Integrand size = 29, antiderivative size = 83 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^4}{4 b d}-\frac {a (A b-a B) \log \left (a+b x^4\right )}{4 b^2 (b c-a d)}-\frac {c (B c-A d) \log \left (c+d x^4\right )}{4 d^2 (b c-a d)} \] Output:
1/4*B*x^4/b/d-1/4*a*(A*b-B*a)*ln(b*x^4+a)/b^2/(-a*d+b*c)-1/4*c*(-A*d+B*c)* ln(d*x^4+c)/d^2/(-a*d+b*c)
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {a (-A b+a B) d^2 \log \left (a+b x^4\right )+b \left (B d (b c-a d) x^4+b c (-B c+A d) \log \left (c+d x^4\right )\right )}{4 b^2 d^2 (b c-a d)} \] Input:
Integrate[(x^7*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x]
Output:
(a*(-(A*b) + a*B)*d^2*Log[a + b*x^4] + b*(B*d*(b*c - a*d)*x^4 + b*c*(-(B*c ) + A*d)*Log[c + d*x^4]))/(4*b^2*d^2*(b*c - a*d))
Time = 0.42 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1043, 159, 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.
\(\displaystyle \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 1043 |
\(\displaystyle \frac {1}{4} \int \frac {x^4 \left (B x^4+A\right )}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^4\) |
\(\Big \downarrow \) 159 |
\(\displaystyle \frac {1}{4} \int \left (\frac {B}{b d}+\frac {a (a B-A b)}{b (b c-a d) \left (b x^4+a\right )}+\frac {c (B c-A d)}{d (a d-b c) \left (d x^4+c\right )}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {a (A b-a B) \log \left (a+b x^4\right )}{b^2 (b c-a d)}-\frac {c (B c-A d) \log \left (c+d x^4\right )}{d^2 (b c-a d)}+\frac {B x^4}{b d}\right )\) |
Input:
Int[(x^7*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x]
Output:
((B*x^4)/(b*d) - (a*(A*b - a*B)*Log[a + b*x^4])/(b^2*(b*c - a*d)) - (c*(B* c - A*d)*Log[c + d*x^4])/(d^2*(b*c - a*d)))/4
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n *(e + f*x)*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && (IGtQ [m, 0] || IntegersQ[m, n])
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 = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {B \,x^{4}}{4 b d}+\frac {a \left (A b -B a \right ) \ln \left (b \,x^{4}+a \right )}{4 \left (a d -c b \right ) b^{2}}-\frac {c \left (A d -B c \right ) \ln \left (d \,x^{4}+c \right )}{4 d^{2} \left (a d -c b \right )}\) | \(78\) |
norman | \(\frac {B \,x^{4}}{4 b d}+\frac {a \left (A b -B a \right ) \ln \left (b \,x^{4}+a \right )}{4 \left (a d -c b \right ) b^{2}}-\frac {c \left (A d -B c \right ) \ln \left (d \,x^{4}+c \right )}{4 d^{2} \left (a d -c b \right )}\) | \(78\) |
parallelrisch | \(\frac {B \,x^{4} a b \,d^{2}-B \,x^{4} b^{2} c d +A \ln \left (b \,x^{4}+a \right ) a b \,d^{2}-A \ln \left (d \,x^{4}+c \right ) b^{2} c d -B \ln \left (b \,x^{4}+a \right ) a^{2} d^{2}+B \ln \left (d \,x^{4}+c \right ) b^{2} c^{2}}{4 b^{2} d^{2} \left (a d -c b \right )}\) | \(105\) |
risch | \(\frac {B \,x^{4}}{4 b d}-\frac {c \ln \left (d \,x^{4}+c \right ) A}{4 d \left (a d -c b \right )}+\frac {c^{2} \ln \left (d \,x^{4}+c \right ) B}{4 d^{2} \left (a d -c b \right )}+\frac {a \ln \left (-b \,x^{4}-a \right ) A}{4 b \left (a d -c b \right )}-\frac {a^{2} \ln \left (-b \,x^{4}-a \right ) B}{4 b^{2} \left (a d -c b \right )}\) | \(124\) |
Input:
int(x^7*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
1/4*B*x^4/b/d+1/4*a*(A*b-B*a)/(a*d-b*c)/b^2*ln(b*x^4+a)-1/4*c*(A*d-B*c)/d^ 2/(a*d-b*c)*ln(d*x^4+c)
Time = 6.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {{\left (B b^{2} c d - B a b d^{2}\right )} x^{4} + {\left (B a^{2} - A a b\right )} d^{2} \log \left (b x^{4} + a\right ) - {\left (B b^{2} c^{2} - A b^{2} c d\right )} \log \left (d x^{4} + c\right )}{4 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}} \] Input:
integrate(x^7*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
Output:
1/4*((B*b^2*c*d - B*a*b*d^2)*x^4 + (B*a^2 - A*a*b)*d^2*log(b*x^4 + a) - (B *b^2*c^2 - A*b^2*c*d)*log(d*x^4 + c))/(b^3*c*d^2 - a*b^2*d^3)
Timed out. \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(x**7*(B*x**4+A)/(b*x**4+a)/(d*x**4+c),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^{4}}{4 \, b d} + \frac {{\left (B a^{2} - A a b\right )} \log \left (b x^{4} + a\right )}{4 \, {\left (b^{3} c - a b^{2} d\right )}} - \frac {{\left (B c^{2} - A c d\right )} \log \left (d x^{4} + c\right )}{4 \, {\left (b c d^{2} - a d^{3}\right )}} \] Input:
integrate(x^7*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
Output:
1/4*B*x^4/(b*d) + 1/4*(B*a^2 - A*a*b)*log(b*x^4 + a)/(b^3*c - a*b^2*d) - 1 /4*(B*c^2 - A*c*d)*log(d*x^4 + c)/(b*c*d^2 - a*d^3)
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^{4}}{4 \, b d} + \frac {{\left (B a^{2} - A a b\right )} \log \left ({\left | b x^{4} + a \right |}\right )}{4 \, {\left (b^{3} c - a b^{2} d\right )}} - \frac {{\left (B c^{2} - A c d\right )} \log \left ({\left | d x^{4} + c \right |}\right )}{4 \, {\left (b c d^{2} - a d^{3}\right )}} \] Input:
integrate(x^7*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
Output:
1/4*B*x^4/(b*d) + 1/4*(B*a^2 - A*a*b)*log(abs(b*x^4 + a))/(b^3*c - a*b^2*d ) - 1/4*(B*c^2 - A*c*d)*log(abs(d*x^4 + c))/(b*c*d^2 - a*d^3)
Time = 9.73 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\ln \left (b\,x^4+a\right )\,\left (B\,a^2-A\,a\,b\right )}{4\,b^3\,c-4\,a\,b^2\,d}+\frac {\ln \left (d\,x^4+c\right )\,\left (B\,c^2-A\,c\,d\right )}{4\,a\,d^3-4\,b\,c\,d^2}+\frac {B\,x^4}{4\,b\,d} \] Input:
int((x^7*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x)
Output:
(log(a + b*x^4)*(B*a^2 - A*a*b))/(4*b^3*c - 4*a*b^2*d) + (log(c + d*x^4)*( B*c^2 - A*c*d))/(4*a*d^3 - 4*b*c*d^2) + (B*x^4)/(4*b*d)
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {x^7 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {-\mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) c -\mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) c +d \,x^{4}}{4 d^{2}} \] Input:
int(x^7*(B*x^4+A)/(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)*c - log(d **(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*c + d*x**4)/(4*d**2)