Integrand size = 13, antiderivative size = 60 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {x^2}{2 c^2}+\frac {a x^2}{4 c^2 \left (a+c x^4\right )}-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 c^{5/2}} \] Output:
1/2*x^2/c^2+1/4*a*x^2/c^2/(c*x^4+a)-3/4*a^(1/2)*arctan(c^(1/2)*x^2/a^(1/2) )/c^(5/2)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {x^2}{2 c^2}+\frac {a x^2}{4 c^2 \left (a+c x^4\right )}-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 c^{5/2}} \] Input:
Integrate[x^9/(a + c*x^4)^2,x]
Output:
x^2/(2*c^2) + (a*x^2)/(4*c^2*(a + c*x^4)) - (3*Sqrt[a]*ArcTan[(Sqrt[c]*x^2 )/Sqrt[a]])/(4*c^(5/2))
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {807, 252, 262, 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.
\(\displaystyle \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{\left (c x^4+a\right )^2}dx^2\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \int \frac {x^4}{c x^4+a}dx^2}{2 c}-\frac {x^6}{2 c \left (a+c x^4\right )}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \left (\frac {x^2}{c}-\frac {a \int \frac {1}{c x^4+a}dx^2}{c}\right )}{2 c}-\frac {x^6}{2 c \left (a+c x^4\right )}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \left (\frac {x^2}{c}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{c^{3/2}}\right )}{2 c}-\frac {x^6}{2 c \left (a+c x^4\right )}\right )\) |
Input:
Int[x^9/(a + c*x^4)^2,x]
Output:
(-1/2*x^6/(c*(a + c*x^4)) + (3*(x^2/c - (Sqrt[a]*ArcTan[(Sqrt[c]*x^2)/Sqrt [a]])/c^(3/2)))/(2*c))/2
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Time = 0.46 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {x^{2}}{2 c^{2}}-\frac {a \left (-\frac {x^{2}}{2 \left (c \,x^{4}+a \right )}+\frac {3 \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \sqrt {a c}}\right )}{2 c^{2}}\) | \(49\) |
risch | \(\frac {x^{2}}{2 c^{2}}+\frac {a \,x^{2}}{4 c^{2} \left (c \,x^{4}+a \right )}+\frac {3 \sqrt {-a c}\, \ln \left (c \,x^{2}-\sqrt {-a c}\right )}{8 c^{3}}-\frac {3 \sqrt {-a c}\, \ln \left (c \,x^{2}+\sqrt {-a c}\right )}{8 c^{3}}\) | \(78\) |
Input:
int(x^9/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
1/2*x^2/c^2-1/2/c^2*a*(-1/2*x^2/(c*x^4+a)+3/2/(a*c)^(1/2)*arctan(c*x^2/(a* c)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.40 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\left [\frac {4 \, c x^{6} + 6 \, a x^{2} + 3 \, {\left (c x^{4} + a\right )} \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{4} - 2 \, c x^{2} \sqrt {-\frac {a}{c}} - a}{c x^{4} + a}\right )}{8 \, {\left (c^{3} x^{4} + a c^{2}\right )}}, \frac {2 \, c x^{6} + 3 \, a x^{2} - 3 \, {\left (c x^{4} + a\right )} \sqrt {\frac {a}{c}} \arctan \left (\frac {c x^{2} \sqrt {\frac {a}{c}}}{a}\right )}{4 \, {\left (c^{3} x^{4} + a c^{2}\right )}}\right ] \] Input:
integrate(x^9/(c*x^4+a)^2,x, algorithm="fricas")
Output:
[1/8*(4*c*x^6 + 6*a*x^2 + 3*(c*x^4 + a)*sqrt(-a/c)*log((c*x^4 - 2*c*x^2*sq rt(-a/c) - a)/(c*x^4 + a)))/(c^3*x^4 + a*c^2), 1/4*(2*c*x^6 + 3*a*x^2 - 3* (c*x^4 + a)*sqrt(a/c)*arctan(c*x^2*sqrt(a/c)/a))/(c^3*x^4 + a*c^2)]
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.53 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {a x^{2}}{4 a c^{2} + 4 c^{3} x^{4}} + \frac {3 \sqrt {- \frac {a}{c^{5}}} \log {\left (- c^{2} \sqrt {- \frac {a}{c^{5}}} + x^{2} \right )}}{8} - \frac {3 \sqrt {- \frac {a}{c^{5}}} \log {\left (c^{2} \sqrt {- \frac {a}{c^{5}}} + x^{2} \right )}}{8} + \frac {x^{2}}{2 c^{2}} \] Input:
integrate(x**9/(c*x**4+a)**2,x)
Output:
a*x**2/(4*a*c**2 + 4*c**3*x**4) + 3*sqrt(-a/c**5)*log(-c**2*sqrt(-a/c**5) + x**2)/8 - 3*sqrt(-a/c**5)*log(c**2*sqrt(-a/c**5) + x**2)/8 + x**2/(2*c** 2)
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {a x^{2}}{4 \, {\left (c^{3} x^{4} + a c^{2}\right )}} + \frac {x^{2}}{2 \, c^{2}} - \frac {3 \, a \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} c^{2}} \] Input:
integrate(x^9/(c*x^4+a)^2,x, algorithm="maxima")
Output:
1/4*a*x^2/(c^3*x^4 + a*c^2) + 1/2*x^2/c^2 - 3/4*a*arctan(c*x^2/sqrt(a*c))/ (sqrt(a*c)*c^2)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {a x^{2}}{4 \, {\left (c x^{4} + a\right )} c^{2}} + \frac {x^{2}}{2 \, c^{2}} - \frac {3 \, a \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, \sqrt {a c} c^{2}} \] Input:
integrate(x^9/(c*x^4+a)^2,x, algorithm="giac")
Output:
1/4*a*x^2/((c*x^4 + a)*c^2) + 1/2*x^2/c^2 - 3/4*a*arctan(c*x^2/sqrt(a*c))/ (sqrt(a*c)*c^2)
Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {x^2}{2\,c^2}+\frac {a\,x^2}{4\,\left (c^3\,x^4+a\,c^2\right )}-\frac {3\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{4\,c^{5/2}} \] Input:
int(x^9/(a + c*x^4)^2,x)
Output:
x^2/(2*c^2) + (a*x^2)/(4*(a*c^2 + c^3*x^4)) - (3*a^(1/2)*atan((c^(1/2)*x^2 )/a^(1/2)))/(4*c^(5/2))
Time = 0.18 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.87 \[ \int \frac {x^9}{\left (a+c x^4\right )^2} \, dx=\frac {3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a +3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c \,x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) a +3 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c \,x^{4}+3 a c \,x^{2}+2 c^{2} x^{6}}{4 c^{3} \left (c \,x^{4}+a \right )} \] Input:
int(x^9/(c*x^4+a)^2,x)
Output:
(3*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4 )*a**(1/4)*sqrt(2)))*a + 3*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c*x**4 + 3*sqrt(c)*sqrt(a)*at an((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))* a + 3*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**( 1/4)*a**(1/4)*sqrt(2)))*c*x**4 + 3*a*c*x**2 + 2*c**2*x**6)/(4*c**3*(a + c* x**4))