Integrand size = 26, antiderivative size = 151 \[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {\arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{\sqrt [8]{2}}-\frac {\arctan \left (\frac {2^{5/8} x \sqrt [4]{x^3+x^5}}{\sqrt [4]{2} x^2-\sqrt {x^3+x^5}}\right )}{2^{5/8}}+\frac {\text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{\sqrt [8]{2}}+\frac {\text {arctanh}\left (\frac {\frac {x^2}{2^{3/8}}+\frac {\sqrt {x^3+x^5}}{2^{5/8}}}{x \sqrt [4]{x^3+x^5}}\right )}{2^{5/8}} \] Output:
1/2*arctan(2^(1/8)*x/(x^5+x^3)^(1/4))*2^(7/8)-1/2*arctan(2^(5/8)*x*(x^5+x^ 3)^(1/4)/(x^2*2^(1/4)-(x^5+x^3)^(1/2)))*2^(3/8)+1/2*arctanh(2^(1/8)*x/(x^5 +x^3)^(1/4))*2^(7/8)+1/2*arctanh((1/2*x^2*2^(5/8)+1/2*(x^5+x^3)^(1/2)*2^(3 /8))/x/(x^5+x^3)^(1/4))*2^(3/8)
Time = 1.07 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.20 \[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {x^{3/4} \sqrt [4]{1+x^2} \left (\sqrt {2} \arctan \left (\frac {\sqrt [8]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )-\arctan \left (\frac {2^{5/8} \sqrt [4]{x} \sqrt [4]{1+x^2}}{\sqrt [4]{2} \sqrt {x}-\sqrt {1+x^2}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [8]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )+\text {arctanh}\left (\frac {2\ 2^{3/8} \sqrt [4]{x} \sqrt [4]{1+x^2}}{2 \sqrt {x}+2^{3/4} \sqrt {1+x^2}}\right )\right )}{2^{5/8} \sqrt [4]{x^3+x^5}} \] Input:
Integrate[(1 - x^4)/((1 + x^4)*(x^3 + x^5)^(1/4)),x]
Output:
(x^(3/4)*(1 + x^2)^(1/4)*(Sqrt[2]*ArcTan[(2^(1/8)*x^(1/4))/(1 + x^2)^(1/4) ] - ArcTan[(2^(5/8)*x^(1/4)*(1 + x^2)^(1/4))/(2^(1/4)*Sqrt[x] - Sqrt[1 + x ^2])] + Sqrt[2]*ArcTanh[(2^(1/8)*x^(1/4))/(1 + x^2)^(1/4)] + ArcTanh[(2*2^ (3/8)*x^(1/4)*(1 + x^2)^(1/4))/(2*Sqrt[x] + 2^(3/4)*Sqrt[1 + x^2])]))/(2^( 5/8)*(x^3 + x^5)^(1/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.55 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2467, 1388, 2035, 7276, 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.
\(\displaystyle \int \frac {1-x^4}{\left (x^4+1\right ) \sqrt [4]{x^5+x^3}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{3/4} \sqrt [4]{x^2+1} \int \frac {1-x^4}{x^{3/4} \sqrt [4]{x^2+1} \left (x^4+1\right )}dx}{\sqrt [4]{x^5+x^3}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {x^{3/4} \sqrt [4]{x^2+1} \int \frac {\left (1-x^2\right ) \left (x^2+1\right )^{3/4}}{x^{3/4} \left (x^4+1\right )}dx}{\sqrt [4]{x^5+x^3}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \int \frac {\left (1-x^2\right ) \left (x^2+1\right )^{3/4}}{x^4+1}d\sqrt [4]{x}}{\sqrt [4]{x^5+x^3}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \int \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (x^2+1\right )^{3/4}}{i-x^2}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (x^2+1\right )^{3/4}}{x^2+i}\right )d\sqrt [4]{x}}{\sqrt [4]{x^5+x^3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt [4]{x} \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},-i x^2,-x^2\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{x} \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},i x^2,-x^2\right )\right )}{\sqrt [4]{x^5+x^3}}\) |
Input:
Int[(1 - x^4)/((1 + x^4)*(x^3 + x^5)^(1/4)),x]
Output:
(4*x^(3/4)*(1 + x^2)^(1/4)*((1/2 - I/2)*x^(1/4)*AppellF1[1/8, 1, -3/4, 9/8 , (-I)*x^2, -x^2] + (1/2 + I/2)*x^(1/4)*AppellF1[1/8, 1, -3/4, 9/8, I*x^2, -x^2]))/(x^3 + x^5)^(1/4)
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
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_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 24.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25
method | result | size |
pseudoelliptic | \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{2}\) | \(37\) |
trager | \(\text {Expression too large to display}\) | \(1515\) |
Input:
int((-x^4+1)/(x^4+1)/(x^5+x^3)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/2*sum(ln((-_R*x+(x^3*(x^2+1))^(1/4))/x)/_R,_R=RootOf(_Z^8-2))
Result contains complex when optimal does not.
Time = 13.36 (sec) , antiderivative size = 1640, normalized size of antiderivative = 10.86 \[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\text {Too large to display} \] Input:
integrate((-x^4+1)/(x^4+1)/(x^5+x^3)^(1/4),x, algorithm="fricas")
Output:
1/8*2^(7/8)*log((2^(7/8)*(x^6 - 2*x^5 + 4*x^4 - 2*x^3 + x^2) + 2*(x^5 + x^ 3)^(3/4)*(2*x^2 - sqrt(2)*(x^2 - 2*x + 1) - 2*x + 2) - 2*sqrt(x^5 + x^3)*( 2^(5/8)*(x^3 - 2*x^2 + x) - 2*2^(1/8)*(x^3 - x^2 + x)) - 2^(3/8)*(x^6 - 4* x^5 + 4*x^4 - 4*x^3 + x^2) - 2*(x^5 + x^3)^(1/4)*(2^(3/4)*(x^4 - 2*x^3 + x ^2) - 2*2^(1/4)*(x^4 - x^3 + x^2)))/(x^6 + x^2)) - 1/8*2^(7/8)*log(-(2^(7/ 8)*(x^6 - 2*x^5 + 4*x^4 - 2*x^3 + x^2) - 2*(x^5 + x^3)^(3/4)*(2*x^2 - sqrt (2)*(x^2 - 2*x + 1) - 2*x + 2) - 2*sqrt(x^5 + x^3)*(2^(5/8)*(x^3 - 2*x^2 + x) - 2*2^(1/8)*(x^3 - x^2 + x)) - 2^(3/8)*(x^6 - 4*x^5 + 4*x^4 - 4*x^3 + x^2) + 2*(x^5 + x^3)^(1/4)*(2^(3/4)*(x^4 - 2*x^3 + x^2) - 2*2^(1/4)*(x^4 - x^3 + x^2)))/(x^6 + x^2)) + 1/8*I*2^(7/8)*log((2^(7/8)*(I*x^6 - 2*I*x^5 + 4*I*x^4 - 2*I*x^3 + I*x^2) + 2*(x^5 + x^3)^(3/4)*(2*x^2 - sqrt(2)*(x^2 - 2*x + 1) - 2*x + 2) - 2*sqrt(x^5 + x^3)*(2^(5/8)*(-I*x^3 + 2*I*x^2 - I*x) + 2*2^(1/8)*(I*x^3 - I*x^2 + I*x)) + 2^(3/8)*(-I*x^6 + 4*I*x^5 - 4*I*x^4 + 4*I*x^3 - I*x^2) + 2*(x^5 + x^3)^(1/4)*(2^(3/4)*(x^4 - 2*x^3 + x^2) - 2*2 ^(1/4)*(x^4 - x^3 + x^2)))/(x^6 + x^2)) - 1/8*I*2^(7/8)*log((2^(7/8)*(-I*x ^6 + 2*I*x^5 - 4*I*x^4 + 2*I*x^3 - I*x^2) + 2*(x^5 + x^3)^(3/4)*(2*x^2 - s qrt(2)*(x^2 - 2*x + 1) - 2*x + 2) - 2*sqrt(x^5 + x^3)*(2^(5/8)*(I*x^3 - 2* I*x^2 + I*x) + 2*2^(1/8)*(-I*x^3 + I*x^2 - I*x)) + 2^(3/8)*(I*x^6 - 4*I*x^ 5 + 4*I*x^4 - 4*I*x^3 + I*x^2) + 2*(x^5 + x^3)^(1/4)*(2^(3/4)*(x^4 - 2*x^3 + x^2) - 2*2^(1/4)*(x^4 - x^3 + x^2)))/(x^6 + x^2)) + (1/8*I + 1/8)*2^...
\[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=- \int \frac {x^{4}}{x^{4} \sqrt [4]{x^{5} + x^{3}} + \sqrt [4]{x^{5} + x^{3}}}\, dx - \int \left (- \frac {1}{x^{4} \sqrt [4]{x^{5} + x^{3}} + \sqrt [4]{x^{5} + x^{3}}}\right )\, dx \] Input:
integrate((-x**4+1)/(x**4+1)/(x**5+x**3)**(1/4),x)
Output:
-Integral(x**4/(x**4*(x**5 + x**3)**(1/4) + (x**5 + x**3)**(1/4)), x) - In tegral(-1/(x**4*(x**5 + x**3)**(1/4) + (x**5 + x**3)**(1/4)), x)
\[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} - 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}} \,d x } \] Input:
integrate((-x^4+1)/(x^4+1)/(x^5+x^3)^(1/4),x, algorithm="maxima")
Output:
-integrate((x^4 - 1)/((x^5 + x^3)^(1/4)*(x^4 + 1)), x)
\[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} - 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}} \,d x } \] Input:
integrate((-x^4+1)/(x^4+1)/(x^5+x^3)^(1/4),x, algorithm="giac")
Output:
integrate(-(x^4 - 1)/((x^5 + x^3)^(1/4)*(x^4 + 1)), x)
Timed out. \[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=-\int \frac {x^4-1}{{\left (x^5+x^3\right )}^{1/4}\,\left (x^4+1\right )} \,d x \] Input:
int(-(x^4 - 1)/((x^3 + x^5)^(1/4)*(x^4 + 1)),x)
Output:
-int((x^4 - 1)/((x^3 + x^5)^(1/4)*(x^4 + 1)), x)
\[ \int \frac {1-x^4}{\left (1+x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=-\left (\int \frac {x^{4}}{x^{\frac {19}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}+x^{\frac {3}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right )+\int \frac {1}{x^{\frac {19}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}+x^{\frac {3}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}}d x \] Input:
int((-x^4+1)/(x^4+1)/(x^5+x^3)^(1/4),x)
Output:
- int(x**4/(x**(3/4)*(x**2 + 1)**(1/4)*x**4 + x**(3/4)*(x**2 + 1)**(1/4)) ,x) + int(1/(x**(3/4)*(x**2 + 1)**(1/4)*x**4 + x**(3/4)*(x**2 + 1)**(1/4)) ,x)