3.2744 \(\int \frac{x^m}{(a+b x^{2+2 m})^2} \, dx\)

Optimal. Leaf size=67 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} (m+1)}+\frac{x^{m+1}}{2 a (m+1) \left (a+b x^{2 (m+1)}\right )} \]

[Out]

x^(1 + m)/(2*a*(1 + m)*(a + b*x^(2*(1 + m)))) + ArcTan[(Sqrt[b]*x^(1 + m))/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]*(1 + m)
)

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Rubi [A]  time = 0.0255506, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {345, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} (m+1)}+\frac{x^{m+1}}{2 a (m+1) \left (a+b x^{2 (m+1)}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m))^2,x]

[Out]

x^(1 + m)/(2*a*(1 + m)*(a + b*x^(2*(1 + m)))) + ArcTan[(Sqrt[b]*x^(1 + m))/Sqrt[a]]/(2*a^(3/2)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [C]  time = 0.0136761, size = 53, normalized size = 0.79 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2 m+2};\frac{m+1}{2 m+2}+1;-\frac{b x^{2 m+2}}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[2, (1 + m)/(2 + 2*m), 1 + (1 + m)/(2 + 2*m), -((b*x^(2 + 2*m))/a)])/(a^2*(1 + m))

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Maple [A]  time = 0.031, size = 95, normalized size = 1.4 \begin{align*}{\frac{x{x}^{m}}{ \left ( 2+2\,m \right ) a \left ( a+b{x}^{2} \left ({x}^{m} \right ) ^{2} \right ) }}-{\frac{1}{ \left ( 4+4\,m \right ) a}\ln \left ({x}^{m}-{\frac{a}{x}{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{ \left ( 4+4\,m \right ) a}\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))^2,x)

[Out]

1/2*x/(1+m)/a*x^m/(a+b*x^2*(x^m)^2)-1/4/(-a*b)^(1/2)/(1+m)/a*ln(x^m-a/x/(-a*b)^(1/2))+1/4/(-a*b)^(1/2)/(1+m)/a
*ln(x^m+a/x/(-a*b)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x x^{m}}{2 \,{\left (a b{\left (m + 1\right )} x^{2} x^{2 \, m} + a^{2}{\left (m + 1\right )}\right )}} + \int \frac{x^{m}}{2 \,{\left (a b x^{2} x^{2 \, m} + a^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.34073, size = 443, normalized size = 6.61 \begin{align*} \left [\frac{2 \, a b x x^{m} -{\left (\sqrt{-a b} b x^{2} x^{2 \, m} + \sqrt{-a b} a\right )} \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 )}{4 \,{\left (a^{3} b m + a^{3} b +{\left (a^{2} b^{2} m + a^{2} b^{2}\right )} x^{2} x^{2 \, m}\right )}}, \frac{a b x x^{m} -{\left (\sqrt{a b} b x^{2} x^{2 \, m} + \sqrt{a b} a\right )} \arctan \left (\frac{\sqrt{a b}}{b x x^{m}}\right )}{2 \,{\left (a^{3} b m + a^{3} b +{\left (a^{2} b^{2} m + a^{2} b^{2}\right )} x^{2} x^{2 \, m}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*a*b*x*x^m - (sqrt(-a*b)*b*x^2*x^(2*m) + sqrt(-a*b)*a)*log((b*x^2*x^(2*m) - 2*sqrt(-a*b)*x*x^m - a)/(b*
x^2*x^(2*m) + a)))/(a^3*b*m + a^3*b + (a^2*b^2*m + a^2*b^2)*x^2*x^(2*m)), 1/2*(a*b*x*x^m - (sqrt(a*b)*b*x^2*x^
(2*m) + sqrt(a*b)*a)*arctan(sqrt(a*b)/(b*x*x^m)))/(a^3*b*m + a^3*b + (a^2*b^2*m + a^2*b^2)*x^2*x^(2*m))]

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Sympy [C]  time = 13.6105, size = 857, normalized size = 12.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(a+b*x**(2+2*m))**2,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))/(8*a*sqrt
(b)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*a*sqrt(b)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(
3/2)*m*x**2*x**(2*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(3/2)*x**2*x**(2*m)*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))/(8*a*sqrt(b)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*a*sqrt(b)*gamma(m/(2*(m + 1)) + 1
+ 1/(2*(m + 1))) + 8*b**(3/2)*m*x**2*x**(2*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(3/2)*x**2*x**(2
*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1)))) + I*sqrt(pi)*a**(-m/(2*(m + 1)))*a**(-1/(2*(m + 1)))*b*x**2*x**(
2*m)*log(1 - sqrt(b)*x*x**m*exp_polar(I*pi/2)/sqrt(a))/(a*(8*a*sqrt(b)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1
))) + 8*a*sqrt(b)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(3/2)*m*x**2*x**(2*m)*gamma(m/(2*(m + 1)) +
1 + 1/(2*(m + 1))) + 8*b**(3/2)*x**2*x**(2*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))))) - I*sqrt(pi)*a**(-m/(
2*(m + 1)))*a**(-1/(2*(m + 1)))*b*x**2*x**(2*m)*log(1 - sqrt(b)*x*x**m*exp_polar(3*I*pi/2)/sqrt(a))/(a*(8*a*sq
rt(b)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*a*sqrt(b)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b*
*(3/2)*m*x**2*x**(2*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(3/2)*x**2*x**(2*m)*gamma(m/(2*(m + 1))
+ 1 + 1/(2*(m + 1))))) + 2*sqrt(pi)*a**(-m/(2*(m + 1)))*a**(-1/(2*(m + 1)))*sqrt(b)*x*x**m/(sqrt(a)*(8*a*sqrt
(b)*m*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*a*sqrt(b)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(
3/2)*m*x**2*x**(2*m)*gamma(m/(2*(m + 1)) + 1 + 1/(2*(m + 1))) + 8*b**(3/2)*x**2*x**(2*m)*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}}{{\left (b x^{2 \, m + 2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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