Integrand size = 22, antiderivative size = 357 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {1}{5 a c x^5}+\frac {b c+a d}{a^2 c^2 x}-\frac {b^{9/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}+\frac {b^{9/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}+\frac {d^{9/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)}-\frac {d^{9/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)}-\frac {b^{9/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}+\frac {d^{9/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)} \] Output:
-1/5/a/c/x^5+(a*d+b*c)/a^2/c^2/x+1/4*b^(9/4)*arctan(-1+2^(1/2)*b^(1/4)*x/a ^(1/4))*2^(1/2)/a^(9/4)/(-a*d+b*c)+1/4*b^(9/4)*arctan(1+2^(1/2)*b^(1/4)*x/ a^(1/4))*2^(1/2)/a^(9/4)/(-a*d+b*c)-1/4*d^(9/4)*arctan(-1+2^(1/2)*d^(1/4)* x/c^(1/4))*2^(1/2)/c^(9/4)/(-a*d+b*c)-1/4*d^(9/4)*arctan(1+2^(1/2)*d^(1/4) *x/c^(1/4))*2^(1/2)/c^(9/4)/(-a*d+b*c)-1/4*b^(9/4)*arctanh(2^(1/2)*a^(1/4) *b^(1/4)*x/(a^(1/2)+b^(1/2)*x^2))*2^(1/2)/a^(9/4)/(-a*d+b*c)+1/4*d^(9/4)*a rctanh(2^(1/2)*c^(1/4)*d^(1/4)*x/(c^(1/2)+d^(1/2)*x^2))*2^(1/2)/c^(9/4)/(- a*d+b*c)
Time = 0.21 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\frac {8 b}{a}-\frac {8 d}{c}-\frac {40 b^2 x^4}{a^2}+\frac {40 d^2 x^4}{c^2}+\frac {10 \sqrt {2} b^{9/4} x^5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{9/4}}-\frac {10 \sqrt {2} b^{9/4} x^5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{9/4}}-\frac {10 \sqrt {2} d^{9/4} x^5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{9/4}}+\frac {10 \sqrt {2} d^{9/4} x^5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{9/4}}-\frac {5 \sqrt {2} b^{9/4} x^5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{9/4}}+\frac {5 \sqrt {2} b^{9/4} x^5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{9/4}}+\frac {5 \sqrt {2} d^{9/4} x^5 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{9/4}}-\frac {5 \sqrt {2} d^{9/4} x^5 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{9/4}}}{40 (-b c+a d) x^5} \] Input:
Integrate[1/(x^6*(a + b*x^4)*(c + d*x^4)),x]
Output:
((8*b)/a - (8*d)/c - (40*b^2*x^4)/a^2 + (40*d^2*x^4)/c^2 + (10*Sqrt[2]*b^( 9/4)*x^5*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(9/4) - (10*Sqrt[2]*b^ (9/4)*x^5*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(9/4) - (10*Sqrt[2]*d ^(9/4)*x^5*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(9/4) + (10*Sqrt[2]* d^(9/4)*x^5*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(9/4) - (5*Sqrt[2]* b^(9/4)*x^5*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/a^(9/4 ) + (5*Sqrt[2]*b^(9/4)*x^5*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[ b]*x^2])/a^(9/4) + (5*Sqrt[2]*d^(9/4)*x^5*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^ (1/4)*x + Sqrt[d]*x^2])/c^(9/4) - (5*Sqrt[2]*d^(9/4)*x^5*Log[Sqrt[c] + Sqr t[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/c^(9/4))/(40*(-(b*c) + a*d)*x^5)
Time = 1.18 (sec) , antiderivative size = 522, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {980, 27, 1053, 1054, 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.
\(\displaystyle \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx\) |
\(\Big \downarrow \) 980 |
\(\displaystyle \frac {\int -\frac {5 \left (b d x^4+b c+a d\right )}{x^2 \left (b x^4+a\right ) \left (d x^4+c\right )}dx}{5 a c}-\frac {1}{5 a c x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {b d x^4+b c+a d}{x^2 \left (b x^4+a\right ) \left (d x^4+c\right )}dx}{a c}-\frac {1}{5 a c x^5}\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle -\frac {-\frac {\int \frac {x^2 \left (b d (b c+a d) x^4+b^2 c^2+a^2 d^2+a b c d\right )}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{5 a c x^5}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle -\frac {-\frac {\int \left (\frac {c^2 x^2 b^3}{(b c-a d) \left (b x^4+a\right )}+\frac {a^2 d^3 x^2}{(a d-b c) \left (d x^4+c\right )}\right )dx}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{5 a c x^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\frac {a^2 d^{9/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {a^2 d^{9/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {a^2 d^{9/4} \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {a^2 d^{9/4} \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {b^{9/4} c^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {b^{9/4} c^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {b^{9/4} c^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {b^{9/4} c^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{5 a c x^5}\) |
Input:
Int[1/(x^6*(a + b*x^4)*(c + d*x^4)),x]
Output:
-1/5*1/(a*c*x^5) - (-((b*c + a*d)/(a*c*x)) - (-1/2*(b^(9/4)*c^2*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(Sqrt[2]*a^(1/4)*(b*c - a*d)) + (b^(9/4)*c^ 2*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(1/4)*(b*c - a*d)) + (a^2*d^(9/4)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(1/4 )*(b*c - a*d)) - (a^2*d^(9/4)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2* Sqrt[2]*c^(1/4)*(b*c - a*d)) + (b^(9/4)*c^2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)* b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(1/4)*(b*c - a*d)) - (b^(9/4)*c^2*L og[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(1/4)* (b*c - a*d)) - (a^2*d^(9/4)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt [d]*x^2])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)) + (a^2*d^(9/4)*Log[Sqrt[c] + Sqr t[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)))/(a *c))/(a*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1)) Int[(e*x)^(m + n)*( a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ ))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b *x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( m + 1)) Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 ) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && LtQ[m, -1]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
Time = 0.18 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.73
method | result | size |
default | \(-\frac {b^{2} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a^{2} \left (a d -c b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {1}{5 a c \,x^{5}}-\frac {-a d -c b}{a^{2} c^{2} x}+\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2} \left (a d -c b \right ) \left (\frac {c}{d}\right )^{\frac {1}{4}}}\) | \(261\) |
risch | \(\text {Expression too large to display}\) | \(1078\) |
Input:
int(1/x^6/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
-1/8*b^2/a^2/(a*d-b*c)/(a/b)^(1/4)*2^(1/2)*(ln((x^2-(a/b)^(1/4)*x*2^(1/2)+ (a/b)^(1/2))/(x^2+(a/b)^(1/4)*x*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/ b)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x-1))-1/5/a/c/x^5-1/a^2/c^2*(-a *d-b*c)/x+1/8*d^2/c^2/(a*d-b*c)/(c/d)^(1/4)*2^(1/2)*(ln((x^2-(c/d)^(1/4)*x *2^(1/2)+(c/d)^(1/2))/(x^2+(c/d)^(1/4)*x*2^(1/2)+(c/d)^(1/2)))+2*arctan(2^ (1/2)/(c/d)^(1/4)*x+1)+2*arctan(2^(1/2)/(c/d)^(1/4)*x-1))
Result contains complex when optimal does not.
Time = 4.64 (sec) , antiderivative size = 1526, normalized size of antiderivative = 4.27 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
Output:
1/20*(5*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^3*d + 6*a^11*b^2*c^2*d^2 - 4*a^1 2*b*c*d^3 + a^13*d^4))^(1/4)*a^2*c^2*x^5*log(b^7*x + (a^7*b^3*c^3 - 3*a^8* b^2*c^2*d + 3*a^9*b*c*d^2 - a^10*d^3)*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^3* d + 6*a^11*b^2*c^2*d^2 - 4*a^12*b*c*d^3 + a^13*d^4))^(3/4)) - 5*(-b^9/(a^9 *b^4*c^4 - 4*a^10*b^3*c^3*d + 6*a^11*b^2*c^2*d^2 - 4*a^12*b*c*d^3 + a^13*d ^4))^(1/4)*a^2*c^2*x^5*log(b^7*x - (a^7*b^3*c^3 - 3*a^8*b^2*c^2*d + 3*a^9* b*c*d^2 - a^10*d^3)*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^3*d + 6*a^11*b^2*c^2 *d^2 - 4*a^12*b*c*d^3 + a^13*d^4))^(3/4)) + 5*I*(-b^9/(a^9*b^4*c^4 - 4*a^1 0*b^3*c^3*d + 6*a^11*b^2*c^2*d^2 - 4*a^12*b*c*d^3 + a^13*d^4))^(1/4)*a^2*c ^2*x^5*log(b^7*x - (I*a^7*b^3*c^3 - 3*I*a^8*b^2*c^2*d + 3*I*a^9*b*c*d^2 - I*a^10*d^3)*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^3*d + 6*a^11*b^2*c^2*d^2 - 4 *a^12*b*c*d^3 + a^13*d^4))^(3/4)) - 5*I*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^ 3*d + 6*a^11*b^2*c^2*d^2 - 4*a^12*b*c*d^3 + a^13*d^4))^(1/4)*a^2*c^2*x^5*l og(b^7*x - (-I*a^7*b^3*c^3 + 3*I*a^8*b^2*c^2*d - 3*I*a^9*b*c*d^2 + I*a^10* d^3)*(-b^9/(a^9*b^4*c^4 - 4*a^10*b^3*c^3*d + 6*a^11*b^2*c^2*d^2 - 4*a^12*b *c*d^3 + a^13*d^4))^(3/4)) - 5*(-d^9/(b^4*c^13 - 4*a*b^3*c^12*d + 6*a^2*b^ 2*c^11*d^2 - 4*a^3*b*c^10*d^3 + a^4*c^9*d^4))^(1/4)*a^2*c^2*x^5*log(d^7*x + (b^3*c^10 - 3*a*b^2*c^9*d + 3*a^2*b*c^8*d^2 - a^3*c^7*d^3)*(-d^9/(b^4*c^ 13 - 4*a*b^3*c^12*d + 6*a^2*b^2*c^11*d^2 - 4*a^3*b*c^10*d^3 + a^4*c^9*d^4) )^(3/4)) + 5*(-d^9/(b^4*c^13 - 4*a*b^3*c^12*d + 6*a^2*b^2*c^11*d^2 - 4*...
Timed out. \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**6/(b*x**4+a)/(d*x**4+c),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {b^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{8 \, {\left (a^{2} b c - a^{3} d\right )}} - \frac {d^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{8 \, {\left (b c^{3} - a c^{2} d\right )}} + \frac {5 \, {\left (b c + a d\right )} x^{4} - a c}{5 \, a^{2} c^{2} x^{5}} \] Input:
integrate(1/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
Output:
1/8*b^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/ 4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arc tan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt( b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(b)*x^2 + sqrt(2)*a ^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a^2*b*c - a^3*d) - 1/8*d^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x + sqrt(2)*c^(1/4)*d^( 1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) + 2*sqrt(2)*a rctan(1/2*sqrt(2)*(2*sqrt(d)*x - sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqr t(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt(2)*log(sqrt(d)*x^2 + sqrt(2) *c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(1/4)*d^(3/4)) + sqrt(2)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(1/4)*d^(3/4)))/(b*c^3 - a*c^2*d ) + 1/5*(5*(b*c + a*d)*x^4 - a*c)/(a^2*c^2*x^5)
Time = 0.13 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{3} b c - \sqrt {2} a^{4} d\right )}} + \frac {\left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{3} b c - \sqrt {2} a^{4} d\right )}} - \frac {\left (c d^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} b c^{4} - \sqrt {2} a c^{3} d\right )}} - \frac {\left (c d^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} b c^{4} - \sqrt {2} a c^{3} d\right )}} - \frac {\left (a b^{3}\right )^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a^{3} b c - \sqrt {2} a^{4} d\right )}} + \frac {\left (a b^{3}\right )^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a^{3} b c - \sqrt {2} a^{4} d\right )}} + \frac {\left (c d^{3}\right )^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{4 \, {\left (\sqrt {2} b c^{4} - \sqrt {2} a c^{3} d\right )}} - \frac {\left (c d^{3}\right )^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{4 \, {\left (\sqrt {2} b c^{4} - \sqrt {2} a c^{3} d\right )}} + \frac {5 \, b c x^{4} + 5 \, a d x^{4} - a c}{5 \, a^{2} c^{2} x^{5}} \] Input:
integrate(1/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
Output:
1/2*(a*b^3)^(3/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/ 4))/(sqrt(2)*a^3*b*c - sqrt(2)*a^4*d) + 1/2*(a*b^3)^(3/4)*arctan(1/2*sqrt( 2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a^3*b*c - sqrt(2)*a^4 *d) - 1/2*(c*d^3)^(3/4)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(c/d)^(1/4))/(c/ d)^(1/4))/(sqrt(2)*b*c^4 - sqrt(2)*a*c^3*d) - 1/2*(c*d^3)^(3/4)*arctan(1/2 *sqrt(2)*(2*x - sqrt(2)*(c/d)^(1/4))/(c/d)^(1/4))/(sqrt(2)*b*c^4 - sqrt(2) *a*c^3*d) - 1/4*(a*b^3)^(3/4)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b)) /(sqrt(2)*a^3*b*c - sqrt(2)*a^4*d) + 1/4*(a*b^3)^(3/4)*log(x^2 - sqrt(2)*x *(a/b)^(1/4) + sqrt(a/b))/(sqrt(2)*a^3*b*c - sqrt(2)*a^4*d) + 1/4*(c*d^3)^ (3/4)*log(x^2 + sqrt(2)*x*(c/d)^(1/4) + sqrt(c/d))/(sqrt(2)*b*c^4 - sqrt(2 )*a*c^3*d) - 1/4*(c*d^3)^(3/4)*log(x^2 - sqrt(2)*x*(c/d)^(1/4) + sqrt(c/d) )/(sqrt(2)*b*c^4 - sqrt(2)*a*c^3*d) + 1/5*(5*b*c*x^4 + 5*a*d*x^4 - a*c)/(a ^2*c^2*x^5)
Time = 4.90 (sec) , antiderivative size = 4547, normalized size of antiderivative = 12.74 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:
int(1/(x^6*(a + b*x^4)*(c + d*x^4)),x)
Output:
- 2*atan((1024*a^11*b^10*c^13*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 10 24*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) + 4* a^11*b^6*d^9*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(1/4) + 1024*a^21*c^3*d^10* x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11* b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) - 4096*a^12*b^9*c^12*d*x*(-b^9/(25 6*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) - 4096*a^20*b*c^4*d^9*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12 *b*c*d^3))^(5/4) + 4*a^8*b^9*c^3*d^6*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c ^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(1/ 4) + 6144*a^13*b^8*c^11*d^2*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024 *a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) - 4096 *a^14*b^7*c^10*d^3*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3 *c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) + 1024*a^15*b^6 *c^9*d^4*x*(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1 536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) + 1024*a^17*b^4*c^7*d^6*x *(-b^9/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b ^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) - 4096*a^18*b^3*c^6*d^7*x*(-b^9/(25 6*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*...
Time = 0.29 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {10 b^{\frac {9}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{3} x^{5}-10 b^{\frac {9}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) c^{3} x^{5}-10 d^{\frac {9}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) a^{3} x^{5}+10 d^{\frac {9}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) a^{3} x^{5}-5 b^{\frac {9}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) c^{3} x^{5}+5 b^{\frac {9}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) c^{3} x^{5}+5 d^{\frac {9}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) a^{3} x^{5}-5 d^{\frac {9}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) a^{3} x^{5}-8 a^{3} c^{2} d +40 a^{3} c \,d^{2} x^{4}+8 a^{2} b \,c^{3}-40 a \,b^{2} c^{3} x^{4}}{40 a^{3} c^{3} x^{5} \left (a d -b c \right )} \] Input:
int(1/x^6/(b*x^4+a)/(d*x^4+c),x)
Output:
(10*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)* x)/(b**(1/4)*a**(1/4)*sqrt(2)))*b**2*c**3*x**5 - 10*b**(1/4)*a**(3/4)*sqrt (2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt (2)))*b**2*c**3*x**5 - 10*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4 )*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*d**2*x**5 + 10* d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/( d**(1/4)*c**(1/4)*sqrt(2)))*a**3*d**2*x**5 - 5*b**(1/4)*a**(3/4)*sqrt(2)*l og( - b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*b**2*c**3*x**5 + 5*b**(1/4)*a**(3/4)*sqrt(2)*log(b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)*x**2)*b**2*c**3*x**5 + 5*d**(1/4)*c**(3/4)*sqrt(2)*log( - d**(1/4 )*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*a**3*d**2*x**5 - 5*d**(1/4) *c**(3/4)*sqrt(2)*log(d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2 )*a**3*d**2*x**5 - 8*a**3*c**2*d + 40*a**3*c*d**2*x**4 + 8*a**2*b*c**3 - 4 0*a*b**2*c**3*x**4)/(40*a**3*c**3*x**5*(a*d - b*c))