3.2745 \(\int \frac {x^m}{(a+b x^{2+2 m})^3} \, dx\)

Optimal. Leaf size=97 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} (m+1)}+\frac {3 x^{m+1}}{8 a^2 (m+1) \left (a+b x^{2 (m+1)}\right )}+\frac {x^{m+1}}{4 a (m+1) \left (a+b x^{2 (m+1)}\right )^2} \]

[Out]

1/4*x^(1+m)/a/(1+m)/(a+b*x^(2+2*m))^2+3/8*x^(1+m)/a^2/(1+m)/(a+b*x^(2+2*m))+3/8*arctan(x^(1+m)*b^(1/2)/a^(1/2)
)/a^(5/2)/(1+m)/b^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {345, 199, 205} \[ \frac {3 x^{m+1}}{8 a^2 (m+1) \left (a+b x^{2 (m+1)}\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} (m+1)}+\frac {x^{m+1}}{4 a (m+1) \left (a+b x^{2 (m+1)}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m))^3,x]

[Out]

x^(1 + m)/(4*a*(1 + m)*(a + b*x^(2*(1 + m)))^2) + (3*x^(1 + m))/(8*a^2*(1 + m)*(a + b*x^(2*(1 + m)))) + (3*Arc
Tan[(Sqrt[b]*x^(1 + m))/Sqrt[a]])/(8*a^(5/2)*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 )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^3} \, dx,x,x^{1+m}\right )}{1+m}\\ &=\frac {x^{1+m}}{4 a (1+m) \left (a+b x^{2 (1+m)}\right )^2}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^2} \, dx,x,x^{1+m}\right )}{4 a (1+m)}\\ &=\frac {x^{1+m}}{4 a (1+m) \left (a+b x^{2 (1+m)}\right )^2}+\frac {3 x^{1+m}}{8 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{1+m}\right )}{8 a^2 (1+m)}\\ &=\frac {x^{1+m}}{4 a (1+m) \left (a+b x^{2 (1+m)}\right )^2}+\frac {3 x^{1+m}}{8 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 35, normalized size = 0.36 \[ \frac {x^{m+1} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};-\frac {b x^{2 m+2}}{a}\right )}{a^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, 3, 3/2, -((b*x^(2 + 2*m))/a)])/(a^3*(1 + m))

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 340, normalized size = 3.51 \[ \left [\frac {6 \, a b^{2} x^{3} x^{3 \, m} + 10 \, a^{2} b x x^{m} - 3 \, {\left (\sqrt {-a b} b^{2} x^{4} x^{4 \, m} + 2 \, \sqrt {-a b} a b x^{2} x^{2 \, m} + \sqrt {-a b} a^{2}\right )} \log \left (\frac {b x^{2} x^{2 \, m} - 2 \, \sqrt {-a b} x x^{m} - a}{b x^{2} x^{2 \, m} + a}\right )}{16 \, {\left (a^{5} b m + a^{5} b + {\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{4} x^{4 \, m} + 2 \, {\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{2} x^{2 \, m}\right )}}, \frac {3 \, a b^{2} x^{3} x^{3 \, m} + 5 \, a^{2} b x x^{m} - 3 \, {\left (\sqrt {a b} b^{2} x^{4} x^{4 \, m} + 2 \, \sqrt {a b} a b x^{2} x^{2 \, m} + \sqrt {a b} a^{2}\right )} \arctan \left (\frac {\sqrt {a b}}{b x x^{m}}\right )}{8 \, {\left (a^{5} b m + a^{5} b + {\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{4} x^{4 \, m} + 2 \, {\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{2} x^{2 \, m}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(6*a*b^2*x^3*x^(3*m) + 10*a^2*b*x*x^m - 3*(sqrt(-a*b)*b^2*x^4*x^(4*m) + 2*sqrt(-a*b)*a*b*x^2*x^(2*m) + s
qrt(-a*b)*a^2)*log((b*x^2*x^(2*m) - 2*sqrt(-a*b)*x*x^m - a)/(b*x^2*x^(2*m) + a)))/(a^5*b*m + a^5*b + (a^3*b^3*
m + a^3*b^3)*x^4*x^(4*m) + 2*(a^4*b^2*m + a^4*b^2)*x^2*x^(2*m)), 1/8*(3*a*b^2*x^3*x^(3*m) + 5*a^2*b*x*x^m - 3*
(sqrt(a*b)*b^2*x^4*x^(4*m) + 2*sqrt(a*b)*a*b*x^2*x^(2*m) + sqrt(a*b)*a^2)*arctan(sqrt(a*b)/(b*x*x^m)))/(a^5*b*
m + a^5*b + (a^3*b^3*m + a^3*b^3)*x^4*x^(4*m) + 2*(a^4*b^2*m + a^4*b^2)*x^2*x^(2*m))]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 110, normalized size = 1.13 \[ \frac {\left (3 b \,x^{2} x^{2 m}+5 a \right ) x \,x^{m}}{8 \left (m +1\right ) \left (b \,x^{2} x^{2 m}+a \right )^{2} a^{2}}-\frac {3 \ln \left (x^{m}-\frac {a}{\sqrt {-a b}\, x}\right )}{16 \sqrt {-a b}\, \left (m +1\right ) a^{2}}+\frac {3 \ln \left (x^{m}+\frac {a}{\sqrt {-a b}\, x}\right )}{16 \sqrt {-a b}\, \left (m +1\right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*x^m*(3*b*x^2*(x^m)^2+5*a)/(m+1)/a^2/(a+b*x^2*(x^m)^2)^2-3/16/(-a*b)^(1/2)/(m+1)/a^2*ln(x^m-1/(-a*b)^(1/2
)*a/x)+3/16/(-a*b)^(1/2)/(m+1)/a^2*ln(x^m+1/(-a*b)^(1/2)*a/x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3 \, b x^{3} x^{3 \, m} + 5 \, a x x^{m}}{8 \, {\left (a^{2} b^{2} {\left (m + 1\right )} x^{4} x^{4 \, m} + 2 \, a^{3} b {\left (m + 1\right )} x^{2} x^{2 \, m} + a^{4} {\left (m + 1\right )}\right )}} + 3 \, \int \frac {x^{m}}{8 \, {\left (a^{2} b x^{2} x^{2 \, m} + a^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b*x^3*x^(3*m) + 5*a*x*x^m)/(a^2*b^2*(m + 1)*x^4*x^(4*m) + 2*a^3*b*(m + 1)*x^2*x^(2*m) + a^4*(m + 1)) +
3*integrate(1/8*x^m/(a^2*b*x^2*x^(2*m) + a^3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m}{{\left (a+b\,x^{2\,m+2}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(2*m + 2))^3,x)

[Out]

int(x^m/(a + b*x^(2*m + 2))^3, x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________