Integrand size = 18, antiderivative size = 86 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=-\frac {a^3}{2 x^2}+\frac {3}{2} a \left (b^2+a c\right ) x^2+\frac {1}{4} b \left (b^2+6 a c\right ) x^4+\frac {1}{2} c \left (b^2+a c\right ) x^6+\frac {3}{8} b c^2 x^8+\frac {c^3 x^{10}}{10}+3 a^2 b \log (x) \] Output:
-1/2*a^3/x^2+3/2*a*(a*c+b^2)*x^2+1/4*b*(6*a*c+b^2)*x^4+1/2*c*(a*c+b^2)*x^6 +3/8*b*c^2*x^8+1/10*c^3*x^10+3*a^2*b*ln(x)
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{40} \left (-\frac {20 a^3}{x^2}+60 a \left (b^2+a c\right ) x^2+10 b \left (b^2+6 a c\right ) x^4+20 c \left (b^2+a c\right ) x^6+15 b c^2 x^8+4 c^3 x^{10}+120 a^2 b \log (x)\right ) \] Input:
Integrate[(a + b*x^2 + c*x^4)^3/x^3,x]
Output:
((-20*a^3)/x^2 + 60*a*(b^2 + a*c)*x^2 + 10*b*(b^2 + 6*a*c)*x^4 + 20*c*(b^2 + a*c)*x^6 + 15*b*c^2*x^8 + 4*c^3*x^10 + 120*a^2*b*Log[x])/40
Time = 0.42 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1434, 1140, 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.
\(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {\left (c x^4+b x^2+a\right )^3}{x^4}dx^2\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \frac {1}{2} \int \left (c^3 x^8+3 b c^2 x^6+3 c \left (b^2+a c\right ) x^4+b \left (b^2+6 a c\right ) x^2+3 a \left (b^2+a c\right )+\frac {3 a^2 b}{x^2}+\frac {a^3}{x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^3}{x^2}+3 a^2 b \log \left (x^2\right )+c x^6 \left (a c+b^2\right )+\frac {1}{2} b x^4 \left (6 a c+b^2\right )+3 a x^2 \left (a c+b^2\right )+\frac {3}{4} b c^2 x^8+\frac {c^3 x^{10}}{5}\right )\) |
Input:
Int[(a + b*x^2 + c*x^4)^3/x^3,x]
Output:
(-(a^3/x^2) + 3*a*(b^2 + a*c)*x^2 + (b*(b^2 + 6*a*c)*x^4)/2 + c*(b^2 + a*c )*x^6 + (3*b*c^2*x^8)/4 + (c^3*x^10)/5 + 3*a^2*b*Log[x^2])/2
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {\left (\frac {3}{2} c \,a^{2}+\frac {3}{2} b^{2} a \right ) x^{4}+\left (\frac {1}{2} c^{2} a +\frac {1}{2} b^{2} c \right ) x^{8}+\left (\frac {3}{2} a b c +\frac {1}{4} b^{3}\right ) x^{6}-\frac {a^{3}}{2}+\frac {c^{3} x^{12}}{10}+\frac {3 b \,c^{2} x^{10}}{8}}{x^{2}}+3 a^{2} b \ln \left (x \right )\) | \(86\) |
default | \(\frac {c^{3} x^{10}}{10}+\frac {3 b \,c^{2} x^{8}}{8}+\frac {a \,c^{2} x^{6}}{2}+\frac {b^{2} c \,x^{6}}{2}+\frac {3 a b c \,x^{4}}{2}+\frac {b^{3} x^{4}}{4}+\frac {3 a^{2} c \,x^{2}}{2}+\frac {3 a \,b^{2} x^{2}}{2}-\frac {a^{3}}{2 x^{2}}+3 a^{2} b \ln \left (x \right )\) | \(87\) |
risch | \(\frac {c^{3} x^{10}}{10}+\frac {3 b \,c^{2} x^{8}}{8}+\frac {a \,c^{2} x^{6}}{2}+\frac {b^{2} c \,x^{6}}{2}+\frac {3 a b c \,x^{4}}{2}+\frac {b^{3} x^{4}}{4}+\frac {3 a^{2} c \,x^{2}}{2}+\frac {3 a \,b^{2} x^{2}}{2}-\frac {a^{3}}{2 x^{2}}+3 a^{2} b \ln \left (x \right )\) | \(87\) |
parallelrisch | \(\frac {4 c^{3} x^{12}+15 b \,c^{2} x^{10}+20 a \,c^{2} x^{8}+20 b^{2} c \,x^{8}+60 a b c \,x^{6}+10 b^{3} x^{6}+60 a^{2} c \,x^{4}+60 b^{2} x^{4} a +120 \ln \left (x \right ) x^{2} a^{2} b -20 a^{3}}{40 x^{2}}\) | \(92\) |
Input:
int((c*x^4+b*x^2+a)^3/x^3,x,method=_RETURNVERBOSE)
Output:
((3/2*c*a^2+3/2*b^2*a)*x^4+(1/2*c^2*a+1/2*b^2*c)*x^8+(3/2*a*b*c+1/4*b^3)*x ^6-1/2*a^3+1/10*c^3*x^12+3/8*b*c^2*x^10)/x^2+3*a^2*b*ln(x)
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=\frac {4 \, c^{3} x^{12} + 15 \, b c^{2} x^{10} + 20 \, {\left (b^{2} c + a c^{2}\right )} x^{8} + 10 \, {\left (b^{3} + 6 \, a b c\right )} x^{6} + 120 \, a^{2} b x^{2} \log \left (x\right ) + 60 \, {\left (a b^{2} + a^{2} c\right )} x^{4} - 20 \, a^{3}}{40 \, x^{2}} \] Input:
integrate((c*x^4+b*x^2+a)^3/x^3,x, algorithm="fricas")
Output:
1/40*(4*c^3*x^12 + 15*b*c^2*x^10 + 20*(b^2*c + a*c^2)*x^8 + 10*(b^3 + 6*a* b*c)*x^6 + 120*a^2*b*x^2*log(x) + 60*(a*b^2 + a^2*c)*x^4 - 20*a^3)/x^2
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=- \frac {a^{3}}{2 x^{2}} + 3 a^{2} b \log {\left (x \right )} + \frac {3 b c^{2} x^{8}}{8} + \frac {c^{3} x^{10}}{10} + x^{6} \left (\frac {a c^{2}}{2} + \frac {b^{2} c}{2}\right ) + x^{4} \cdot \left (\frac {3 a b c}{2} + \frac {b^{3}}{4}\right ) + x^{2} \cdot \left (\frac {3 a^{2} c}{2} + \frac {3 a b^{2}}{2}\right ) \] Input:
integrate((c*x**4+b*x**2+a)**3/x**3,x)
Output:
-a**3/(2*x**2) + 3*a**2*b*log(x) + 3*b*c**2*x**8/8 + c**3*x**10/10 + x**6* (a*c**2/2 + b**2*c/2) + x**4*(3*a*b*c/2 + b**3/4) + x**2*(3*a**2*c/2 + 3*a *b**2/2)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{10} \, c^{3} x^{10} + \frac {3}{8} \, b c^{2} x^{8} + \frac {1}{2} \, {\left (b^{2} c + a c^{2}\right )} x^{6} + \frac {1}{4} \, {\left (b^{3} + 6 \, a b c\right )} x^{4} + \frac {3}{2} \, a^{2} b \log \left (x^{2}\right ) + \frac {3}{2} \, {\left (a b^{2} + a^{2} c\right )} x^{2} - \frac {a^{3}}{2 \, x^{2}} \] Input:
integrate((c*x^4+b*x^2+a)^3/x^3,x, algorithm="maxima")
Output:
1/10*c^3*x^10 + 3/8*b*c^2*x^8 + 1/2*(b^2*c + a*c^2)*x^6 + 1/4*(b^3 + 6*a*b *c)*x^4 + 3/2*a^2*b*log(x^2) + 3/2*(a*b^2 + a^2*c)*x^2 - 1/2*a^3/x^2
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{10} \, c^{3} x^{10} + \frac {3}{8} \, b c^{2} x^{8} + \frac {1}{2} \, b^{2} c x^{6} + \frac {1}{2} \, a c^{2} x^{6} + \frac {1}{4} \, b^{3} x^{4} + \frac {3}{2} \, a b c x^{4} + \frac {3}{2} \, a b^{2} x^{2} + \frac {3}{2} \, a^{2} c x^{2} + \frac {3}{2} \, a^{2} b \log \left (x^{2}\right ) - \frac {3 \, a^{2} b x^{2} + a^{3}}{2 \, x^{2}} \] Input:
integrate((c*x^4+b*x^2+a)^3/x^3,x, algorithm="giac")
Output:
1/10*c^3*x^10 + 3/8*b*c^2*x^8 + 1/2*b^2*c*x^6 + 1/2*a*c^2*x^6 + 1/4*b^3*x^ 4 + 3/2*a*b*c*x^4 + 3/2*a*b^2*x^2 + 3/2*a^2*c*x^2 + 3/2*a^2*b*log(x^2) - 1 /2*(3*a^2*b*x^2 + a^3)/x^2
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=x^4\,\left (\frac {b^3}{4}+\frac {3\,a\,c\,b}{2}\right )-\frac {a^3}{2\,x^2}+\frac {c^3\,x^{10}}{10}+\frac {3\,b\,c^2\,x^8}{8}+3\,a^2\,b\,\ln \left (x\right )+\frac {3\,a\,x^2\,\left (b^2+a\,c\right )}{2}+\frac {c\,x^6\,\left (b^2+a\,c\right )}{2} \] Input:
int((a + b*x^2 + c*x^4)^3/x^3,x)
Output:
x^4*(b^3/4 + (3*a*b*c)/2) - a^3/(2*x^2) + (c^3*x^10)/10 + (3*b*c^2*x^8)/8 + 3*a^2*b*log(x) + (3*a*x^2*(a*c + b^2))/2 + (c*x^6*(a*c + b^2))/2
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x^3} \, dx=\frac {120 \,\mathrm {log}\left (x \right ) a^{2} b \,x^{2}-20 a^{3}+60 a^{2} c \,x^{4}+60 a \,b^{2} x^{4}+60 a b c \,x^{6}+20 a \,c^{2} x^{8}+10 b^{3} x^{6}+20 b^{2} c \,x^{8}+15 b \,c^{2} x^{10}+4 c^{3} x^{12}}{40 x^{2}} \] Input:
int((c*x^4+b*x^2+a)^3/x^3,x)
Output:
(120*log(x)*a**2*b*x**2 - 20*a**3 + 60*a**2*c*x**4 + 60*a*b**2*x**4 + 60*a *b*c*x**6 + 20*a*c**2*x**8 + 10*b**3*x**6 + 20*b**2*c*x**8 + 15*b*c**2*x** 10 + 4*c**3*x**12)/(40*x**2)