Integrand size = 26, antiderivative size = 191 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\frac {a x \sqrt [3]{x+x^3}}{2 c}-\frac {a \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 \sqrt {3} c}-\frac {a \log \left (-x+\sqrt [3]{x+x^3}\right )}{6 c}+\frac {a \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{12 c}+\frac {(-b c+a d) \text {RootSum}\left [c-d+3 d \text {$\#$1}^3-3 d \text {$\#$1}^6+d \text {$\#$1}^9\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [3]{x+x^3}-x \text {$\#$1}\right ) \text {$\#$1}}{-1+\text {$\#$1}^3}\&\right ]}{6 c d} \] Output:
Unintegrable
Time = 0.00 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\frac {\sqrt [3]{x+x^3} \left (a d \left (6 x^{4/3} \sqrt [3]{1+x^2}-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )-2 \log \left (c \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )\right )+\log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )+(-2 b c+2 a d) \text {RootSum}\left [c-d+3 d \text {$\#$1}^3-3 d \text {$\#$1}^6+d \text {$\#$1}^9\&,\frac {-2 \log \left (\sqrt [3]{x}\right ) \text {$\#$1}+\log \left (\sqrt [3]{1+x^2}-x^{2/3} \text {$\#$1}\right ) \text {$\#$1}}{-1+\text {$\#$1}^3}\&\right ]\right )}{12 c d \sqrt [3]{x} \sqrt [3]{1+x^2}} \] Input:
Integrate[((x + x^3)^(1/3)*(b + a*x^6))/(d + c*x^6),x]
Output:
((x + x^3)^(1/3)*(a*d*(6*x^(4/3)*(1 + x^2)^(1/3) - 2*Sqrt[3]*ArcTan[(Sqrt[ 3]*x^(2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))] - 2*Log[c*(-x^(2/3) + (1 + x^2) ^(1/3))] + Log[x^(4/3) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)]) + (-2 *b*c + 2*a*d)*RootSum[c - d + 3*d*#1^3 - 3*d*#1^6 + d*#1^9 & , (-2*Log[x^( 1/3)]*#1 + Log[(1 + x^2)^(1/3) - x^(2/3)*#1]*#1)/(-1 + #1^3) & ]))/(12*c*d *x^(1/3)*(1 + x^2)^(1/3))
Leaf count is larger than twice the leaf count of optimal. \(1249\) vs. \(2(191)=382\).
Time = 5.16 (sec) , antiderivative size = 1249, normalized size of antiderivative = 6.54, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2467, 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.
| \(\displaystyle \int \frac {\sqrt [3]{x^3+x} \left (a x^6+b\right )}{c x^6+d} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x^3+x} \int \frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \left (a x^6+b\right )}{c x^6+d}dx}{\sqrt [3]{x} \sqrt [3]{x^2+1}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {3 \sqrt [3]{x^3+x} \int \frac {x \sqrt [3]{x^2+1} \left (a x^6+b\right )}{c x^6+d}d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{x^2+1}}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {3 \sqrt [3]{x^3+x} \int \left (\frac {a \sqrt [3]{x^2+1} x}{c}+\frac {(b c-a d) \sqrt [3]{x^2+1} x}{c \left (c x^6+d\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{x} \sqrt [3]{x^2+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \sqrt [3]{x^3+x} \left (\frac {a \sqrt [3]{x^2+1} x^{4/3}}{6 c}-\frac {(-1)^{2/3} (b c-a d) \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{2/3}}+\frac {\sqrt [3]{-1} (b c-a d) \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{2/3}}-\frac {(b c-a d) \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{2/3}}-\frac {a \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{6 \sqrt {3} c}-\frac {(-1)^{2/3} \sqrt [3]{\sqrt [3]{-1} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \arctan \left (\frac {1-\frac {2 \sqrt [3]{\sqrt [3]{-1} \sqrt [3]{-c}-\sqrt [3]{d}} x^{2/3}}{\sqrt [9]{d} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{7/9}}+\frac {\sqrt [3]{-1} \sqrt [3]{-(-1)^{2/3} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \arctan \left (\frac {1-\frac {2 \sqrt [3]{-(-1)^{2/3} \sqrt [3]{-c}-\sqrt [3]{d}} x^{2/3}}{\sqrt [9]{d} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{7/9}}+\frac {\sqrt [3]{\sqrt [3]{-c}+\sqrt [3]{d}} (b c-a d) \arctan \left (\frac {\frac {2 \sqrt [3]{\sqrt [3]{-c}+\sqrt [3]{d}} x^{2/3}}{\sqrt [9]{d} \sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{18 \sqrt {3} (-c)^{4/3} d^{7/9}}-\frac {\sqrt [3]{\sqrt [3]{-c}+\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{d}-\sqrt [3]{-c} x^2\right )}{108 (-c)^{4/3} d^{7/9}}-\frac {\sqrt [3]{-1} \sqrt [3]{-(-1)^{2/3} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{-c} x^2+\sqrt [3]{-1} \sqrt [3]{d}\right )}{108 (-c)^{4/3} d^{7/9}}+\frac {(-1)^{2/3} \sqrt [3]{\sqrt [3]{-1} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{-c} x^2-(-1)^{2/3} \sqrt [3]{d}\right )}{108 (-c)^{4/3} d^{7/9}}-\frac {(-1)^{2/3} (b c-a d) \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{2/3}}+\frac {\sqrt [3]{-1} (b c-a d) \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{2/3}}-\frac {(b c-a d) \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{2/3}}-\frac {a \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{12 c}+\frac {\sqrt [3]{\sqrt [3]{-c}+\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{\sqrt [3]{-c}+\sqrt [3]{d}} x^{2/3}-\sqrt [9]{d} \sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{7/9}}-\frac {(-1)^{2/3} \sqrt [3]{\sqrt [3]{-1} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{\sqrt [3]{-1} \sqrt [3]{-c}-\sqrt [3]{d}} x^{2/3}+\sqrt [9]{d} \sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{7/9}}+\frac {\sqrt [3]{-1} \sqrt [3]{-(-1)^{2/3} \sqrt [3]{-c}-\sqrt [3]{d}} (b c-a d) \log \left (\sqrt [3]{-(-1)^{2/3} \sqrt [3]{-c}-\sqrt [3]{d}} x^{2/3}+\sqrt [9]{d} \sqrt [3]{x^2+1}\right )}{36 (-c)^{4/3} d^{7/9}}\right )}{\sqrt [3]{x} \sqrt [3]{x^2+1}}\) |
Input:
Int[((x + x^3)^(1/3)*(b + a*x^6))/(d + c*x^6),x]
Output:
(3*(x + x^3)^(1/3)*((a*x^(4/3)*(1 + x^2)^(1/3))/(6*c) - (a*ArcTan[(1 + (2* x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(6*Sqrt[3]*c) - ((b*c - a*d)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(18*Sqrt[3]*(-c)^(4/3)*d^(2/3)) + ((-1)^(1/3)*(b*c - a*d)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3] ])/(18*Sqrt[3]*(-c)^(4/3)*d^(2/3)) - ((-1)^(2/3)*(b*c - a*d)*ArcTan[(1 + ( 2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(18*Sqrt[3]*(-c)^(4/3)*d^(2/3)) - (( -1)^(2/3)*((-1)^(1/3)*(-c)^(1/3) - d^(1/3))^(1/3)*(b*c - a*d)*ArcTan[(1 - (2*((-1)^(1/3)*(-c)^(1/3) - d^(1/3))^(1/3)*x^(2/3))/(d^(1/9)*(1 + x^2)^(1/ 3)))/Sqrt[3]])/(18*Sqrt[3]*(-c)^(4/3)*d^(7/9)) + ((-1)^(1/3)*(-((-1)^(2/3) *(-c)^(1/3)) - d^(1/3))^(1/3)*(b*c - a*d)*ArcTan[(1 - (2*(-((-1)^(2/3)*(-c )^(1/3)) - d^(1/3))^(1/3)*x^(2/3))/(d^(1/9)*(1 + x^2)^(1/3)))/Sqrt[3]])/(1 8*Sqrt[3]*(-c)^(4/3)*d^(7/9)) + (((-c)^(1/3) + d^(1/3))^(1/3)*(b*c - a*d)* ArcTan[(1 + (2*((-c)^(1/3) + d^(1/3))^(1/3)*x^(2/3))/(d^(1/9)*(1 + x^2)^(1 /3)))/Sqrt[3]])/(18*Sqrt[3]*(-c)^(4/3)*d^(7/9)) - (((-c)^(1/3) + d^(1/3))^ (1/3)*(b*c - a*d)*Log[d^(1/3) - (-c)^(1/3)*x^2])/(108*(-c)^(4/3)*d^(7/9)) - ((-1)^(1/3)*(-((-1)^(2/3)*(-c)^(1/3)) - d^(1/3))^(1/3)*(b*c - a*d)*Log[( -1)^(1/3)*d^(1/3) + (-c)^(1/3)*x^2])/(108*(-c)^(4/3)*d^(7/9)) + ((-1)^(2/3 )*((-1)^(1/3)*(-c)^(1/3) - d^(1/3))^(1/3)*(b*c - a*d)*Log[-((-1)^(2/3)*d^( 1/3)) + (-c)^(1/3)*x^2])/(108*(-c)^(4/3)*d^(7/9)) - (a*Log[x^(2/3) - (1 + x^2)^(1/3)])/(12*c) - ((b*c - a*d)*Log[x^(2/3) - (1 + x^2)^(1/3)])/(36*...
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]
Time = 0.00 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.08
| method | result | size |
| pseudoelliptic | \(\frac {\left (\left (-a d +b c \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{9} d -3 \textit {\_Z}^{6} d +3 \textit {\_Z}^{3} d +c -d \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}{x}\right ) \textit {\_R}}{\textit {\_R}^{3}-1}\right )+d a \left (-\sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )-3 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right )-\frac {\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{2}\right )\right ) x}{6 \left (-{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) c \left ({\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+x \left (x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )\right ) d}\) | \(206\) |
Input:
int((x^3+x)^(1/3)*(a*x^6+b)/(c*x^6+d),x,method=_RETURNVERBOSE)
Output:
1/6*((-a*d+b*c)*sum(ln((-_R*x+(x*(x^2+1))^(1/3))/x)*_R/(_R^3-1),_R=RootOf( _Z^9*d-3*_Z^6*d+3*_Z^3*d+c-d))+d*a*(-3^(1/2)*arctan(1/3*(2*(x*(x^2+1))^(1/ 3)+x)*3^(1/2)/x)-3*(x*(x^2+1))^(1/3)*x+ln(((x*(x^2+1))^(1/3)-x)/x)-1/2*ln( ((x*(x^2+1))^(2/3)+(x*(x^2+1))^(1/3)*x+x^2)/x^2)))*x/(-(x*(x^2+1))^(1/3)+x )/c/((x*(x^2+1))^(2/3)+x*(x+(x*(x^2+1))^(1/3)))/d
Exception generated. \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((x^3+x)^(1/3)*(a*x^6+b)/(c*x^6+d),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
Timed out. \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\text {Timed out} \] Input:
integrate((x**3+x)**(1/3)*(a*x**6+b)/(c*x**6+d),x)
Output:
Timed out
Not integrable
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\int { \frac {{\left (a x^{6} + b\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}}{c x^{6} + d} \,d x } \] Input:
integrate((x^3+x)^(1/3)*(a*x^6+b)/(c*x^6+d),x, algorithm="maxima")
Output:
integrate((a*x^6 + b)*(x^3 + x)^(1/3)/(c*x^6 + d), x)
Not integrable
Time = 5.26 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.02 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\int { \frac {{\left (a x^{6} + b\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}}{c x^{6} + d} \,d x } \] Input:
integrate((x^3+x)^(1/3)*(a*x^6+b)/(c*x^6+d),x, algorithm="giac")
Output:
sage0*x
Not integrable
Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\int \frac {\left (a\,x^6+b\right )\,{\left (x^3+x\right )}^{1/3}}{c\,x^6+d} \,d x \] Input:
int(((b + a*x^6)*(x + x^3)^(1/3))/(d + c*x^6),x)
Output:
int(((b + a*x^6)*(x + x^3)^(1/3))/(d + c*x^6), x)
Not integrable
Time = 3.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.26 \[ \int \frac {\sqrt [3]{x+x^3} \left (b+a x^6\right )}{d+c x^6} \, dx=\left (\int \frac {x^{\frac {19}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{c \,x^{6}+d}d x \right ) a +\left (\int \frac {x^{\frac {1}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}{c \,x^{6}+d}d x \right ) b \] Input:
int((x^3+x)^(1/3)*(a*x^6+b)/(c*x^6+d),x)
Output:
int((x**(1/3)*(x**2 + 1)**(1/3)*x**6)/(c*x**6 + d),x)*a + int((x**(1/3)*(x **2 + 1)**(1/3))/(c*x**6 + d),x)*b