Integrand size = 17, antiderivative size = 97 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\frac {x^{1+m}}{4 a (1+m) \left (a+b x^{2 (1+m)}\right )^2}+\frac {3 x^{1+m}}{8 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )}+\frac {3 \arctan \left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} (1+m)} \] Output:
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(b^(1/2)*x^(1+m)/a^(1/2))/a^(5/2)/b^(1/2)/(1+m)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.36 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},-\frac {b x^{2+2 m}}{a}\right )}{a^3 (1+m)} \] Input:
Integrate[x^m/(a + b*x^(2 + 2*m))^3,x]
Output:
(x^(1 + m)*Hypergeometric2F1[1/2, 3, 3/2, -((b*x^(2 + 2*m))/a)])/(a^3*(1 + m))
Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {868, 215, 215, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^m}{\left (a+b x^{2 m+2}\right )^3} \, dx\) |
\(\Big \downarrow \) 868 |
\(\displaystyle \frac {\int \frac {1}{\left (b x^{2 m+2}+a\right )^3}dx^{m+1}}{m+1}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left (b x^{2 m+2}+a\right )^2}dx^{m+1}}{4 a}+\frac {x^{m+1}}{4 a \left (a+b x^{2 m+2}\right )^2}}{m+1}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{b x^{2 m+2}+a}dx^{m+1}}{2 a}+\frac {x^{m+1}}{2 a \left (a+b x^{2 m+2}\right )}\right )}{4 a}+\frac {x^{m+1}}{4 a \left (a+b x^{2 m+2}\right )^2}}{m+1}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x^{m+1}}{2 a \left (a+b x^{2 m+2}\right )}\right )}{4 a}+\frac {x^{m+1}}{4 a \left (a+b x^{2 m+2}\right )^2}}{m+1}\) |
Input:
Int[x^m/(a + b*x^(2 + 2*m))^3,x]
Output:
(x^(1 + m)/(4*a*(a + b*x^(2 + 2*m))^2) + (3*(x^(1 + m)/(2*a*(a + b*x^(2 + 2*m))) + ArcTan[(Sqrt[b]*x^(1 + m))/Sqrt[a]]/(2*a^(3/2)*Sqrt[b])))/(4*a))/ (1 + m)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Time = 0.83 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {x \,x^{m} \left (3 b \,x^{2} x^{2 m}+5 a \right )}{8 \left (1+m \right ) a^{2} \left (a +b \,x^{2} x^{2 m}\right )^{2}}-\frac {3 \ln \left (x^{m}-\frac {a}{x \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, \left (1+m \right ) a^{2}}+\frac {3 \ln \left (x^{m}+\frac {a}{x \sqrt {-a b}}\right )}{16 \sqrt {-a b}\, \left (1+m \right ) a^{2}}\) | \(110\) |
Input:
int(x^m/(a+b*x^(2+2*m))^3,x,method=_RETURNVERBOSE)
Output:
1/8*x*x^m*(3*b*x^2*(x^m)^2+5*a)/(1+m)/a^2/(a+b*x^2*(x^m)^2)^2-3/16/(-a*b)^ (1/2)/(1+m)/a^2*ln(x^m-a/x/(-a*b)^(1/2))+3/16/(-a*b)^(1/2)/(1+m)/a^2*ln(x^ m+a/x/(-a*b)^(1/2))
Time = 0.08 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.46 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\left [\frac {6 \, a b^{2} x^{3} x^{3 \, m} + 10 \, a^{2} b x x^{m} - 3 \, {\left (\sqrt {-a b} b^{2} x^{4} x^{4 \, m} + 2 \, \sqrt {-a b} a b x^{2} x^{2 \, m} + \sqrt {-a b} a^{2}\right )} \log \left (\frac {b x^{2} x^{2 \, m} - 2 \, \sqrt {-a b} x x^{m} - a}{b x^{2} x^{2 \, m} + a}\right )}{16 \, {\left (a^{5} b m + a^{5} b + {\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{4} x^{4 \, m} + 2 \, {\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{2} x^{2 \, m}\right )}}, \frac {3 \, a b^{2} x^{3} x^{3 \, m} + 5 \, a^{2} b x x^{m} + 3 \, {\left (\sqrt {a b} b^{2} x^{4} x^{4 \, m} + 2 \, \sqrt {a b} a b x^{2} x^{2 \, m} + \sqrt {a b} a^{2}\right )} \arctan \left (\frac {\sqrt {a b} x x^{m}}{a}\right )}{8 \, {\left (a^{5} b m + a^{5} b + {\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{4} x^{4 \, m} + 2 \, {\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{2} x^{2 \, m}\right )}}\right ] \] Input:
integrate(x^m/(a+b*x^(2+2*m))^3,x, algorithm="fricas")
Output:
[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) + sqrt(-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)*x*x^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))]
Timed out. \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x**m/(a+b*x**(2+2*m))**3,x)
Output:
Timed out
\[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\int { \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{3}} \,d x } \] Input:
integrate(x^m/(a+b*x^(2+2*m))^3,x, algorithm="maxima")
Output:
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)
\[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\int { \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{3}} \,d x } \] Input:
integrate(x^m/(a+b*x^(2+2*m))^3,x, algorithm="giac")
Output:
integrate(x^m/(b*x^(2*m + 2) + a)^3, x)
Timed out. \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\int \frac {x^m}{{\left (a+b\,x^{2\,m+2}\right )}^3} \,d x \] Input:
int(x^m/(a + b*x^(2*m + 2))^3,x)
Output:
int(x^m/(a + b*x^(2*m + 2))^3, x)
Time = 0.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.87 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^3} \, dx=\frac {3 x^{4 m} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{m} b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}+6 x^{2 m} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{m} b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{m} b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+3 x^{3 m} a \,b^{2} x^{3}+5 x^{m} a^{2} b x}{8 a^{3} b \left (x^{4 m} b^{2} m \,x^{4}+x^{4 m} b^{2} x^{4}+2 x^{2 m} a b m \,x^{2}+2 x^{2 m} a b \,x^{2}+a^{2} m +a^{2}\right )} \] Input:
int(x^m/(a+b*x^(2+2*m))^3,x)
Output:
(3*x**(4*m)*sqrt(b)*sqrt(a)*atan((x**m*b*x)/(sqrt(b)*sqrt(a)))*b**2*x**4 + 6*x**(2*m)*sqrt(b)*sqrt(a)*atan((x**m*b*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 + 3*sqrt(b)*sqrt(a)*atan((x**m*b*x)/(sqrt(b)*sqrt(a)))*a**2 + 3*x**(3*m)*a*b **2*x**3 + 5*x**m*a**2*b*x)/(8*a**3*b*(x**(4*m)*b**2*m*x**4 + x**(4*m)*b** 2*x**4 + 2*x**(2*m)*a*b*m*x**2 + 2*x**(2*m)*a*b*x**2 + a**2*m + a**2))