Integrand size = 31, antiderivative size = 247 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\frac {1}{2} x \sqrt [4]{b x^2+a x^4}+\frac {\left (4 a^2-b\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{4 a^{3/4}}+\frac {\left (-4 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{4 a^{3/4}}+\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a b \log (x)-a b \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )-a^2 \log (x) \text {$\#$1}^4+b \log (x) \text {$\#$1}^4+a^2 \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-b \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.21 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\frac {x^{3/2} \left (b+a x^2\right )^{3/4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{b+a x^2}+4 a^2 \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-b \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\left (-4 a^2+b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-a^{3/4} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+2 a b \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-2 a^2 \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 a^{3/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \]
(x^(3/2)*(b + a*x^2)^(3/4)*(2*a^(3/4)*x^(3/2)*(b + a*x^2)^(1/4) + 4*a^2*Ar cTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] - b*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + (-4*a^2 + b)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4) ] - a^(3/4)*RootSum[b - a*#1^4 + #1^8 & , (-(a*b*Log[x]) + 2*a*b*Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1] + a^2*Log[x]*#1^4 - b*Log[x]*#1^4 - 2*a^2*Log[( b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1^4 + 2*b*Log[(b + a*x^2)^(1/4) - Sqrt[x]* #1]*#1^4)/(a*#1^3 - 2*#1^7) & ]))/(4*a^(3/4)*(x^2*(b + a*x^2))^(3/4))
Leaf count is larger than twice the leaf count of optimal. \(1130\) vs. \(2(247)=494\).
Time = 2.55 (sec) , antiderivative size = 1130, normalized size of antiderivative = 4.57, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {2467, 1592, 1840, 25, 959, 854, 827, 216, 219, 7279, 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.
| \(\displaystyle \int \frac {x^4 \sqrt [4]{a x^4+b x^2}}{a x^2+b+x^4} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{a x^4+b x^2} \int \frac {x^{9/2} \sqrt [4]{a x^2+b}}{x^4+a x^2+b}dx}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 1592 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \int \frac {x^5 \sqrt [4]{a x^2+b}}{x^4+a x^2+b}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 1840 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\int -\frac {x \left (a^2-x^2 a-b\right )}{\left (a x^2+b\right )^{3/4}}d\sqrt {x}-\int -\frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}-\int \frac {x \left (a^2-x^2 a-b\right )}{\left (a x^2+b\right )^{3/4}}d\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (-\frac {1}{4} \left (4 a^2-b\right ) \int \frac {x}{\left (a x^2+b\right )^{3/4}}d\sqrt {x}+\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (-\frac {1}{4} \left (4 a^2-b\right ) \int \frac {x}{1-a x^2}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}+\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (-\frac {1}{4} \left (4 a^2-b\right ) \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} x+1}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}\right )+\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}-\frac {1}{4} \left (4 a^2-b\right ) \left (\frac {\int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\int \frac {x \left (a \left (a^2-2 b\right ) x^2+\left (a^2-b\right ) b\right )}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}d\sqrt {x}-\frac {1}{4} \left (4 a^2-b\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\int \left (\frac {a \left (a^2-2 b\right ) x^3}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}+\frac {\left (a^2-b\right ) b x}{\left (a x^2+b\right )^{3/4} \left (x^4+a x^2+b\right )}\right )d\sqrt {x}-\frac {1}{4} \left (4 a^2-b\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{4} x^{3/2} \sqrt [4]{a x^2+b}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (\frac {1}{4} \sqrt [4]{a x^2+b} x^{3/2}-\frac {\left (a^2-b\right ) b \arctan \left (\frac {\sqrt [4]{a^2-\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \left (a^2-\sqrt {a^2-4 b} a-2 b\right )^{3/4}}+\frac {a \left (a-\sqrt {a^2-4 b}\right )^{3/4} \left (a^2-2 b\right ) \arctan \left (\frac {\sqrt [4]{a^2-\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{2 \sqrt {a^2-4 b} \left (a^2-\sqrt {a^2-4 b} a-2 b\right )^{3/4}}+\frac {\left (a^2-b\right ) b \arctan \left (\frac {\sqrt [4]{a^2+\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \left (a^2+\sqrt {a^2-4 b} a-2 b\right )^{3/4}}-\frac {a \left (a+\sqrt {a^2-4 b}\right )^{3/4} \left (a^2-2 b\right ) \arctan \left (\frac {\sqrt [4]{a^2+\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{2 \sqrt {a^2-4 b} \left (a^2+\sqrt {a^2-4 b} a-2 b\right )^{3/4}}-\frac {1}{4} \left (4 a^2-b\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4}}\right )+\frac {\left (a^2-b\right ) b \text {arctanh}\left (\frac {\sqrt [4]{a^2-\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \left (a^2-\sqrt {a^2-4 b} a-2 b\right )^{3/4}}-\frac {a \left (a-\sqrt {a^2-4 b}\right )^{3/4} \left (a^2-2 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a^2-\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{2 \sqrt {a^2-4 b} \left (a^2-\sqrt {a^2-4 b} a-2 b\right )^{3/4}}-\frac {\left (a^2-b\right ) b \text {arctanh}\left (\frac {\sqrt [4]{a^2+\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \left (a^2+\sqrt {a^2-4 b} a-2 b\right )^{3/4}}+\frac {a \left (a+\sqrt {a^2-4 b}\right )^{3/4} \left (a^2-2 b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a^2+\sqrt {a^2-4 b} a-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{2 \sqrt {a^2-4 b} \left (a^2+\sqrt {a^2-4 b} a-2 b\right )^{3/4}}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\) |
(2*(b*x^2 + a*x^4)^(1/4)*((x^(3/2)*(b + a*x^2)^(1/4))/4 + (a*(a - Sqrt[a^2 - 4*b])^(3/4)*(a^2 - 2*b)*ArcTan[((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*S qrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/(2*Sqrt[a^2 - 4* b]*(a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)) - ((a^2 - b)*b*ArcTan[((a^2 - a* Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a* x^2)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b]*(a^2 - a*Sqrt[a ^2 - 4*b] - 2*b)^(3/4)) - (a*(a + Sqrt[a^2 - 4*b])^(3/4)*(a^2 - 2*b)*ArcTa n[((a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a + Sqrt[a^2 - 4*b])^( 1/4)*(b + a*x^2)^(1/4))])/(2*Sqrt[a^2 - 4*b]*(a^2 + a*Sqrt[a^2 - 4*b] - 2* b)^(3/4)) + ((a^2 - b)*b*ArcTan[((a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqr t[x])/((a + Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/((a + Sqrt[a^2 - 4 *b])^(1/4)*Sqrt[a^2 - 4*b]*(a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)) - ((4*a^ 2 - b)*(-1/2*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/a^(3/4) + ArcTanh [(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/(2*a^(3/4))))/4 - (a*(a - Sqrt[a^2 - 4*b])^(3/4)*(a^2 - 2*b)*ArcTanh[((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sq rt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/(2*Sqrt[a^2 - 4*b ]*(a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(3/4)) + ((a^2 - b)*b*ArcTanh[((a^2 - a* Sqrt[a^2 - 4*b] - 2*b)^(1/4)*Sqrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a* x^2)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b]*(a^2 - a*Sqrt[a ^2 - 4*b] - 2*b)^(3/4)) + (a*(a + Sqrt[a^2 - 4*b])^(3/4)*(a^2 - 2*b)*Ar...
3.28.7.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/f Subst[ Int[x^(k*(m + 1) - 1)*(d + e*(x^(2*k)/f^2))^q*(a + b*(x^(2*k)/f^k) + c*(x^( 4*k)/f^4))^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x ] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^ (n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[f^(2*n)/c^2 Int[(f*x)^(m - 2 *n)*(c*d - b*e + c*e*x^n)*(d + e*x^n)^(q - 1), x], x] - Simp[f^(2*n)/c^2 Int[(f*x)^(m - 2*n)*(d + e*x^n)^(q - 1)*(Simp[a*(c*d - b*e) + (b*c*d - b^2* e + a*c*e)*x^n, x]/(a + b*x^n + c*x^(2*n))), x], x] /; FreeQ[{a, b, c, d, e , f}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && !IntegerQ [q] && GtQ[q, 0] && GtQ[m, 2*n - 1]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.00 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.01
| method | result | size |
| pseudoelliptic | \(\frac {4 x \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} a^{\frac {3}{4}}-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\left (-\frac {\left (\left (a^{2}-b \right ) \textit {\_R}^{4}-a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (-2 \textit {\_R}^{4}+a \right )}\right )\right ) a^{\frac {3}{4}}-4 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right ) a^{2}-8 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{2}+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b}{8 a^{\frac {3}{4}}}\) | \(250\) |
1/8*(4*x*(x^2*(a*x^2+b))^(1/4)*a^(3/4)-4*sum(-((a^2-b)*_R^4-a*b)*ln((-_R*x +(x^2*(a*x^2+b))^(1/4))/x)/_R^3/(-2*_R^4+a),_R=RootOf(_Z^8-_Z^4*a+b))*a^(3 /4)-4*ln((-a^(1/4)*x-(x^2*(a*x^2+b))^(1/4))/(a^(1/4)*x-(x^2*(a*x^2+b))^(1/ 4)))*a^2-8*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4))*a^2+ln((-a^(1/4)*x-(x ^2*(a*x^2+b))^(1/4))/(a^(1/4)*x-(x^2*(a*x^2+b))^(1/4)))*b+2*arctan(1/a^(1/ 4)/x*(x^2*(a*x^2+b))^(1/4))*b)/a^(3/4)
Timed out. \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\text {Timed out} \]
Not integrable
Time = 3.97 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.11 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\int \frac {x^{4} \sqrt [4]{x^{2} \left (a x^{2} + b\right )}}{a x^{2} + b + x^{4}}\, dx \]
Not integrable
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.13 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} x^{4}}{x^{4} + a x^{2} + b} \,d x } \]
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 8.98 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\frac {1}{2} \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} x^{2} + \frac {\sqrt {2} {\left (4 \, a^{2} - b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{8 \, \left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (4 \, a^{2} - b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{8 \, \left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (4 \, a^{2} - b\right )} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{16 \, \left (-a\right )^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (4 \, a^{2} - b\right )} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{16 \, \left (-a\right )^{\frac {3}{4}}} \]
1/2*(a + b/x^2)^(1/4)*x^2 + 1/8*sqrt(2)*(4*a^2 - b)*arctan(1/2*sqrt(2)*(sq rt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 1/8*sqrt( 2)*(4*a^2 - b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/ 4))/(-a)^(1/4))/(-a)^(3/4) + 1/16*sqrt(2)*(4*a^2 - b)*log(sqrt(2)*(-a)^(1/ 4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3/4) - 1/16*sqrt( 2)*(4*a^2 - b)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt (a + b/x^2))/(-a)^(3/4)
Not integrable
Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.13 \[ \int \frac {x^4 \sqrt [4]{b x^2+a x^4}}{b+a x^2+x^4} \, dx=\int \frac {x^4\,{\left (a\,x^4+b\,x^2\right )}^{1/4}}{x^4+a\,x^2+b} \,d x \]