Integrand size = 20, antiderivative size = 142 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=-\frac {a^5 d}{2 x^2}+\frac {5}{2} a^4 c d x^2+\frac {5}{4} a^4 c e x^4+\frac {5}{3} a^3 c^2 d x^6+\frac {5}{4} a^3 c^2 e x^8+a^2 c^3 d x^{10}+\frac {5}{6} a^2 c^3 e x^{12}+\frac {5}{14} a c^4 d x^{14}+\frac {5}{16} a c^4 e x^{16}+\frac {1}{18} c^5 d x^{18}+\frac {1}{20} c^5 e x^{20}+a^5 e \log (x) \]
-1/2*a^5*d/x^2+5/2*a^4*c*d*x^2+5/4*a^4*c*e*x^4+5/3*a^3*c^2*d*x^6+5/4*a^3*c ^2*e*x^8+a^2*c^3*d*x^10+5/6*a^2*c^3*e*x^12+5/14*a*c^4*d*x^14+5/16*a*c^4*e* x^16+1/18*c^5*d*x^18+1/20*c^5*e*x^20+a^5*e*ln(x)
Time = 0.01 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=-\frac {a^5 d}{2 x^2}+\frac {5}{2} a^4 c d x^2+\frac {5}{4} a^4 c e x^4+\frac {5}{3} a^3 c^2 d x^6+\frac {5}{4} a^3 c^2 e x^8+a^2 c^3 d x^{10}+\frac {5}{6} a^2 c^3 e x^{12}+\frac {5}{14} a c^4 d x^{14}+\frac {5}{16} a c^4 e x^{16}+\frac {1}{18} c^5 d x^{18}+\frac {1}{20} c^5 e x^{20}+a^5 e \log (x) \]
-1/2*(a^5*d)/x^2 + (5*a^4*c*d*x^2)/2 + (5*a^4*c*e*x^4)/4 + (5*a^3*c^2*d*x^ 6)/3 + (5*a^3*c^2*e*x^8)/4 + a^2*c^3*d*x^10 + (5*a^2*c^3*e*x^12)/6 + (5*a* c^4*d*x^14)/14 + (5*a*c^4*e*x^16)/16 + (c^5*d*x^18)/18 + (c^5*e*x^20)/20 + a^5*e*Log[x]
Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1579, 522, 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.
\(\displaystyle \int \frac {\left (a+c x^4\right )^5 \left (d+e x^2\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (e x^2+d\right ) \left (c x^4+a\right )^5}{x^4}dx^2\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {1}{2} \int \left (c^5 e x^{18}+c^5 d x^{16}+5 a c^4 e x^{14}+5 a c^4 d x^{12}+10 a^2 c^3 e x^{10}+10 a^2 c^3 d x^8+10 a^3 c^2 e x^6+10 a^3 c^2 d x^4+5 a^4 c e x^2+5 a^4 c d+\frac {a^5 e}{x^2}+\frac {a^5 d}{x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^5 d}{x^2}+a^5 e \log \left (x^2\right )+5 a^4 c d x^2+\frac {5}{2} a^4 c e x^4+\frac {10}{3} a^3 c^2 d x^6+\frac {5}{2} a^3 c^2 e x^8+2 a^2 c^3 d x^{10}+\frac {5}{3} a^2 c^3 e x^{12}+\frac {5}{7} a c^4 d x^{14}+\frac {5}{8} a c^4 e x^{16}+\frac {1}{9} c^5 d x^{18}+\frac {1}{10} c^5 e x^{20}\right )\) |
(-((a^5*d)/x^2) + 5*a^4*c*d*x^2 + (5*a^4*c*e*x^4)/2 + (10*a^3*c^2*d*x^6)/3 + (5*a^3*c^2*e*x^8)/2 + 2*a^2*c^3*d*x^10 + (5*a^2*c^3*e*x^12)/3 + (5*a*c^ 4*d*x^14)/7 + (5*a*c^4*e*x^16)/8 + (c^5*d*x^18)/9 + (c^5*e*x^20)/10 + a^5* e*Log[x^2])/2
3.1.7.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {a^{5} d}{2 x^{2}}+\frac {5 a^{4} c d \,x^{2}}{2}+\frac {5 a^{4} c e \,x^{4}}{4}+\frac {5 a^{3} c^{2} d \,x^{6}}{3}+\frac {5 a^{3} c^{2} e \,x^{8}}{4}+a^{2} c^{3} d \,x^{10}+\frac {5 a^{2} c^{3} e \,x^{12}}{6}+\frac {5 a \,c^{4} d \,x^{14}}{14}+\frac {5 a \,c^{4} e \,x^{16}}{16}+\frac {c^{5} d \,x^{18}}{18}+\frac {c^{5} e \,x^{20}}{20}+a^{5} e \ln \left (x \right )\) | \(123\) |
risch | \(-\frac {a^{5} d}{2 x^{2}}+\frac {5 a^{4} c d \,x^{2}}{2}+\frac {5 a^{4} c e \,x^{4}}{4}+\frac {5 a^{3} c^{2} d \,x^{6}}{3}+\frac {5 a^{3} c^{2} e \,x^{8}}{4}+a^{2} c^{3} d \,x^{10}+\frac {5 a^{2} c^{3} e \,x^{12}}{6}+\frac {5 a \,c^{4} d \,x^{14}}{14}+\frac {5 a \,c^{4} e \,x^{16}}{16}+\frac {c^{5} d \,x^{18}}{18}+\frac {c^{5} e \,x^{20}}{20}+a^{5} e \ln \left (x \right )\) | \(123\) |
norman | \(\frac {a^{2} c^{3} d \,x^{12}-\frac {1}{2} d \,a^{5}+\frac {1}{18} c^{5} d \,x^{20}+\frac {1}{20} c^{5} e \,x^{22}+\frac {5}{14} a \,c^{4} d \,x^{16}+\frac {5}{16} a \,c^{4} e \,x^{18}+\frac {5}{6} a^{2} c^{3} e \,x^{14}+\frac {5}{3} a^{3} c^{2} d \,x^{8}+\frac {5}{4} a^{3} c^{2} e \,x^{10}+\frac {5}{2} a^{4} c d \,x^{4}+\frac {5}{4} a^{4} c e \,x^{6}}{x^{2}}+a^{5} e \ln \left (x \right )\) | \(125\) |
parallelrisch | \(\frac {252 c^{5} e \,x^{22}+280 c^{5} d \,x^{20}+1575 a \,c^{4} e \,x^{18}+1800 a \,c^{4} d \,x^{16}+4200 a^{2} c^{3} e \,x^{14}+5040 a^{2} c^{3} d \,x^{12}+6300 a^{3} c^{2} e \,x^{10}+8400 a^{3} c^{2} d \,x^{8}+6300 a^{4} c e \,x^{6}+12600 a^{4} c d \,x^{4}+5040 e \,a^{5} \ln \left (x \right ) x^{2}-2520 d \,a^{5}}{5040 x^{2}}\) | \(130\) |
-1/2*a^5*d/x^2+5/2*a^4*c*d*x^2+5/4*a^4*c*e*x^4+5/3*a^3*c^2*d*x^6+5/4*a^3*c ^2*e*x^8+a^2*c^3*d*x^10+5/6*a^2*c^3*e*x^12+5/14*a*c^4*d*x^14+5/16*a*c^4*e* x^16+1/18*c^5*d*x^18+1/20*c^5*e*x^20+a^5*e*ln(x)
Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=\frac {252 \, c^{5} e x^{22} + 280 \, c^{5} d x^{20} + 1575 \, a c^{4} e x^{18} + 1800 \, a c^{4} d x^{16} + 4200 \, a^{2} c^{3} e x^{14} + 5040 \, a^{2} c^{3} d x^{12} + 6300 \, a^{3} c^{2} e x^{10} + 8400 \, a^{3} c^{2} d x^{8} + 6300 \, a^{4} c e x^{6} + 12600 \, a^{4} c d x^{4} + 5040 \, a^{5} e x^{2} \log \left (x\right ) - 2520 \, a^{5} d}{5040 \, x^{2}} \]
1/5040*(252*c^5*e*x^22 + 280*c^5*d*x^20 + 1575*a*c^4*e*x^18 + 1800*a*c^4*d *x^16 + 4200*a^2*c^3*e*x^14 + 5040*a^2*c^3*d*x^12 + 6300*a^3*c^2*e*x^10 + 8400*a^3*c^2*d*x^8 + 6300*a^4*c*e*x^6 + 12600*a^4*c*d*x^4 + 5040*a^5*e*x^2 *log(x) - 2520*a^5*d)/x^2
Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=- \frac {a^{5} d}{2 x^{2}} + a^{5} e \log {\left (x \right )} + \frac {5 a^{4} c d x^{2}}{2} + \frac {5 a^{4} c e x^{4}}{4} + \frac {5 a^{3} c^{2} d x^{6}}{3} + \frac {5 a^{3} c^{2} e x^{8}}{4} + a^{2} c^{3} d x^{10} + \frac {5 a^{2} c^{3} e x^{12}}{6} + \frac {5 a c^{4} d x^{14}}{14} + \frac {5 a c^{4} e x^{16}}{16} + \frac {c^{5} d x^{18}}{18} + \frac {c^{5} e x^{20}}{20} \]
-a**5*d/(2*x**2) + a**5*e*log(x) + 5*a**4*c*d*x**2/2 + 5*a**4*c*e*x**4/4 + 5*a**3*c**2*d*x**6/3 + 5*a**3*c**2*e*x**8/4 + a**2*c**3*d*x**10 + 5*a**2* c**3*e*x**12/6 + 5*a*c**4*d*x**14/14 + 5*a*c**4*e*x**16/16 + c**5*d*x**18/ 18 + c**5*e*x**20/20
Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=\frac {1}{20} \, c^{5} e x^{20} + \frac {1}{18} \, c^{5} d x^{18} + \frac {5}{16} \, a c^{4} e x^{16} + \frac {5}{14} \, a c^{4} d x^{14} + \frac {5}{6} \, a^{2} c^{3} e x^{12} + a^{2} c^{3} d x^{10} + \frac {5}{4} \, a^{3} c^{2} e x^{8} + \frac {5}{3} \, a^{3} c^{2} d x^{6} + \frac {5}{4} \, a^{4} c e x^{4} + \frac {5}{2} \, a^{4} c d x^{2} + \frac {1}{2} \, a^{5} e \log \left (x^{2}\right ) - \frac {a^{5} d}{2 \, x^{2}} \]
1/20*c^5*e*x^20 + 1/18*c^5*d*x^18 + 5/16*a*c^4*e*x^16 + 5/14*a*c^4*d*x^14 + 5/6*a^2*c^3*e*x^12 + a^2*c^3*d*x^10 + 5/4*a^3*c^2*e*x^8 + 5/3*a^3*c^2*d* x^6 + 5/4*a^4*c*e*x^4 + 5/2*a^4*c*d*x^2 + 1/2*a^5*e*log(x^2) - 1/2*a^5*d/x ^2
Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=\frac {1}{20} \, c^{5} e x^{20} + \frac {1}{18} \, c^{5} d x^{18} + \frac {5}{16} \, a c^{4} e x^{16} + \frac {5}{14} \, a c^{4} d x^{14} + \frac {5}{6} \, a^{2} c^{3} e x^{12} + a^{2} c^{3} d x^{10} + \frac {5}{4} \, a^{3} c^{2} e x^{8} + \frac {5}{3} \, a^{3} c^{2} d x^{6} + \frac {5}{4} \, a^{4} c e x^{4} + \frac {5}{2} \, a^{4} c d x^{2} + \frac {1}{2} \, a^{5} e \log \left (x^{2}\right ) - \frac {a^{5} e x^{2} + a^{5} d}{2 \, x^{2}} \]
1/20*c^5*e*x^20 + 1/18*c^5*d*x^18 + 5/16*a*c^4*e*x^16 + 5/14*a*c^4*d*x^14 + 5/6*a^2*c^3*e*x^12 + a^2*c^3*d*x^10 + 5/4*a^3*c^2*e*x^8 + 5/3*a^3*c^2*d* x^6 + 5/4*a^4*c*e*x^4 + 5/2*a^4*c*d*x^2 + 1/2*a^5*e*log(x^2) - 1/2*(a^5*e* x^2 + a^5*d)/x^2
Time = 0.06 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.86 \[ \int \frac {\left (d+e x^2\right ) \left (a+c x^4\right )^5}{x^3} \, dx=\frac {c^5\,d\,x^{18}}{18}-\frac {a^5\,d}{2\,x^2}+\frac {c^5\,e\,x^{20}}{20}+a^5\,e\,\ln \left (x\right )+\frac {5\,a^3\,c^2\,d\,x^6}{3}+a^2\,c^3\,d\,x^{10}+\frac {5\,a^3\,c^2\,e\,x^8}{4}+\frac {5\,a^2\,c^3\,e\,x^{12}}{6}+\frac {5\,a^4\,c\,d\,x^2}{2}+\frac {5\,a\,c^4\,d\,x^{14}}{14}+\frac {5\,a^4\,c\,e\,x^4}{4}+\frac {5\,a\,c^4\,e\,x^{16}}{16} \]