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]

1/2*x^(1+m)/a/(1+m)/(a+b*x^(2+2*m))+1/2*arctan(x^(1+m)*b^(1/2)/a^(1/2))/a^(3/2)/(1+m)/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 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]

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}{\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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.60, size = 212, normalized size = 3.16 \[ \left [\frac {2 \, a b x x^{m} - {\left (\sqrt {-a b} b x^{2} x^{2 \, m} + \sqrt {-a b} a\right )} \log \left (\frac {b x^{2} x^{2 \, m} - 2 \, \sqrt {-a b} x x^{m} - a}{b x^{2} x^{2 \, m} + a}\right )}{4 \, {\left (a^{3} b m + a^{3} b + {\left (a^{2} b^{2} m + a^{2} b^{2}\right )} x^{2} x^{2 \, m}\right )}}, \frac {a b x x^{m} - {\left (\sqrt {a b} b x^{2} x^{2 \, m} + \sqrt {a b} a\right )} \arctan \left (\frac {\sqrt {a b}}{b x x^{m}}\right )}{2 \, {\left (a^{3} b m + a^{3} b + {\left (a^{2} b^{2} m + a^{2} b^{2}\right )} x^{2} x^{2 \, m}\right )}}\right ] \]

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

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)

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maple [A]  time = 0.05, size = 95, normalized size = 1.42 \[ \frac {x \,x^{m}}{2 \left (m +1\right ) \left (b \,x^{2} x^{2 m}+a \right ) a}-\frac {\ln \left (x^{m}-\frac {a}{\sqrt {-a b}\, x}\right )}{4 \sqrt {-a b}\, \left (m +1\right ) a}+\frac {\ln \left (x^{m}+\frac {a}{\sqrt {-a b}\, x}\right )}{4 \sqrt {-a b}\, \left (m +1\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x x^{m}}{2 \, {\left (a b {\left (m + 1\right )} x^{2} x^{2 \, m} + a^{2} {\left (m + 1\right )}\right )}} + \int \frac {x^{m}}{2 \, {\left (a b x^{2} x^{2 \, m} + a^{2}\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m}{{\left (a+b\,x^{2\,m+2}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(2*m + 2))^2,x)

[Out]

int(x^m/(a + b*x^(2*m + 2))^2, x)

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sympy [C]  time = 8.62, size = 865, normalized size = 12.91 \[ - \frac {i \sqrt {\pi } a^{- \frac {m}{2 m + 2}} a^{- \frac {1}{2 m + 2}} \log {\left (1 - \frac {\sqrt {b} x x^{m} e^{\frac {i \pi }{2}}}{\sqrt {a}} \right )}}{- 8 a \sqrt {b} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 a \sqrt {b} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} m x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} + \frac {i \sqrt {\pi } a^{- \frac {m}{2 m + 2}} a^{- \frac {1}{2 m + 2}} \log {\left (1 - \frac {\sqrt {b} x x^{m} e^{\frac {3 i \pi }{2}}}{\sqrt {a}} \right )}}{- 8 a \sqrt {b} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 a \sqrt {b} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} m x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} - \frac {i \sqrt {\pi } a^{- \frac {m}{2 m + 2}} a^{- \frac {1}{2 m + 2}} b x^{2} x^{2 m} \log {\left (1 - \frac {\sqrt {b} x x^{m} e^{\frac {i \pi }{2}}}{\sqrt {a}} \right )}}{a \left (- 8 a \sqrt {b} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 a \sqrt {b} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} m x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )\right )} + \frac {i \sqrt {\pi } a^{- \frac {m}{2 m + 2}} a^{- \frac {1}{2 m + 2}} b x^{2} x^{2 m} \log {\left (1 - \frac {\sqrt {b} x x^{m} e^{\frac {3 i \pi }{2}}}{\sqrt {a}} \right )}}{a \left (- 8 a \sqrt {b} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 a \sqrt {b} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} m x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )\right )} - \frac {2 \sqrt {\pi } a^{- \frac {m}{2 m + 2}} a^{- \frac {1}{2 m + 2}} \sqrt {b} x x^{m}}{\sqrt {a} \left (- 8 a \sqrt {b} m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 a \sqrt {b} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} m x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) - 8 b^{\frac {3}{2}} x^{2} x^{2 m} \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )\right )} \]

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 + 2))*a**(-1/(2*m + 2))*log(1 - sqrt(b)*x*x**m*exp_polar(I*pi/2)/sqrt(a))/(-8*a*sqrt(b
)*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*a*sqrt(b)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*m*x**
2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)
)) + I*sqrt(pi)*a**(-m/(2*m + 2))*a**(-1/(2*m + 2))*log(1 - sqrt(b)*x*x**m*exp_polar(3*I*pi/2)/sqrt(a))/(-8*a*
sqrt(b)*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*a*sqrt(b)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)
*m*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*
m + 2))) - I*sqrt(pi)*a**(-m/(2*m + 2))*a**(-1/(2*m + 2))*b*x**2*x**(2*m)*log(1 - sqrt(b)*x*x**m*exp_polar(I*p
i/2)/sqrt(a))/(a*(-8*a*sqrt(b)*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*a*sqrt(b)*gamma(m/(2*m + 2) + 1 + 1/
(2*m + 2)) - 8*b**(3/2)*m*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*x**2*x**(2*m)*gamma(
m/(2*m + 2) + 1 + 1/(2*m + 2)))) + I*sqrt(pi)*a**(-m/(2*m + 2))*a**(-1/(2*m + 2))*b*x**2*x**(2*m)*log(1 - sqrt
(b)*x*x**m*exp_polar(3*I*pi/2)/sqrt(a))/(a*(-8*a*sqrt(b)*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*a*sqrt(b)*
gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*m*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**
(3/2)*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)))) - 2*sqrt(pi)*a**(-m/(2*m + 2))*a**(-1/(2*m + 2))*sq
rt(b)*x*x**m/(sqrt(a)*(-8*a*sqrt(b)*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*a*sqrt(b)*gamma(m/(2*m + 2) + 1
+ 1/(2*m + 2)) - 8*b**(3/2)*m*x**2*x**(2*m)*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) - 8*b**(3/2)*x**2*x**(2*m)*g
amma(m/(2*m + 2) + 1 + 1/(2*m + 2))))

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