Integrand size = 17, antiderivative size = 85 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {3 b x}{c^4}+\frac {x^3}{3 c^3}+\frac {b^3 x}{4 c^4 \left (b+c x^2\right )^2}-\frac {13 b^2 x}{8 c^4 \left (b+c x^2\right )}+\frac {35 b^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{9/2}} \] Output:
-3*b*x/c^4+1/3*x^3/c^3+1/4*b^3*x/c^4/(c*x^2+b)^2-13/8*b^2*x/c^4/(c*x^2+b)+ 35/8*b^(3/2)*arctan(c^(1/2)*x/b^(1/2))/c^(9/2)
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {105 b^3 x+175 b^2 c x^3+56 b c^2 x^5-8 c^3 x^7}{24 c^4 \left (b+c x^2\right )^2}+\frac {35 b^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{9/2}} \] Input:
Integrate[x^14/(b*x^2 + c*x^4)^3,x]
Output:
-1/24*(105*b^3*x + 175*b^2*c*x^3 + 56*b*c^2*x^5 - 8*c^3*x^7)/(c^4*(b + c*x ^2)^2) + (35*b^(3/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*c^(9/2))
Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13, 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, 254, 2009}
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.
\(\displaystyle \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^8}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {7 \int \frac {x^6}{\left (c x^2+b\right )^2}dx}{4 c}-\frac {x^7}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {7 \left (\frac {5 \int \frac {x^4}{c x^2+b}dx}{2 c}-\frac {x^5}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^7}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {7 \left (\frac {5 \int \left (\frac {b^2}{c^2 \left (c x^2+b\right )}-\frac {b}{c^2}+\frac {x^2}{c}\right )dx}{2 c}-\frac {x^5}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^7}{4 c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{5/2}}-\frac {b x}{c^2}+\frac {x^3}{3 c}\right )}{2 c}-\frac {x^5}{2 c \left (b+c x^2\right )}\right )}{4 c}-\frac {x^7}{4 c \left (b+c x^2\right )^2}\) |
Input:
Int[x^14/(b*x^2 + c*x^4)^3,x]
Output:
-1/4*x^7/(c*(b + c*x^2)^2) + (7*(-1/2*x^5/(c*(b + c*x^2)) + (5*(-((b*x)/c^ 2) + x^3/(3*c) + (b^(3/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(5/2)))/(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[((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[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {-\frac {1}{3} c \,x^{3}+3 b x}{c^{4}}+\frac {b^{2} \left (\frac {-\frac {13}{8} c \,x^{3}-\frac {11}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {35 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{4}}\) | \(63\) |
risch | \(\frac {x^{3}}{3 c^{3}}-\frac {3 b x}{c^{4}}+\frac {-\frac {13}{8} b^{2} c \,x^{3}-\frac {11}{8} b^{3} x}{c^{4} \left (c \,x^{2}+b \right )^{2}}+\frac {35 \sqrt {-b c}\, b \ln \left (-\sqrt {-b c}\, x +b \right )}{16 c^{5}}-\frac {35 \sqrt {-b c}\, b \ln \left (\sqrt {-b c}\, x +b \right )}{16 c^{5}}\) | \(93\) |
Input:
int(x^14/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
-1/c^4*(-1/3*c*x^3+3*b*x)+b^2/c^4*((-13/8*c*x^3-11/8*b*x)/(c*x^2+b)^2+35/8 /(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.71 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {16 \, c^{3} x^{7} - 112 \, b c^{2} x^{5} - 350 \, b^{2} c x^{3} - 210 \, b^{3} x + 105 \, {\left (b c^{2} x^{4} + 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right )}{48 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}, \frac {8 \, c^{3} x^{7} - 56 \, b c^{2} x^{5} - 175 \, b^{2} c x^{3} - 105 \, b^{3} x + 105 \, {\left (b c^{2} x^{4} + 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right )}{24 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}\right ] \] Input:
integrate(x^14/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
[1/48*(16*c^3*x^7 - 112*b*c^2*x^5 - 350*b^2*c*x^3 - 210*b^3*x + 105*(b*c^2 *x^4 + 2*b^2*c*x^2 + b^3)*sqrt(-b/c)*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)/(c *x^2 + b)))/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4), 1/24*(8*c^3*x^7 - 56*b*c^2* x^5 - 175*b^2*c*x^3 - 105*b^3*x + 105*(b*c^2*x^4 + 2*b^2*c*x^2 + b^3)*sqrt (b/c)*arctan(c*x*sqrt(b/c)/b))/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4)]
Time = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.56 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {3 b x}{c^{4}} - \frac {35 \sqrt {- \frac {b^{3}}{c^{9}}} \log {\left (x - \frac {c^{4} \sqrt {- \frac {b^{3}}{c^{9}}}}{b} \right )}}{16} + \frac {35 \sqrt {- \frac {b^{3}}{c^{9}}} \log {\left (x + \frac {c^{4} \sqrt {- \frac {b^{3}}{c^{9}}}}{b} \right )}}{16} + \frac {- 11 b^{3} x - 13 b^{2} c x^{3}}{8 b^{2} c^{4} + 16 b c^{5} x^{2} + 8 c^{6} x^{4}} + \frac {x^{3}}{3 c^{3}} \] Input:
integrate(x**14/(c*x**4+b*x**2)**3,x)
Output:
-3*b*x/c**4 - 35*sqrt(-b**3/c**9)*log(x - c**4*sqrt(-b**3/c**9)/b)/16 + 35 *sqrt(-b**3/c**9)*log(x + c**4*sqrt(-b**3/c**9)/b)/16 + (-11*b**3*x - 13*b **2*c*x**3)/(8*b**2*c**4 + 16*b*c**5*x**2 + 8*c**6*x**4) + x**3/(3*c**3)
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {13 \, b^{2} c x^{3} + 11 \, b^{3} x}{8 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{4}} + \frac {c x^{3} - 9 \, b x}{3 \, c^{4}} \] Input:
integrate(x^14/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
-1/8*(13*b^2*c*x^3 + 11*b^3*x)/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4) + 35/8*b^ 2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/3*(c*x^3 - 9*b*x)/c^4
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{4}} - \frac {13 \, b^{2} c x^{3} + 11 \, b^{3} x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{4}} + \frac {c^{6} x^{3} - 9 \, b c^{5} x}{3 \, c^{9}} \] Input:
integrate(x^14/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
35/8*b^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) - 1/8*(13*b^2*c*x^3 + 11*b^ 3*x)/((c*x^2 + b)^2*c^4) + 1/3*(c^6*x^3 - 9*b*c^5*x)/c^9
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x^3}{3\,c^3}-\frac {\frac {11\,b^3\,x}{8}+\frac {13\,c\,b^2\,x^3}{8}}{b^2\,c^4+2\,b\,c^5\,x^2+c^6\,x^4}+\frac {35\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,c^{9/2}}-\frac {3\,b\,x}{c^4} \] Input:
int(x^14/(b*x^2 + c*x^4)^3,x)
Output:
x^3/(3*c^3) - ((11*b^3*x)/8 + (13*b^2*c*x^3)/8)/(b^2*c^4 + c^6*x^4 + 2*b*c ^5*x^2) + (35*b^(3/2)*atan((c^(1/2)*x)/b^(1/2)))/(8*c^(9/2)) - (3*b*x)/c^4
Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.55 \[ \int \frac {x^{14}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3}+210 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c \,x^{2}+105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b \,c^{2} x^{4}-105 b^{3} c x -175 b^{2} c^{2} x^{3}-56 b \,c^{3} x^{5}+8 c^{4} x^{7}}{24 c^{5} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^14/(c*x^4+b*x^2)^3,x)
Output:
(105*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**3 + 210*sqrt(c)*sqrt (b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c*x**2 + 105*sqrt(c)*sqrt(b)*atan(( c*x)/(sqrt(c)*sqrt(b)))*b*c**2*x**4 - 105*b**3*c*x - 175*b**2*c**2*x**3 - 56*b*c**3*x**5 + 8*c**4*x**7)/(24*c**5*(b**2 + 2*b*c*x**2 + c**2*x**4))