Integrand size = 20, antiderivative size = 250 \[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=-\frac {\sqrt [3]{(1-2 x)^2} \left (7+7^{2/3} \sqrt [3]{-1+2 x}\right ) \left (7-7^{2/3} \sqrt [3]{-1+2 x}+\sqrt [3]{7} (-1+2 x)^{2/3}\right )^2 \left (\frac {3 (-1+2 x)^{2/3}}{2^{2/3}}+\sqrt [3]{2} \sqrt {3} 7^{2/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-1+2 x}}{\sqrt {3} \sqrt [3]{7}}\right )+\sqrt [3]{2} 7^{2/3} \log \left (7+7^{2/3} \sqrt [3]{-1+2 x}\right )-\left (\frac {7}{2}\right )^{2/3} \log \left (-7+7^{2/3} \sqrt [3]{-1+2 x}-\sqrt [3]{7} (-1+2 x)^{2/3}\right )\right )}{14 (3+x) \sqrt [3]{-1+2 x} \left (\sqrt [3]{7}-2 \sqrt [3]{7} x-7 \sqrt [3]{-1+2 x}+(-7+14 x)^{2/3}\right )} \]
-1/14*((1-2*x)^2)^(1/3)*(7+7^(2/3)*(-1+2*x)^(1/3))*(7-7^(2/3)*(-1+2*x)^(1/ 3)+7^(1/3)*(-1+2*x)^(2/3))^2*(3/2*(-1+2*x)^(2/3)*2^(1/3)-2^(1/3)*3^(1/2)*7 ^(2/3)*arctan(-1/3*3^(1/2)+2/21*(-1+2*x)^(1/3)*3^(1/2)*7^(2/3))+2^(1/3)*7^ (2/3)*ln(7+7^(2/3)*(-1+2*x)^(1/3))-1/2*2^(1/3)*7^(2/3)*ln(-7+7^(2/3)*(-1+2 *x)^(1/3)-7^(1/3)*(-1+2*x)^(2/3)))/(3+x)/(-1+2*x)^(1/3)/(7^(1/3)-2*7^(1/3) *x-7*(-1+2*x)^(1/3)+(-7+14*x)^(2/3))
Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=\frac {\sqrt [3]{(1-2 x)^2} \left (3 (-1+2 x)^{2/3}+2 \sqrt {3} 7^{2/3} \arctan \left (\frac {7-2\ 7^{2/3} \sqrt [3]{-1+2 x}}{7 \sqrt {3}}\right )+2\ 7^{2/3} \log \left (7+7^{2/3} \sqrt [3]{-1+2 x}\right )-7^{2/3} \log \left (-7+7^{2/3} \sqrt [3]{-1+2 x}-\sqrt [3]{7} (-1+2 x)^{2/3}\right )\right )}{(-2+4 x)^{2/3}} \]
(((1 - 2*x)^2)^(1/3)*(3*(-1 + 2*x)^(2/3) + 2*Sqrt[3]*7^(2/3)*ArcTan[(7 - 2 *7^(2/3)*(-1 + 2*x)^(1/3))/(7*Sqrt[3])] + 2*7^(2/3)*Log[7 + 7^(2/3)*(-1 + 2*x)^(1/3)] - 7^(2/3)*Log[-7 + 7^(2/3)*(-1 + 2*x)^(1/3) - 7^(1/3)*(-1 + 2* x)^(2/3)]))/(-2 + 4*x)^(2/3)
Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.48, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1102, 27, 60, 68, 16, 1082, 217}
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.
| \(\displaystyle \int \frac {\sqrt [3]{8 x^2-8 x+2}}{x+3} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {\sqrt [3]{4 x^2-4 x+1} \int \frac {2 \sqrt [3]{2} (2 x-1)^{2/3}}{x+3}dx}{2 (2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \int \frac {(2 x-1)^{2/3}}{x+3}dx}{(2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \left (\frac {3}{2} (2 x-1)^{2/3}-7 \int \frac {1}{(x+3) \sqrt [3]{2 x-1}}dx\right )}{(2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 68 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \left (\frac {3}{2} (2 x-1)^{2/3}-7 \left (-\frac {3 \int \frac {1}{\sqrt [3]{2 x-1}+\sqrt [3]{7}}d\sqrt [3]{2 x-1}}{2 \sqrt [3]{7}}+\frac {3}{2} \int \frac {1}{(2 x-1)^{2/3}-\sqrt [3]{7} \sqrt [3]{2 x-1}+7^{2/3}}d\sqrt [3]{2 x-1}+\frac {\log (x+3)}{2 \sqrt [3]{7}}\right )\right )}{(2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \left (\frac {3}{2} (2 x-1)^{2/3}-7 \left (\frac {3}{2} \int \frac {1}{(2 x-1)^{2/3}-\sqrt [3]{7} \sqrt [3]{2 x-1}+7^{2/3}}d\sqrt [3]{2 x-1}+\frac {\log (x+3)}{2 \sqrt [3]{7}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{7}\right )}{2 \sqrt [3]{7}}\right )\right )}{(2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \left (\frac {3}{2} (2 x-1)^{2/3}-7 \left (\frac {3 \int \frac {1}{-(2 x-1)^{2/3}-3}d\left (1-\frac {2 \sqrt [3]{2 x-1}}{\sqrt [3]{7}}\right )}{\sqrt [3]{7}}+\frac {\log (x+3)}{2 \sqrt [3]{7}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{7}\right )}{2 \sqrt [3]{7}}\right )\right )}{(2 x-1)^{2/3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{4 x^2-4 x+1} \left (\frac {3}{2} (2 x-1)^{2/3}-7 \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2 x-1}}{\sqrt [3]{7}}}{\sqrt {3}}\right )}{\sqrt [3]{7}}+\frac {\log (x+3)}{2 \sqrt [3]{7}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{7}\right )}{2 \sqrt [3]{7}}\right )\right )}{(2 x-1)^{2/3}}\) |
(2^(1/3)*(1 - 4*x + 4*x^2)^(1/3)*((3*(-1 + 2*x)^(2/3))/2 - 7*(-((Sqrt[3]*A rcTan[(1 - (2*(-1 + 2*x)^(1/3))/7^(1/3))/Sqrt[3]])/7^(1/3)) + Log[3 + x]/( 2*7^(1/3)) - (3*Log[7^(1/3) + (-1 + 2*x)^(1/3)])/(2*7^(1/3)))))/(-1 + 2*x) ^(2/3)
3.28.21.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.44
| method | result | size |
| risch | \(\frac {3 \,2^{\frac {1}{3}} \left (\left (-1+2 x \right )^{2}\right )^{\frac {1}{3}}}{2}+\frac {\left (7^{\frac {2}{3}} \ln \left (\left (-1+2 x \right )^{\frac {1}{3}}+7^{\frac {1}{3}}\right )-\frac {7^{\frac {2}{3}} \ln \left (\left (-1+2 x \right )^{\frac {2}{3}}-7^{\frac {1}{3}} \left (-1+2 x \right )^{\frac {1}{3}}+7^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, 7^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,7^{\frac {2}{3}} \left (-1+2 x \right )^{\frac {1}{3}}}{7}-1\right )}{3}\right )\right ) 2^{\frac {1}{3}} \left (\left (-1+2 x \right )^{2}\right )^{\frac {1}{3}}}{\left (-1+2 x \right )^{\frac {2}{3}}}\) | \(110\) |
| trager | \(\text {Expression too large to display}\) | \(1484\) |
3/2*2^(1/3)*((-1+2*x)^2)^(1/3)+(7^(2/3)*ln((-1+2*x)^(1/3)+7^(1/3))-1/2*7^( 2/3)*ln((-1+2*x)^(2/3)-7^(1/3)*(-1+2*x)^(1/3)+7^(2/3))-3^(1/2)*7^(2/3)*arc tan(1/3*3^(1/2)*(2/7*7^(2/3)*(-1+2*x)^(1/3)-1)))*2^(1/3)*((-1+2*x)^2)^(1/3 )/(-1+2*x)^(2/3)
Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=98^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {98^{\frac {2}{3}} \sqrt {3} {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} - 7 \, \sqrt {3} {\left (2 \, x - 1\right )}}{21 \, {\left (2 \, x - 1\right )}}\right ) - \frac {1}{2} \cdot 98^{\frac {1}{3}} \log \left (\frac {98^{\frac {2}{3}} {\left (4 \, x^{2} - 4 \, x + 1\right )} - 7 \cdot 98^{\frac {1}{3}} {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} {\left (2 \, x - 1\right )} + 49 \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {2}{3}}}{4 \, x^{2} - 4 \, x + 1}\right ) + 98^{\frac {1}{3}} \log \left (\frac {98^{\frac {1}{3}} {\left (2 \, x - 1\right )} + 7 \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{2 \, x - 1}\right ) + \frac {3}{2} \, {\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}} \]
98^(1/3)*sqrt(3)*arctan(1/21*(98^(2/3)*sqrt(3)*(8*x^2 - 8*x + 2)^(1/3) - 7 *sqrt(3)*(2*x - 1))/(2*x - 1)) - 1/2*98^(1/3)*log((98^(2/3)*(4*x^2 - 4*x + 1) - 7*98^(1/3)*(8*x^2 - 8*x + 2)^(1/3)*(2*x - 1) + 49*(8*x^2 - 8*x + 2)^ (2/3))/(4*x^2 - 4*x + 1)) + 98^(1/3)*log((98^(1/3)*(2*x - 1) + 7*(8*x^2 - 8*x + 2)^(1/3))/(2*x - 1)) + 3/2*(8*x^2 - 8*x + 2)^(1/3)
\[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=\sqrt [3]{2} \int \frac {\sqrt [3]{4 x^{2} - 4 x + 1}}{x + 3}\, dx \]
\[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=\int { \frac {{\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{x + 3} \,d x } \]
\[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=\int { \frac {{\left (8 \, x^{2} - 8 \, x + 2\right )}^{\frac {1}{3}}}{x + 3} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{2-8 x+8 x^2}}{3+x} \, dx=\int \frac {{\left (8\,x^2-8\,x+2\right )}^{1/3}}{x+3} \,d x \]