Integrand size = 22, antiderivative size = 90 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {(b c+a d) x^4}{4 b^2 d^2}+\frac {x^8}{8 b d}-\frac {a^3 \log \left (a+b x^4\right )}{4 b^3 (b c-a d)}+\frac {c^3 \log \left (c+d x^4\right )}{4 d^3 (b c-a d)} \] Output:
-1/4*(a*d+b*c)*x^4/b^2/d^2+1/8*x^8/b/d-1/4*a^3*ln(b*x^4+a)/b^3/(-a*d+b*c)+ 1/4*c^3*ln(d*x^4+c)/d^3/(-a*d+b*c)
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {b d (b c-a d) x^4 \left (-2 b c-2 a d+b d x^4\right )-2 a^3 d^3 \log \left (a+b x^4\right )+2 b^3 c^3 \log \left (c+d x^4\right )}{8 b^3 d^3 (b c-a d)} \] Input:
Integrate[x^15/((a + b*x^4)*(c + d*x^4)),x]
Output:
(b*d*(b*c - a*d)*x^4*(-2*b*c - 2*a*d + b*d*x^4) - 2*a^3*d^3*Log[a + b*x^4] + 2*b^3*c^3*Log[c + d*x^4])/(8*b^3*d^3*(b*c - a*d))
Time = 0.44 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 93, 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.
\(\displaystyle \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {x^{12}}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^4\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \frac {1}{4} \int \left (\frac {x^4}{b d}+\frac {-b c-a d}{b^2 d^2}-\frac {a^3}{b^2 (b c-a d) \left (b x^4+a\right )}-\frac {c^3}{d^2 (a d-b c) \left (d x^4+c\right )}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {a^3 \log \left (a+b x^4\right )}{b^3 (b c-a d)}-\frac {x^4 (a d+b c)}{b^2 d^2}+\frac {c^3 \log \left (c+d x^4\right )}{d^3 (b c-a d)}+\frac {x^8}{2 b d}\right )\) |
Input:
Int[x^15/((a + b*x^4)*(c + d*x^4)),x]
Output:
(-(((b*c + a*d)*x^4)/(b^2*d^2)) + x^8/(2*b*d) - (a^3*Log[a + b*x^4])/(b^3* (b*c - a*d)) + (c^3*Log[c + d*x^4])/(d^3*(b*c - a*d)))/4
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\left (-d b \,x^{4}+a d +c b \right )^{2}}{8 b^{3} d^{3}}+\frac {a^{3} \ln \left (b \,x^{4}+a \right )}{4 b^{3} \left (a d -c b \right )}-\frac {c^{3} \ln \left (d \,x^{4}+c \right )}{4 d^{3} \left (a d -c b \right )}\) | \(78\) |
norman | \(\frac {x^{8}}{8 b d}-\frac {\left (a d +c b \right ) x^{4}}{4 b^{2} d^{2}}+\frac {a^{3} \ln \left (b \,x^{4}+a \right )}{4 b^{3} \left (a d -c b \right )}-\frac {c^{3} \ln \left (d \,x^{4}+c \right )}{4 d^{3} \left (a d -c b \right )}\) | \(83\) |
parallelrisch | \(\frac {x^{8} a \,b^{2} d^{3}-x^{8} b^{3} c \,d^{2}-2 x^{4} a^{2} b \,d^{3}+2 x^{4} b^{3} c^{2} d +2 a^{3} \ln \left (b \,x^{4}+a \right ) d^{3}-2 c^{3} \ln \left (d \,x^{4}+c \right ) b^{3}}{8 b^{3} d^{3} \left (a d -c b \right )}\) | \(99\) |
risch | \(\frac {x^{8}}{8 b d}-\frac {a \,x^{4}}{4 b^{2} d}-\frac {c \,x^{4}}{4 b \,d^{2}}+\frac {a^{2}}{8 b^{3} d}+\frac {a c}{4 b^{2} d^{2}}+\frac {c^{2}}{8 b \,d^{3}}+\frac {a^{3} \ln \left (-b \,x^{4}-a \right )}{4 b^{3} \left (a d -c b \right )}-\frac {c^{3} \ln \left (d \,x^{4}+c \right )}{4 d^{3} \left (a d -c b \right )}\) | \(124\) |
Input:
int(x^15/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
1/8*(-b*d*x^4+a*d+b*c)^2/b^3/d^3+1/4*a^3/b^3/(a*d-b*c)*ln(b*x^4+a)-1/4*c^3 /d^3/(a*d-b*c)*ln(d*x^4+c)
Time = 2.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {{\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{8} - 2 \, a^{3} d^{3} \log \left (b x^{4} + a\right ) + 2 \, b^{3} c^{3} \log \left (d x^{4} + c\right ) - 2 \, {\left (b^{3} c^{2} d - a^{2} b d^{3}\right )} x^{4}}{8 \, {\left (b^{4} c d^{3} - a b^{3} d^{4}\right )}} \] Input:
integrate(x^15/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
Output:
1/8*((b^3*c*d^2 - a*b^2*d^3)*x^8 - 2*a^3*d^3*log(b*x^4 + a) + 2*b^3*c^3*lo g(d*x^4 + c) - 2*(b^3*c^2*d - a^2*b*d^3)*x^4)/(b^4*c*d^3 - a*b^3*d^4)
Timed out. \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(x**15/(b*x**4+a)/(d*x**4+c),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {a^{3} \log \left (b x^{4} + a\right )}{4 \, {\left (b^{4} c - a b^{3} d\right )}} + \frac {c^{3} \log \left (d x^{4} + c\right )}{4 \, {\left (b c d^{3} - a d^{4}\right )}} + \frac {b d x^{8} - 2 \, {\left (b c + a d\right )} x^{4}}{8 \, b^{2} d^{2}} \] Input:
integrate(x^15/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
Output:
-1/4*a^3*log(b*x^4 + a)/(b^4*c - a*b^3*d) + 1/4*c^3*log(d*x^4 + c)/(b*c*d^ 3 - a*d^4) + 1/8*(b*d*x^8 - 2*(b*c + a*d)*x^4)/(b^2*d^2)
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {a^{3} \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, {\left (b^{4} c - a b^{3} d\right )}} + \frac {c^{3} \log \left ({\left | d x^{4} + c \right |}\right )}{4 \, {\left (b c d^{3} - a d^{4}\right )}} + \frac {b d x^{8} - 2 \, b c x^{4} - 2 \, a d x^{4}}{8 \, b^{2} d^{2}} \] Input:
integrate(x^15/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
Output:
-1/4*a^3*log(abs(b*x^4 + a))/(b^4*c - a*b^3*d) + 1/4*c^3*log(abs(d*x^4 + c ))/(b*c*d^3 - a*d^4) + 1/8*(b*d*x^8 - 2*b*c*x^4 - 2*a*d*x^4)/(b^2*d^2)
Time = 4.88 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {x^8}{8\,b\,d}-\frac {c^3\,\ln \left (d\,x^4+c\right )}{4\,\left (a\,d^4-b\,c\,d^3\right )}-\frac {a^3\,\ln \left (b\,x^4+a\right )}{4\,\left (b^4\,c-a\,b^3\,d\right )}-\frac {x^4\,\left (a\,d+b\,c\right )}{4\,b^2\,d^2} \] Input:
int(x^15/((a + b*x^4)*(c + d*x^4)),x)
Output:
x^8/(8*b*d) - (c^3*log(c + d*x^4))/(4*(a*d^4 - b*c*d^3)) - (a^3*log(a + b* x^4))/(4*(b^4*c - a*b^3*d)) - (x^4*(a*d + b*c))/(4*b^2*d^2)
Time = 0.25 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.00 \[ \int \frac {x^{15}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {2 \,\mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) a^{3} d^{3}-2 \,\mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) b^{3} c^{3}+2 \,\mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) a^{3} d^{3}-2 \,\mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) b^{3} c^{3}-2 a^{2} b \,d^{3} x^{4}+a \,b^{2} d^{3} x^{8}+2 b^{3} c^{2} d \,x^{4}-b^{3} c \,d^{2} x^{8}}{8 b^{3} d^{3} \left (a d -b c \right )} \] Input:
int(x^15/(b*x^4+a)/(d*x^4+c),x)
Output:
(2*log( - b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*a**3*d**3 - 2*log( - d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*b**3*c**3 + 2*log(b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*a**3*d**3 - 2*log(d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*b**3*c**3 - 2 *a**2*b*d**3*x**4 + a*b**2*d**3*x**8 + 2*b**3*c**2*d*x**4 - b**3*c*d**2*x* *8)/(8*b**3*d**3*(a*d - b*c))