Integrand size = 17, antiderivative size = 71 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x}{c^3}-\frac {b^2 x}{4 c^3 \left (b+c x^2\right )^2}+\frac {9 b x}{8 c^3 \left (b+c x^2\right )}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{7/2}} \] Output:
x/c^3-1/4*b^2*x/c^3/(c*x^2+b)^2+9/8*b*x/c^3/(c*x^2+b)-15/8*b^(1/2)*arctan( c^(1/2)*x/b^(1/2))/c^(7/2)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {15 b^2 x+25 b c x^3+8 c^2 x^5}{8 c^3 \left (b+c x^2\right )^2}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{7/2}} \] Input:
Integrate[x^12/(b*x^2 + c*x^4)^3,x]
Output:
(15*b^2*x + 25*b*c*x^3 + 8*c^2*x^5)/(8*c^3*(b + c*x^2)^2) - (15*Sqrt[b]*Ar cTan[(Sqrt[c]*x)/Sqrt[b]])/(8*c^(7/2))
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {9, 252, 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.
\(\displaystyle \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^6}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \int \frac {x^4}{\left (c x^2+b\right )^2}dx}{4 c}-\frac {x^5}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2}{c x^2+b}dx}{2 c}-\frac {x^3}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^5}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {x}{c}-\frac {b \int \frac {1}{c x^2+b}dx}{c}\right )}{2 c}-\frac {x^3}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^5}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {x}{c}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{3/2}}\right )}{2 c}-\frac {x^3}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^5}{4 c \left (b+c x^2\right )^2}\) |
Input:
Int[x^12/(b*x^2 + c*x^4)^3,x]
Output:
-1/4*x^5/(c*(b + c*x^2)^2) + (5*(-1/2*x^3/(c*(b + c*x^2)) + (3*(x/c - (Sqr t[b]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(3/2)))/(2*c)))/(4*c)
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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]
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {x}{c^{3}}-\frac {b \left (\frac {-\frac {9}{8} c \,x^{3}-\frac {7}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{3}}\) | \(51\) |
risch | \(\frac {x}{c^{3}}+\frac {\frac {9}{8} b c \,x^{3}+\frac {7}{8} b^{2} x}{c^{3} \left (c \,x^{2}+b \right )^{2}}+\frac {15 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right )}{16 c^{4}}-\frac {15 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right )}{16 c^{4}}\) | \(83\) |
Input:
int(x^12/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
x/c^3-b/c^3*((-9/8*c*x^3-7/8*b*x)/(c*x^2+b)^2+15/8/(b*c)^(1/2)*arctan(c*x/ (b*c)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.85 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {16 \, c^{2} x^{5} + 50 \, b c x^{3} + 30 \, b^{2} x + 15 \, {\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right )}{16 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}}, \frac {8 \, c^{2} x^{5} + 25 \, b c x^{3} + 15 \, b^{2} x - 15 \, {\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right )}{8 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}}\right ] \] Input:
integrate(x^12/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
[1/16*(16*c^2*x^5 + 50*b*c*x^3 + 30*b^2*x + 15*(c^2*x^4 + 2*b*c*x^2 + b^2) *sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)))/(c^5*x^4 + 2* b*c^4*x^2 + b^2*c^3), 1/8*(8*c^2*x^5 + 25*b*c*x^3 + 15*b^2*x - 15*(c^2*x^4 + 2*b*c*x^2 + b^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b))/(c^5*x^4 + 2*b*c^4* x^2 + b^2*c^3)]
Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.51 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {15 \sqrt {- \frac {b}{c^{7}}} \log {\left (- c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{16} - \frac {15 \sqrt {- \frac {b}{c^{7}}} \log {\left (c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{16} + \frac {7 b^{2} x + 9 b c x^{3}}{8 b^{2} c^{3} + 16 b c^{4} x^{2} + 8 c^{5} x^{4}} + \frac {x}{c^{3}} \] Input:
integrate(x**12/(c*x**4+b*x**2)**3,x)
Output:
15*sqrt(-b/c**7)*log(-c**3*sqrt(-b/c**7) + x)/16 - 15*sqrt(-b/c**7)*log(c* *3*sqrt(-b/c**7) + x)/16 + (7*b**2*x + 9*b*c*x**3)/(8*b**2*c**3 + 16*b*c** 4*x**2 + 8*c**5*x**4) + x/c**3
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {9 \, b c x^{3} + 7 \, b^{2} x}{8 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} - \frac {15 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} + \frac {x}{c^{3}} \] Input:
integrate(x^12/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
1/8*(9*b*c*x^3 + 7*b^2*x)/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3) - 15/8*b*arcta n(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + x/c^3
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {15 \, b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} + \frac {x}{c^{3}} + \frac {9 \, b c x^{3} + 7 \, b^{2} x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{3}} \] Input:
integrate(x^12/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
-15/8*b*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + x/c^3 + 1/8*(9*b*c*x^3 + 7 *b^2*x)/((c*x^2 + b)^2*c^3)
Time = 17.54 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {7\,b^2\,x}{8}+\frac {9\,c\,b\,x^3}{8}}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {x}{c^3}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,c^{7/2}} \] Input:
int(x^12/(b*x^2 + c*x^4)^3,x)
Output:
((7*b^2*x)/8 + (9*b*c*x^3)/8)/(b^2*c^3 + c^5*x^4 + 2*b*c^4*x^2) + x/c^3 - (15*b^(1/2)*atan((c^(1/2)*x)/b^(1/2)))/(8*c^(7/2))
Time = 0.17 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int \frac {x^{12}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2}-30 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b c \,x^{2}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) c^{2} x^{4}+15 b^{2} c x +25 b \,c^{2} x^{3}+8 c^{3} x^{5}}{8 c^{4} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^12/(c*x^4+b*x^2)^3,x)
Output:
( - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2 - 30*sqrt(c)*sqr t(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b*c*x**2 - 15*sqrt(c)*sqrt(b)*atan((c*x )/(sqrt(c)*sqrt(b)))*c**2*x**4 + 15*b**2*c*x + 25*b*c**2*x**3 + 8*c**3*x** 5)/(8*c**4*(b**2 + 2*b*c*x**2 + c**2*x**4))