a*b*m + a*b)]

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Sympy [C]  time = 2.62913, size = 197, normalized size = 5.97 \begin{align*} \frac{i \sqrt{\pi } a^{- \frac{m}{2 \left (m + 1\right )}} a^{- \frac{1}{2 \left (m + 1\right )}} \log{\left (1 - \frac{\sqrt{b} x x^{m} e^{\frac{i \pi }{2}}}{\sqrt{a}} \right )}}{4 \sqrt{b} m \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right ) + 4 \sqrt{b} \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right )} - \frac{i \sqrt{\pi } a^{- \frac{m}{2 \left (m + 1\right )}} a^{- \frac{1}{2 \left (m + 1\right )}} \log{\left (1 - \frac{\sqrt{b} x x^{m} e^{\frac{3 i \pi }{2}}}{\sqrt{a}} \right )}}{4 \sqrt{b} m \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right ) + 4 \sqrt{b} \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)),x)

[Out]

I*sqrt(pi)*a**(-m/(2*(m + 1)))*a**(-1/(2*(m + 1)))*log(1 - sqrt(b)*x*x**m*exp_polar(I*pi/2)/sqrt(a))/(4*sqrt(b
)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 4*sqrt(b)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1)))) - I*sqrt(pi
)*a**(-m/(2*(m + 1)))*a**(-1/(2*(m + 1)))*log(1 - sqrt(b)*x*x**m*exp_polar(3*I*pi/2)/sqrt(a))/(4*sqrt(b)*m*gam
ma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 4*sqrt(b)*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}}{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)),x, algorithm="giac")

[Out]

integrate(x^m/(b*x^(2*m + 2) + a), x)