Integrand size = 18, antiderivative size = 85 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\frac {3}{2} a^2 b x^2+\frac {3}{4} a \left (b^2+a c\right ) x^4+\frac {1}{6} b \left (b^2+6 a c\right ) x^6+\frac {3}{8} c \left (b^2+a c\right ) x^8+\frac {3}{10} b c^2 x^{10}+\frac {c^3 x^{12}}{12}+a^3 \log (x) \] Output:
3/2*a^2*b*x^2+3/4*a*(a*c+b^2)*x^4+1/6*b*(6*a*c+b^2)*x^6+3/8*c*(a*c+b^2)*x^ 8+3/10*b*c^2*x^10+1/12*c^3*x^12+a^3*ln(x)
Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\frac {3}{2} a^2 b x^2+\frac {3}{4} a \left (b^2+a c\right ) x^4+\frac {1}{6} b \left (b^2+6 a c\right ) x^6+\frac {3}{8} c \left (b^2+a c\right ) x^8+\frac {3}{10} b c^2 x^{10}+\frac {c^3 x^{12}}{12}+a^3 \log (x) \] Input:
Integrate[(a + b*x^2 + c*x^4)^3/x,x]
Output:
(3*a^2*b*x^2)/2 + (3*a*(b^2 + a*c)*x^4)/4 + (b*(b^2 + 6*a*c)*x^6)/6 + (3*c *(b^2 + a*c)*x^8)/8 + (3*b*c^2*x^10)/10 + (c^3*x^12)/12 + a^3*Log[x]
Time = 0.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {\left (c x^4+b x^2+a\right )^3}{x^2}dx^2\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \frac {1}{2} \int \left (c^3 x^{10}+3 b c^2 x^8+3 c \left (b^2+a c\right ) x^6+b \left (b^2+6 a c\right ) x^4+3 a \left (b^2+a c\right ) x^2+3 a^2 b+\frac {a^3}{x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (a^3 \log \left (x^2\right )+3 a^2 b x^2+\frac {3}{4} c x^8 \left (a c+b^2\right )+\frac {1}{3} b x^6 \left (6 a c+b^2\right )+\frac {3}{2} a x^4 \left (a c+b^2\right )+\frac {3}{5} b c^2 x^{10}+\frac {c^3 x^{12}}{6}\right )\) |
Input:
Int[(a + b*x^2 + c*x^4)^3/x,x]
Output:
(3*a^2*b*x^2 + (3*a*(b^2 + a*c)*x^4)/2 + (b*(b^2 + 6*a*c)*x^6)/3 + (3*c*(b ^2 + a*c)*x^8)/4 + (3*b*c^2*x^10)/5 + (c^3*x^12)/6 + a^3*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 = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\left (\frac {3}{4} c \,a^{2}+\frac {3}{4} b^{2} a \right ) x^{4}+\left (\frac {3}{8} c^{2} a +\frac {3}{8} b^{2} c \right ) x^{8}+\left (a b c +\frac {1}{6} b^{3}\right ) x^{6}+\frac {c^{3} x^{12}}{12}+\frac {3 a^{2} b \,x^{2}}{2}+\frac {3 b \,c^{2} x^{10}}{10}+a^{3} \ln \left (x \right )\) | \(82\) |
default | \(\frac {c^{3} x^{12}}{12}+\frac {3 b \,c^{2} x^{10}}{10}+\frac {3 a \,c^{2} x^{8}}{8}+\frac {3 b^{2} c \,x^{8}}{8}+a b c \,x^{6}+\frac {b^{3} x^{6}}{6}+\frac {3 a^{2} c \,x^{4}}{4}+\frac {3 b^{2} x^{4} a}{4}+\frac {3 a^{2} b \,x^{2}}{2}+a^{3} \ln \left (x \right )\) | \(85\) |
parallelrisch | \(\frac {c^{3} x^{12}}{12}+\frac {3 b \,c^{2} x^{10}}{10}+\frac {3 a \,c^{2} x^{8}}{8}+\frac {3 b^{2} c \,x^{8}}{8}+a b c \,x^{6}+\frac {b^{3} x^{6}}{6}+\frac {3 a^{2} c \,x^{4}}{4}+\frac {3 b^{2} x^{4} a}{4}+\frac {3 a^{2} b \,x^{2}}{2}+a^{3} \ln \left (x \right )\) | \(85\) |
risch | \(\frac {3 a^{2} b \,x^{2}}{2}+a b c \,x^{6}+\frac {3 b^{2} x^{4} a}{4}+\frac {3 b \,c^{2} x^{10}}{10}+\frac {c^{3} x^{12}}{12}+\frac {3 b^{2} c \,x^{8}}{8}+\frac {3 a \,c^{2} x^{8}}{8}-\frac {a \,b^{4}}{8 c^{2}}+\frac {b^{3} x^{6}}{6}+\frac {b^{6}}{120 c^{3}}+\frac {3 a^{2} c \,x^{4}}{4}+\frac {3 a^{2} b^{2}}{4 c}+a^{3} \ln \left (x \right )\) | \(113\) |
Input:
int((c*x^4+b*x^2+a)^3/x,x,method=_RETURNVERBOSE)
Output:
(3/4*c*a^2+3/4*b^2*a)*x^4+(3/8*c^2*a+3/8*b^2*c)*x^8+(a*b*c+1/6*b^3)*x^6+1/ 12*c^3*x^12+3/2*a^2*b*x^2+3/10*b*c^2*x^10+a^3*ln(x)
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\frac {1}{12} \, c^{3} x^{12} + \frac {3}{10} \, b c^{2} x^{10} + \frac {3}{8} \, {\left (b^{2} c + a c^{2}\right )} x^{8} + \frac {1}{6} \, {\left (b^{3} + 6 \, a b c\right )} x^{6} + \frac {3}{2} \, a^{2} b x^{2} + \frac {3}{4} \, {\left (a b^{2} + a^{2} c\right )} x^{4} + a^{3} \log \left (x\right ) \] Input:
integrate((c*x^4+b*x^2+a)^3/x,x, algorithm="fricas")
Output:
1/12*c^3*x^12 + 3/10*b*c^2*x^10 + 3/8*(b^2*c + a*c^2)*x^8 + 1/6*(b^3 + 6*a *b*c)*x^6 + 3/2*a^2*b*x^2 + 3/4*(a*b^2 + a^2*c)*x^4 + a^3*log(x)
Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=a^{3} \log {\left (x \right )} + \frac {3 a^{2} b x^{2}}{2} + \frac {3 b c^{2} x^{10}}{10} + \frac {c^{3} x^{12}}{12} + x^{8} \cdot \left (\frac {3 a c^{2}}{8} + \frac {3 b^{2} c}{8}\right ) + x^{6} \left (a b c + \frac {b^{3}}{6}\right ) + x^{4} \cdot \left (\frac {3 a^{2} c}{4} + \frac {3 a b^{2}}{4}\right ) \] Input:
integrate((c*x**4+b*x**2+a)**3/x,x)
Output:
a**3*log(x) + 3*a**2*b*x**2/2 + 3*b*c**2*x**10/10 + c**3*x**12/12 + x**8*( 3*a*c**2/8 + 3*b**2*c/8) + x**6*(a*b*c + b**3/6) + x**4*(3*a**2*c/4 + 3*a* b**2/4)
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\frac {1}{12} \, c^{3} x^{12} + \frac {3}{10} \, b c^{2} x^{10} + \frac {3}{8} \, {\left (b^{2} c + a c^{2}\right )} x^{8} + \frac {1}{6} \, {\left (b^{3} + 6 \, a b c\right )} x^{6} + \frac {3}{2} \, a^{2} b x^{2} + \frac {3}{4} \, {\left (a b^{2} + a^{2} c\right )} x^{4} + \frac {1}{2} \, a^{3} \log \left (x^{2}\right ) \] Input:
integrate((c*x^4+b*x^2+a)^3/x,x, algorithm="maxima")
Output:
1/12*c^3*x^12 + 3/10*b*c^2*x^10 + 3/8*(b^2*c + a*c^2)*x^8 + 1/6*(b^3 + 6*a *b*c)*x^6 + 3/2*a^2*b*x^2 + 3/4*(a*b^2 + a^2*c)*x^4 + 1/2*a^3*log(x^2)
Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\frac {1}{12} \, c^{3} x^{12} + \frac {3}{10} \, b c^{2} x^{10} + \frac {3}{8} \, b^{2} c x^{8} + \frac {3}{8} \, a c^{2} x^{8} + \frac {1}{6} \, b^{3} x^{6} + a b c x^{6} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{4} \, a^{2} c x^{4} + \frac {3}{2} \, a^{2} b x^{2} + \frac {1}{2} \, a^{3} \log \left (x^{2}\right ) \] Input:
integrate((c*x^4+b*x^2+a)^3/x,x, algorithm="giac")
Output:
1/12*c^3*x^12 + 3/10*b*c^2*x^10 + 3/8*b^2*c*x^8 + 3/8*a*c^2*x^8 + 1/6*b^3* x^6 + a*b*c*x^6 + 3/4*a*b^2*x^4 + 3/4*a^2*c*x^4 + 3/2*a^2*b*x^2 + 1/2*a^3* log(x^2)
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=a^3\,\ln \left (x\right )+x^6\,\left (\frac {b^3}{6}+a\,c\,b\right )+\frac {c^3\,x^{12}}{12}+\frac {3\,a^2\,b\,x^2}{2}+\frac {3\,b\,c^2\,x^{10}}{10}+\frac {3\,a\,x^4\,\left (b^2+a\,c\right )}{4}+\frac {3\,c\,x^8\,\left (b^2+a\,c\right )}{8} \] Input:
int((a + b*x^2 + c*x^4)^3/x,x)
Output:
a^3*log(x) + x^6*(b^3/6 + a*b*c) + (c^3*x^12)/12 + (3*a^2*b*x^2)/2 + (3*b* c^2*x^10)/10 + (3*a*x^4*(a*c + b^2))/4 + (3*c*x^8*(a*c + b^2))/8
Time = 0.17 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2+c x^4\right )^3}{x} \, dx=\mathrm {log}\left (x \right ) a^{3}+\frac {3 a^{2} b \,x^{2}}{2}+\frac {3 a^{2} c \,x^{4}}{4}+\frac {3 a \,b^{2} x^{4}}{4}+a b c \,x^{6}+\frac {3 a \,c^{2} x^{8}}{8}+\frac {b^{3} x^{6}}{6}+\frac {3 b^{2} c \,x^{8}}{8}+\frac {3 b \,c^{2} x^{10}}{10}+\frac {c^{3} x^{12}}{12} \] Input:
int((c*x^4+b*x^2+a)^3/x,x)
Output:
(120*log(x)*a**3 + 180*a**2*b*x**2 + 90*a**2*c*x**4 + 90*a*b**2*x**4 + 120 *a*b*c*x**6 + 45*a*c**2*x**8 + 20*b**3*x**6 + 45*b**2*c*x**8 + 36*b*c**2*x **10 + 10*c**3*x**12)/120