Integrand size = 29, antiderivative size = 109 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^2}{2 b d}-\frac {\sqrt {a} (A b-a B) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2} (b c-a d)}-\frac {\sqrt {c} (B c-A d) \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{3/2} (b c-a d)} \] Output:
1/2*B*x^2/b/d-1/2*a^(1/2)*(A*b-B*a)*arctan(b^(1/2)*x^2/a^(1/2))/b^(3/2)/(- a*d+b*c)-1/2*c^(1/2)*(-A*d+B*c)*arctan(d^(1/2)*x^2/c^(1/2))/d^(3/2)/(-a*d+ b*c)
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {1}{2} \left (\frac {B x^2}{b d}+\frac {\sqrt {a} (-A b+a B) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{b^{3/2} (b c-a d)}+\frac {\sqrt {c} (B c-A d) \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{d^{3/2} (-b c+a d)}\right ) \] Input:
Integrate[(x^5*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x]
Output:
((B*x^2)/(b*d) + (Sqrt[a]*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(b ^(3/2)*(b*c - a*d)) + (Sqrt[c]*(B*c - A*d)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]])/ (d^(3/2)*(-(b*c) + a*d)))/2
Time = 0.47 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1045, 444, 397, 218}
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.
\(\displaystyle \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 1045 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (B x^4+A\right )}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^2\) |
\(\Big \downarrow \) 444 |
\(\displaystyle \frac {1}{2} \left (\frac {B x^2}{b d}-\frac {\int \frac {(b B c-A b d+a B d) x^4+a B c}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx^2}{b d}\right )\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {1}{2} \left (\frac {B x^2}{b d}-\frac {\frac {a d (A b-a B) \int \frac {1}{b x^4+a}dx^2}{b c-a d}+\frac {b c (B c-A d) \int \frac {1}{d x^4+c}dx^2}{b c-a d}}{b d}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {B x^2}{b d}-\frac {\frac {\sqrt {a} d (A b-a B) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {b} (b c-a d)}+\frac {b \sqrt {c} (B c-A d) \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{\sqrt {d} (b c-a d)}}{b d}\right )\) |
Input:
Int[(x^5*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x]
Output:
((B*x^2)/(b*d) - ((Sqrt[a]*(A*b - a*B)*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(S qrt[b]*(b*c - a*d)) + (b*Sqrt[c]*(B*c - A*d)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]] )/(Sqrt[d]*(b*c - a*d)))/(b*d))/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && GtQ[m, 1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. )*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Si mp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q*(e + f*x^(n/k))^r, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IGtQ[n, 0] && IntegerQ[m]
Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {B \,x^{2}}{2 b d}+\frac {a \left (A b -B a \right ) \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \left (a d -c b \right ) b \sqrt {a b}}-\frac {c \left (A d -B c \right ) \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 d \left (a d -c b \right ) \sqrt {c d}}\) | \(94\) |
risch | \(\text {Expression too large to display}\) | \(2570\) |
Input:
int(x^5*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
1/2*B*x^2/b/d+1/2*a*(A*b-B*a)/(a*d-b*c)/b/(a*b)^(1/2)*arctan(b*x^2/(a*b)^( 1/2))-1/2*c*(A*d-B*c)/d/(a*d-b*c)/(c*d)^(1/2)*arctan(d*x^2/(c*d)^(1/2))
Time = 9.37 (sec) , antiderivative size = 483, normalized size of antiderivative = 4.43 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\left [\frac {{\left (B a - A b\right )} d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} + 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) + 2 \, {\left (B b c - B a d\right )} x^{2} + {\left (B b c - A b d\right )} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} - 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right )}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {2 \, {\left (B a - A b\right )} d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) + 2 \, {\left (B b c - B a d\right )} x^{2} + {\left (B b c - A b d\right )} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} - 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right )}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {{\left (B a - A b\right )} d \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} + 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) + 2 \, {\left (B b c - B a d\right )} x^{2} - 2 \, {\left (B b c - A b d\right )} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right )}{4 \, {\left (b^{2} c d - a b d^{2}\right )}}, \frac {{\left (B a - A b\right )} d \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (B b c - B a d\right )} x^{2} - {\left (B b c - A b d\right )} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right )}{2 \, {\left (b^{2} c d - a b d^{2}\right )}}\right ] \] Input:
integrate(x^5*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
Output:
[1/4*((B*a - A*b)*d*sqrt(-a/b)*log((b*x^4 + 2*b*x^2*sqrt(-a/b) - a)/(b*x^4 + a)) + 2*(B*b*c - B*a*d)*x^2 + (B*b*c - A*b*d)*sqrt(-c/d)*log((d*x^4 - 2 *d*x^2*sqrt(-c/d) - c)/(d*x^4 + c)))/(b^2*c*d - a*b*d^2), 1/4*(2*(B*a - A* b)*d*sqrt(a/b)*arctan(b*x^2*sqrt(a/b)/a) + 2*(B*b*c - B*a*d)*x^2 + (B*b*c - A*b*d)*sqrt(-c/d)*log((d*x^4 - 2*d*x^2*sqrt(-c/d) - c)/(d*x^4 + c)))/(b^ 2*c*d - a*b*d^2), 1/4*((B*a - A*b)*d*sqrt(-a/b)*log((b*x^4 + 2*b*x^2*sqrt( -a/b) - a)/(b*x^4 + a)) + 2*(B*b*c - B*a*d)*x^2 - 2*(B*b*c - A*b*d)*sqrt(c /d)*arctan(d*x^2*sqrt(c/d)/c))/(b^2*c*d - a*b*d^2), 1/2*((B*a - A*b)*d*sqr t(a/b)*arctan(b*x^2*sqrt(a/b)/a) + (B*b*c - B*a*d)*x^2 - (B*b*c - A*b*d)*s qrt(c/d)*arctan(d*x^2*sqrt(c/d)/c))/(b^2*c*d - a*b*d^2)]
Timed out. \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(x**5*(B*x**4+A)/(b*x**4+a)/(d*x**4+c),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^{2}}{2 \, b d} + \frac {{\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {{\left (B c^{2} - A c d\right )} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} \] Input:
integrate(x^5*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
Output:
1/2*B*x^2/(b*d) + 1/2*(B*a^2 - A*a*b)*arctan(b*x^2/sqrt(a*b))/((b^2*c - a* b*d)*sqrt(a*b)) - 1/2*(B*c^2 - A*c*d)*arctan(d*x^2/sqrt(c*d))/((b*c*d - a* d^2)*sqrt(c*d))
Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^{2}}{2 \, b d} + \frac {{\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{2} c - a b d\right )} \sqrt {a b}} - \frac {{\left (B c^{2} - A c d\right )} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d - a d^{2}\right )} \sqrt {c d}} \] Input:
integrate(x^5*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
Output:
1/2*B*x^2/(b*d) + 1/2*(B*a^2 - A*a*b)*arctan(b*x^2/sqrt(a*b))/((b^2*c - a* b*d)*sqrt(a*b)) - 1/2*(B*c^2 - A*c*d)*arctan(d*x^2/sqrt(c*d))/((b*c*d - a* d^2)*sqrt(c*d))
Time = 7.80 (sec) , antiderivative size = 15787, normalized size of antiderivative = 144.83 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:
int((x^5*(A + B*x^4))/((a + b*x^4)*(c + d*x^4)),x)
Output:
(atan((((A*d - B*c)*(-c*d^3)^(1/2)*(((A*d - B*c)*(-c*d^3)^(1/2)*((16*(B^4* a*b^8*c^9 + B^4*a^9*c*d^8 + 2*A^4*a^3*b^6*c^3*d^6 + 2*B^4*a^5*b^4*c^5*d^4 + A^4*a*b^8*c^5*d^4 + A^4*a^5*b^4*c*d^8 - 4*A*B^3*a^4*b^5*c^5*d^4 - 4*A*B^ 3*a^5*b^4*c^4*d^5 + 6*A^2*B^2*a*b^8*c^7*d^2 + 6*A^2*B^2*a^7*b^2*c*d^8 - 4* A^3*B*a^3*b^6*c^4*d^5 - 4*A^3*B*a^4*b^5*c^3*d^6 - 4*A*B^3*a*b^8*c^8*d - 4* A*B^3*a^8*b*c*d^8 + 2*A^2*B^2*a^3*b^6*c^5*d^4 + 8*A^2*B^2*a^4*b^5*c^4*d^5 + 2*A^2*B^2*a^5*b^4*c^3*d^6 - 4*A^3*B*a*b^8*c^6*d^3 - 4*A^3*B*a^6*b^3*c*d^ 8))/(b^2*d^2) + ((A*d - B*c)*(((A*d - B*c)*((16*(32*B^2*a^4*b^7*c^5*d^6 - 32*A^2*a^5*b^6*c^2*d^9 - 32*B^2*a^2*b^9*c^7*d^4 - 32*B^2*a^3*b^8*c^6*d^5 - 32*A^2*a^2*b^9*c^5*d^6 + 32*B^2*a^5*b^6*c^4*d^7 - 32*B^2*a^6*b^5*c^3*d^8 - 32*B^2*a^7*b^4*c^2*d^9 + 32*A^2*a*b^10*c^6*d^5 + 32*A^2*a^6*b^5*c*d^10 + 32*B^2*a*b^10*c^8*d^3 + 32*B^2*a^8*b^3*c*d^10 - 64*A*B*a*b^10*c^7*d^4 - 6 4*A*B*a^7*b^4*c*d^10 + 64*A*B*a^2*b^9*c^6*d^5 + 64*A*B*a^3*b^8*c^5*d^6 - 1 28*A*B*a^4*b^7*c^4*d^7 + 64*A*B*a^5*b^6*c^3*d^8 + 64*A*B*a^6*b^5*c^2*d^9)) /(b^2*d^2) + ((A*d - B*c)*((4*x^2*(256*A*a*b^11*c^6*d^6 + 256*A*a^6*b^6*c* d^11 - 768*A*a^2*b^10*c^5*d^7 + 512*A*a^3*b^9*c^4*d^8 + 512*A*a^4*b^8*c^3* d^9 - 768*A*a^5*b^7*c^2*d^10))/(b^2*d^2) - (4*(A*d - B*c)*(-c*d^3)^(1/2)*( 512*a^2*b^11*c^6*d^7 - 256*a^7*b^6*c*d^12 - 256*a*b^12*c^7*d^6 + 256*a^3*b ^10*c^5*d^8 - 1024*a^4*b^9*c^4*d^9 + 256*a^5*b^8*c^3*d^10 + 512*a^6*b^7*c^ 2*d^11))/(b^2*d^2*(a*d^4 - b*c*d^3)))*(-c*d^3)^(1/2))/(4*(a*d^4 - b*c*d...
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {x^5 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right )+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right )+d \,x^{2}}{2 d^{2}} \] Input:
int(x^5*(B*x^4+A)/(b*x^4+a)/(d*x^4+c),x)
Output:
(sqrt(d)*sqrt(c)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)* c**(1/4)*sqrt(2))) + sqrt(d)*sqrt(c)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*s qrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2))) + d*x**2)/(2*d**2)