Integrand size = 22, antiderivative size = 173 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=-2 \arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )+\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right )}{2^{3/4}}+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^4}}\right )}{2^{3/4}}-\frac {1}{4} \text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+2 \log \left (\sqrt [4]{x^3+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{x^3+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+\text {$\#$1}^7}\&\right ] \] Output:
Unintegrable
Time = 0.39 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=-\frac {x^{9/4} (1+x)^{3/4} \left (8 \left (4 \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )-\sqrt [4]{2} \arctan \left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )-4 \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )+\sqrt [4]{2} \text {arctanh}\left (\sqrt [4]{2} \sqrt [4]{\frac {x}{1+x}}\right )\right )+\text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+8 \log \left (\sqrt [4]{1+x}-\sqrt [4]{x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-4 \log \left (\sqrt [4]{1+x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+\text {$\#$1}^7}\&\right ]\right )}{16 \left (x^3 (1+x)\right )^{3/4}} \] Input:
Integrate[(x^2*(x^3 + x^4)^(1/4))/(-1 + x^4),x]
Output:
-1/16*(x^(9/4)*(1 + x)^(3/4)*(8*(4*ArcTan[(x/(1 + x))^(1/4)] - 2^(1/4)*Arc Tan[2^(1/4)*(x/(1 + x))^(1/4)] - 4*ArcTanh[(x/(1 + x))^(1/4)] + 2^(1/4)*Ar cTanh[2^(1/4)*(x/(1 + x))^(1/4)]) + RootSum[2 - 2*#1^4 + #1^8 & , (-2*Log[ x] + 8*Log[(1 + x)^(1/4) - x^(1/4)*#1] + Log[x]*#1^4 - 4*Log[(1 + x)^(1/4) - x^(1/4)*#1]*#1^4)/(-#1^3 + #1^7) & ]))/(x^3*(1 + x))^(3/4)
Result contains complex when optimal does not.
Time = 1.56 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2467, 25, 2019, 2035, 2461, 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.
\(\displaystyle \int \frac {x^2 \sqrt [4]{x^4+x^3}}{x^4-1} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4+x^3} \int -\frac {x^{11/4} \sqrt [4]{x+1}}{1-x^4}dx}{x^{3/4} \sqrt [4]{x+1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \int \frac {x^{11/4} \sqrt [4]{x+1}}{1-x^4}dx}{x^{3/4} \sqrt [4]{x+1}}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \int \frac {x^{11/4}}{(x+1)^{3/4} \left (-x^3+x^2-x+1\right )}dx}{x^{3/4} \sqrt [4]{x+1}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4+x^3} \int \frac {x^{7/2}}{(x+1)^{3/4} \left (-x^3+x^2-x+1\right )}d\sqrt [4]{x}}{x^{3/4} \sqrt [4]{x+1}}\) |
\(\Big \downarrow \) 2461 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4+x^3} \int \left (-\frac {x^{7/2}}{4 \left (\sqrt {x}-1\right ) (x+1)^{3/4}}+\frac {x^{7/2}}{4 \left (\sqrt {x}+1\right ) (x+1)^{3/4}}+\frac {\sqrt [4]{x+1} x^{7/2}}{2 \left (x^2+1\right )}\right )d\sqrt [4]{x}}{x^{3/4} \sqrt [4]{x+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4+x^3} \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-\frac {1}{16} (1-i)^{5/4} \arctan \left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 (1-i)^{3/4}}-\frac {1}{16} (1+i)^{5/4} \arctan \left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 (1+i)^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{4\ 2^{3/4}}-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )+\frac {1}{16} (1-i)^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 (1-i)^{3/4}}+\frac {1}{16} (1+i)^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{8 (1+i)^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x+1}}\right )}{4\ 2^{3/4}}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
Input:
Int[(x^2*(x^3 + x^4)^(1/4))/(-1 + x^4),x]
Output:
(-4*(x^3 + x^4)^(1/4)*(ArcTan[x^(1/4)/(1 + x)^(1/4)]/2 - ArcTan[((1 - I)^( 1/4)*x^(1/4))/(1 + x)^(1/4)]/(8*(1 - I)^(3/4)) - ((1 - I)^(5/4)*ArcTan[((1 - I)^(1/4)*x^(1/4))/(1 + x)^(1/4)])/16 - ArcTan[((1 + I)^(1/4)*x^(1/4))/( 1 + x)^(1/4)]/(8*(1 + I)^(3/4)) - ((1 + I)^(5/4)*ArcTan[((1 + I)^(1/4)*x^( 1/4))/(1 + x)^(1/4)])/16 - ArcTan[(2^(1/4)*x^(1/4))/(1 + x)^(1/4)]/(4*2^(3 /4)) - ArcTanh[x^(1/4)/(1 + x)^(1/4)]/2 + ArcTanh[((1 - I)^(1/4)*x^(1/4))/ (1 + x)^(1/4)]/(8*(1 - I)^(3/4)) + ((1 - I)^(5/4)*ArcTanh[((1 - I)^(1/4)*x ^(1/4))/(1 + x)^(1/4)])/16 + ArcTanh[((1 + I)^(1/4)*x^(1/4))/(1 + x)^(1/4) ]/(8*(1 + I)^(3/4)) + ((1 + I)^(5/4)*ArcTanh[((1 + I)^(1/4)*x^(1/4))/(1 + x)^(1/4)])/16 + ArcTanh[(2^(1/4)*x^(1/4))/(1 + x)^(1/4)]/(4*2^(3/4))))/(x^ (3/4)*(1 + x)^(1/4))
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, Int[ExpandIntegrand[u, (Qx /. x -> x^2)^p, x], x] /; !SumQ[NonfreeFactors[ Qx, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
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]
Time = 11.55 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97
method | result | size |
pseudoelliptic | \(-\ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )+\ln \left (\frac {x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )-\frac {2^{\frac {1}{4}} \ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}\right )}{4}-\frac {2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{2 x}\right )}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (\textit {\_R}^{4}-1\right )}\right )}{4}\) | \(168\) |
trager | \(\text {Expression too large to display}\) | \(3936\) |
Input:
int(x^2*(x^4+x^3)^(1/4)/(x^4-1),x,method=_RETURNVERBOSE)
Output:
-ln(((x^3*(1+x))^(1/4)-x)/x)+2*arctan((x^3*(1+x))^(1/4)/x)+ln((x+(x^3*(1+x ))^(1/4))/x)-1/4*2^(1/4)*ln((2^(1/4)*x+(x^3*(1+x))^(1/4))/(-2^(1/4)*x+(x^3 *(1+x))^(1/4)))-1/2*2^(1/4)*arctan(1/2*2^(3/4)*(x^3*(1+x))^(1/4)/x)+1/4*su m((_R^4-2)*ln((-_R*x+(x^3*(1+x))^(1/4))/x)/_R^3/(_R^4-1),_R=RootOf(_Z^8-2* _Z^4+2))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.10 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.24 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\frac {1}{8} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {8^{\frac {3}{4}} {\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{4 \, {\left (x^{3} + x^{2}\right )}}\right ) - \frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (\frac {8^{\frac {3}{4}} x + 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (-\frac {8^{\frac {3}{4}} x - 4 \, {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {i + 1}} \log \left (\frac {x \sqrt {-\sqrt {i + 1}} + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \sqrt {-\sqrt {i + 1}} \log \left (-\frac {x \sqrt {-\sqrt {i + 1}} - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {-i + 1}} \log \left (\frac {x \sqrt {-\sqrt {-i + 1}} + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \sqrt {-\sqrt {-i + 1}} \log \left (-\frac {x \sqrt {-\sqrt {-i + 1}} - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \left (i + 1\right )^{\frac {1}{4}} \log \left (\frac {\left (i + 1\right )^{\frac {1}{4}} x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \left (i + 1\right )^{\frac {1}{4}} \log \left (-\frac {\left (i + 1\right )^{\frac {1}{4}} x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \left (-i + 1\right )^{\frac {1}{4}} \log \left (\frac {\left (-i + 1\right )^{\frac {1}{4}} x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \left (-i + 1\right )^{\frac {1}{4}} \log \left (-\frac {\left (-i + 1\right )^{\frac {1}{4}} x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 2 \, \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {3}{4}}}{x^{3} + x^{2}}\right ) + \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate(x^2*(x^4+x^3)^(1/4)/(x^4-1),x, algorithm="fricas")
Output:
1/8*8^(3/4)*arctan(1/4*8^(3/4)*(x^4 + x^3)^(3/4)/(x^3 + x^2)) - 1/16*8^(3/ 4)*log((8^(3/4)*x + 4*(x^4 + x^3)^(1/4))/x) + 1/16*8^(3/4)*log(-(8^(3/4)*x - 4*(x^4 + x^3)^(1/4))/x) - 1/4*sqrt(-sqrt(I + 1))*log((x*sqrt(-sqrt(I + 1)) + (x^4 + x^3)^(1/4))/x) + 1/4*sqrt(-sqrt(I + 1))*log(-(x*sqrt(-sqrt(I + 1)) - (x^4 + x^3)^(1/4))/x) - 1/4*sqrt(-sqrt(-I + 1))*log((x*sqrt(-sqrt( -I + 1)) + (x^4 + x^3)^(1/4))/x) + 1/4*sqrt(-sqrt(-I + 1))*log(-(x*sqrt(-s qrt(-I + 1)) - (x^4 + x^3)^(1/4))/x) - 1/4*(I + 1)^(1/4)*log(((I + 1)^(1/4 )*x + (x^4 + x^3)^(1/4))/x) + 1/4*(I + 1)^(1/4)*log(-((I + 1)^(1/4)*x - (x ^4 + x^3)^(1/4))/x) - 1/4*(-I + 1)^(1/4)*log(((-I + 1)^(1/4)*x + (x^4 + x^ 3)^(1/4))/x) + 1/4*(-I + 1)^(1/4)*log(-((-I + 1)^(1/4)*x - (x^4 + x^3)^(1/ 4))/x) - 2*arctan((x^4 + x^3)^(3/4)/(x^3 + x^2)) + log((x + (x^4 + x^3)^(1 /4))/x) - log(-(x - (x^4 + x^3)^(1/4))/x)
Not integrable
Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.15 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\int \frac {x^{2} \sqrt [4]{x^{3} \left (x + 1\right )}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \] Input:
integrate(x**2*(x**4+x**3)**(1/4)/(x**4-1),x)
Output:
Integral(x**2*(x**3*(x + 1))**(1/4)/((x - 1)*(x + 1)*(x**2 + 1)), x)
Not integrable
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.13 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\int { \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} x^{2}}{x^{4} - 1} \,d x } \] Input:
integrate(x^2*(x^4+x^3)^(1/4)/(x^4-1),x, algorithm="maxima")
Output:
integrate((x^4 + x^3)^(1/4)*x^2/(x^4 - 1), x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.27 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.49 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\text {Too large to display} \] Input:
integrate(x^2*(x^4+x^3)^(1/4)/(x^4-1),x, algorithm="giac")
Output:
-1/2*2^(1/4)*arctan(1/2*2^(3/4)*(1/x + 1)^(1/4)) - 1/4*2^(1/4)*log(2^(1/4) + (1/x + 1)^(1/4)) + 2*(-1/4096*I + 1/4096)^(1/4)*log(I*(9807971461541688 6934934209737619787751599303819750539264*I - 98079714615416886934934209737 619787751599303819750539264)^(1/4) - (70368744177664*I - 70368744177664)*( 1/x + 1)^(1/4)) - 2*(-1/4096*I + 1/4096)^(1/4)*log(I*(98079714615416886934 934209737619787751599303819750539264*I - 980797146154168869349342097376197 87751599303819750539264)^(1/4) + (70368744177664*I - 70368744177664)*(1/x + 1)^(1/4)) - 2*I*(1/4096*I + 1/4096)^(1/4)*log(I*(-9807971461541688693493 4209737619787751599303819750539264*I - 98079714615416886934934209737619787 751599303819750539264)^(1/4) - (70368744177664*I - 70368744177664)*(1/x + 1)^(1/4)) + 2*I*(1/4096*I + 1/4096)^(1/4)*log(I*(-980797146154168869349342 09737619787751599303819750539264*I - 9807971461541688693493420973761978775 1599303819750539264)^(1/4) + (70368744177664*I - 70368744177664)*(1/x + 1) ^(1/4)) + 2*(1/4096*I + 1/4096)^(1/4)*log(-I*(-980797146154168869349342097 37619787751599303819750539264*I - 9807971461541688693493420973761978775159 9303819750539264)^(1/4) + (70368744177664*I + 70368744177664)*(1/x + 1)^(1 /4)) - 2*(1/4096*I + 1/4096)^(1/4)*log(-I*(-980797146154168869349342097376 19787751599303819750539264*I - 9807971461541688693493420973761978775159930 3819750539264)^(1/4) - (70368744177664*I + 70368744177664)*(1/x + 1)^(1/4) ) + 16*I*(-1/16777216*I + 1/16777216)^(1/4)*log(-I*(1503067252975253265...
Not integrable
Time = 8.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.13 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\int \frac {x^2\,{\left (x^4+x^3\right )}^{1/4}}{x^4-1} \,d x \] Input:
int((x^2*(x^3 + x^4)^(1/4))/(x^4 - 1),x)
Output:
int((x^2*(x^3 + x^4)^(1/4))/(x^4 - 1), x)
Not integrable
Time = 1.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.10 \[ \int \frac {x^2 \sqrt [4]{x^3+x^4}}{-1+x^4} \, dx=\int \frac {x^{\frac {11}{4}} \left (x +1\right )^{\frac {1}{4}}}{x^{4}-1}d x \] Input:
int(x^2*(x^4+x^3)^(1/4)/(x^4-1),x)
Output:
int((x**(3/4)*(x + 1)**(1/4)*x**2)/(x**4 - 1),x)