Integrand size = 29, antiderivative size = 107 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {A}{4 a c x^4}+\frac {(a B c-A (b c+a d)) \log (x)}{a^2 c^2}+\frac {b (A b-a B) \log \left (a+b x^4\right )}{4 a^2 (b c-a d)}+\frac {d (B c-A d) \log \left (c+d x^4\right )}{4 c^2 (b c-a d)} \] Output:
-1/4*A/a/c/x^4+(a*B*c-A*(a*d+b*c))*ln(x)/a^2/c^2+1/4*b*(A*b-B*a)*ln(b*x^4+ a)/a^2/(-a*d+b*c)+1/4*d*(-A*d+B*c)*ln(d*x^4+c)/c^2/(-a*d+b*c)
Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {A}{4 a c x^4}+\frac {(-A b c+a B c-a A d) \log (x)}{a^2 c^2}+\frac {\left (-A b^2+a b B\right ) \log \left (a+b x^4\right )}{4 a^2 (-b c+a d)}-\frac {\left (-B c d+A d^2\right ) \log \left (c+d x^4\right )}{4 c^2 (b c-a d)} \] Input:
Integrate[(A + B*x^4)/(x^5*(a + b*x^4)*(c + d*x^4)),x]
Output:
-1/4*A/(a*c*x^4) + ((-(A*b*c) + a*B*c - a*A*d)*Log[x])/(a^2*c^2) + ((-(A*b ^2) + a*b*B)*Log[a + b*x^4])/(4*a^2*(-(b*c) + a*d)) - ((-(B*c*d) + A*d^2)* Log[c + d*x^4])/(4*c^2*(b*c - a*d))
Time = 0.52 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1043, 165, 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.
\(\displaystyle \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 1043 |
\(\displaystyle \frac {1}{4} \int \frac {B x^4+A}{x^8 \left (b x^4+a\right ) \left (d x^4+c\right )}dx^4\) |
\(\Big \downarrow \) 165 |
\(\displaystyle \frac {1}{4} \int \left (\frac {(a B-A b) b^2}{a^2 (a d-b c) \left (b x^4+a\right )}+\frac {d^2 (B c-A d)}{c^2 (b c-a d) \left (d x^4+c\right )}+\frac {a B c-A (b c+a d)}{a^2 c^2 x^4}+\frac {A}{a c x^8}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right ) (a B c-A (a d+b c))}{a^2 c^2}+\frac {b (A b-a B) \log \left (a+b x^4\right )}{a^2 (b c-a d)}+\frac {d (B c-A d) \log \left (c+d x^4\right )}{c^2 (b c-a d)}-\frac {A}{a c x^4}\right )\) |
Input:
Int[(A + B*x^4)/(x^5*(a + b*x^4)*(c + d*x^4)),x]
Output:
(-(A/(a*c*x^4)) + ((a*B*c - A*(b*c + a*d))*Log[x^4])/(a^2*c^2) + (b*(A*b - a*B)*Log[a + b*x^4])/(a^2*(b*c - a*d)) + (d*(B*c - A*d)*Log[c + d*x^4])/( c^2*(b*c - a*d)))/4
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d* x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. )*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simp lify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] / ; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/ n]]
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {b \left (A b -B a \right ) \ln \left (b \,x^{4}+a \right )}{4 a^{2} \left (a d -c b \right )}-\frac {A}{4 a c \,x^{4}}+\frac {\left (-A a d -A b c +a B c \right ) \ln \left (x \right )}{a^{2} c^{2}}+\frac {d \left (A d -B c \right ) \ln \left (d \,x^{4}+c \right )}{4 c^{2} \left (a d -c b \right )}\) | \(102\) |
norman | \(-\frac {A}{4 a c \,x^{4}}-\frac {\left (A a d +A b c -a B c \right ) \ln \left (x \right )}{a^{2} c^{2}}-\frac {b \left (A b -B a \right ) \ln \left (b \,x^{4}+a \right )}{4 a^{2} \left (a d -c b \right )}+\frac {d \left (A d -B c \right ) \ln \left (d \,x^{4}+c \right )}{4 c^{2} \left (a d -c b \right )}\) | \(102\) |
risch | \(-\frac {A}{4 a c \,x^{4}}-\frac {\ln \left (x \right ) A d}{a \,c^{2}}-\frac {\ln \left (x \right ) A b}{a^{2} c}+\frac {\ln \left (x \right ) B}{a c}+\frac {d^{2} \ln \left (d \,x^{4}+c \right ) A}{4 c^{2} \left (a d -c b \right )}-\frac {d \ln \left (d \,x^{4}+c \right ) B}{4 c \left (a d -c b \right )}-\frac {b^{2} \ln \left (-b \,x^{4}-a \right ) A}{4 \left (a d -c b \right ) a^{2}}+\frac {b \ln \left (-b \,x^{4}-a \right ) B}{4 \left (a d -c b \right ) a}\) | \(158\) |
parallelrisch | \(-\frac {4 A \ln \left (x \right ) x^{4} a^{2} d^{2}-4 A \ln \left (x \right ) x^{4} b^{2} c^{2}+A \ln \left (b \,x^{4}+a \right ) x^{4} b^{2} c^{2}-A \ln \left (d \,x^{4}+c \right ) x^{4} a^{2} d^{2}-4 B \ln \left (x \right ) x^{4} a^{2} c d +4 B \ln \left (x \right ) x^{4} a b \,c^{2}-B \ln \left (b \,x^{4}+a \right ) x^{4} a b \,c^{2}+B \ln \left (d \,x^{4}+c \right ) x^{4} a^{2} c d +A \,a^{2} c d -A a b \,c^{2}}{4 a^{2} c^{2} x^{4} \left (a d -c b \right )}\) | \(168\) |
Input:
int((B*x^4+A)/x^5/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
-1/4*b*(A*b-B*a)/a^2/(a*d-b*c)*ln(b*x^4+a)-1/4*A/a/c/x^4+1/a^2/c^2*(-A*a*d -A*b*c+B*a*c)*ln(x)+1/4*d*(A*d-B*c)/c^2/(a*d-b*c)*ln(d*x^4+c)
Time = 47.80 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {{\left (B a b - A b^{2}\right )} c^{2} x^{4} \log \left (b x^{4} + a\right ) - {\left (B a^{2} c d - A a^{2} d^{2}\right )} x^{4} \log \left (d x^{4} + c\right ) + 4 \, {\left (B a^{2} c d - A a^{2} d^{2} - {\left (B a b - A b^{2}\right )} c^{2}\right )} x^{4} \log \left (x\right ) + A a b c^{2} - A a^{2} c d}{4 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{4}} \] Input:
integrate((B*x^4+A)/x^5/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
Output:
-1/4*((B*a*b - A*b^2)*c^2*x^4*log(b*x^4 + a) - (B*a^2*c*d - A*a^2*d^2)*x^4 *log(d*x^4 + c) + 4*(B*a^2*c*d - A*a^2*d^2 - (B*a*b - A*b^2)*c^2)*x^4*log( x) + A*a*b*c^2 - A*a^2*c*d)/((a^2*b*c^3 - a^3*c^2*d)*x^4)
Timed out. \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate((B*x**4+A)/x**5/(b*x**4+a)/(d*x**4+c),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {{\left (B a b - A b^{2}\right )} \log \left (b x^{4} + a\right )}{4 \, {\left (a^{2} b c - a^{3} d\right )}} + \frac {{\left (B c d - A d^{2}\right )} \log \left (d x^{4} + c\right )}{4 \, {\left (b c^{3} - a c^{2} d\right )}} - \frac {{\left (A a d - {\left (B a - A b\right )} c\right )} \log \left (x^{4}\right )}{4 \, a^{2} c^{2}} - \frac {A}{4 \, a c x^{4}} \] Input:
integrate((B*x^4+A)/x^5/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
Output:
-1/4*(B*a*b - A*b^2)*log(b*x^4 + a)/(a^2*b*c - a^3*d) + 1/4*(B*c*d - A*d^2 )*log(d*x^4 + c)/(b*c^3 - a*c^2*d) - 1/4*(A*a*d - (B*a - A*b)*c)*log(x^4)/ (a^2*c^2) - 1/4*A/(a*c*x^4)
Time = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.41 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {{\left (B a b^{2} - A b^{3}\right )} \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, {\left (a^{2} b^{2} c - a^{3} b d\right )}} + \frac {{\left (B c d^{2} - A d^{3}\right )} \log \left ({\left | d x^{4} + c \right |}\right )}{4 \, {\left (b c^{3} d - a c^{2} d^{2}\right )}} + \frac {{\left (B a c - A b c - A a d\right )} \log \left (x^{4}\right )}{4 \, a^{2} c^{2}} - \frac {B a c x^{4} - A b c x^{4} - A a d x^{4} + A a c}{4 \, a^{2} c^{2} x^{4}} \] Input:
integrate((B*x^4+A)/x^5/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
Output:
-1/4*(B*a*b^2 - A*b^3)*log(abs(b*x^4 + a))/(a^2*b^2*c - a^3*b*d) + 1/4*(B* c*d^2 - A*d^3)*log(abs(d*x^4 + c))/(b*c^3*d - a*c^2*d^2) + 1/4*(B*a*c - A* b*c - A*a*d)*log(x^4)/(a^2*c^2) - 1/4*(B*a*c*x^4 - A*b*c*x^4 - A*a*d*x^4 + A*a*c)/(a^2*c^2*x^4)
Time = 11.81 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {\ln \left (b\,x^4+a\right )\,\left (A\,b^2-B\,a\,b\right )}{4\,a^3\,d-4\,a^2\,b\,c}-\frac {\ln \left (d\,x^4+c\right )\,\left (A\,d^2-B\,c\,d\right )}{4\,b\,c^3-4\,a\,c^2\,d}-\frac {\ln \left (x\right )\,\left (A\,a\,d+A\,b\,c-B\,a\,c\right )}{a^2\,c^2}-\frac {A}{4\,a\,c\,x^4} \] Input:
int((A + B*x^4)/(x^5*(a + b*x^4)*(c + d*x^4)),x)
Output:
- (log(a + b*x^4)*(A*b^2 - B*a*b))/(4*a^3*d - 4*a^2*b*c) - (log(c + d*x^4) *(A*d^2 - B*c*d))/(4*b*c^3 - 4*a*c^2*d) - (log(x)*(A*a*d + A*b*c - B*a*c)) /(a^2*c^2) - A/(4*a*c*x^4)
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^4}{x^5 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) d \,x^{4}+\mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) d \,x^{4}-4 \,\mathrm {log}\left (x \right ) d \,x^{4}-c}{4 c^{2} x^{4}} \] Input:
int((B*x^4+A)/x^5/(b*x^4+a)/(d*x^4+c),x)
Output:
(log( - d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*d*x**4 + log (d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*d*x**4 - 4*log(x)*d *x**4 - c)/(4*c**2*x**4)