Integrand size = 38, antiderivative size = 308 \[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {5 \sqrt {3} \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}}{12 \sqrt [3]{10}-2 \sqrt [3]{10} x-2 \sqrt [3]{10} x^2+5 \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}}\right )}{5 \sqrt [3]{10}}+\frac {\log \left (-6 \sqrt [3]{10}+\sqrt [3]{10} x+\sqrt [3]{10} x^2+5 \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (36\ 10^{2/3}-12\ 10^{2/3} x-11\ 10^{2/3} x^2+2\ 10^{2/3} x^3+10^{2/3} x^4+\left (30 \sqrt [3]{10}-5 \sqrt [3]{10} x-5 \sqrt [3]{10} x^2\right ) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}+25 \left (27+27 x+36 x^2+28 x^3+9 x^4+x^5\right )^{2/3}\right )}{10 \sqrt [3]{10}} \]
1/50*3^(1/2)*arctan(5*3^(1/2)*(x^5+9*x^4+28*x^3+36*x^2+27*x+27)^(1/3)/(12* 10^(1/3)-2*10^(1/3)*x-2*10^(1/3)*x^2+5*(x^5+9*x^4+28*x^3+36*x^2+27*x+27)^( 1/3)))*10^(2/3)+1/50*ln(-6*10^(1/3)+10^(1/3)*x+10^(1/3)*x^2+5*(x^5+9*x^4+2 8*x^3+36*x^2+27*x+27)^(1/3))*10^(2/3)-1/100*ln(36*10^(2/3)-12*10^(2/3)*x-1 1*10^(2/3)*x^2+2*10^(2/3)*x^3+10^(2/3)*x^4+(30*10^(1/3)-5*10^(1/3)*x-5*10^ (1/3)*x^2)*(x^5+9*x^4+28*x^3+36*x^2+27*x+27)^(1/3)+25*(x^5+9*x^4+28*x^3+36 *x^2+27*x+27)^(2/3))*10^(2/3)
Time = 0.07 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.57 \[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=-\frac {(3+x) \sqrt [3]{1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {4 \sqrt [3]{10}-2 \sqrt [3]{10} x+5 \sqrt [3]{1+x^2}}{5 \sqrt {3} \sqrt [3]{1+x^2}}\right )-2 \log \left (-2 \sqrt [3]{10}+\sqrt [3]{10} x+5 \sqrt [3]{1+x^2}\right )+\log \left (4\ 10^{2/3}-4\ 10^{2/3} x+10^{2/3} x^2-5 \sqrt [3]{10} (-2+x) \sqrt [3]{1+x^2}+25 \left (1+x^2\right )^{2/3}\right )\right )}{10 \sqrt [3]{10} \sqrt [3]{(3+x)^3 \left (1+x^2\right )}} \]
-1/10*((3 + x)*(1 + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(4*10^(1/3) - 2*10^(1/3)* x + 5*(1 + x^2)^(1/3))/(5*Sqrt[3]*(1 + x^2)^(1/3))] - 2*Log[-2*10^(1/3) + 10^(1/3)*x + 5*(1 + x^2)^(1/3)] + Log[4*10^(2/3) - 4*10^(2/3)*x + 10^(2/3) *x^2 - 5*10^(1/3)*(-2 + x)*(1 + x^2)^(1/3) + 25*(1 + x^2)^(2/3)]))/(10^(1/ 3)*((3 + x)^3*(1 + x^2))^(1/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.12 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {7239, 7270, 7293, 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.
| \(\displaystyle \int \frac {x+1}{(2 x+1) \sqrt [3]{x^5+9 x^4+28 x^3+36 x^2+27 x+27}} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x+1}{(2 x+1) \sqrt [3]{(x+3)^3 \left (x^2+1\right )}}dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {(x+3) \sqrt [3]{x^2+1} \int \frac {x+1}{(x+3) (2 x+1) \sqrt [3]{x^2+1}}dx}{\sqrt [3]{(x+3)^3 \left (x^2+1\right )}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {(x+3) \sqrt [3]{x^2+1} \int \left (\frac {2}{5 (x+3) \sqrt [3]{x^2+1}}+\frac {1}{5 (2 x+1) \sqrt [3]{x^2+1}}\right )dx}{\sqrt [3]{(x+3)^3 \left (x^2+1\right )}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(x+3) \sqrt [3]{x^2+1} \left (\frac {2}{15} x \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},\frac {x^2}{9},-x^2\right )+\frac {1}{5} x \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},4 x^2,-x^2\right )+\frac {\sqrt {3} \arctan \left (\frac {2^{2/3} \sqrt [3]{x^2+1}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )}{5 \sqrt [3]{10}}+\frac {\sqrt {3} \arctan \left (\frac {2\ 2^{2/3} \sqrt [3]{x^2+1}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )}{10 \sqrt [3]{10}}-\frac {\log \left (1-4 x^2\right )}{20 \sqrt [3]{10}}-\frac {\log \left (9-x^2\right )}{10 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-2 \sqrt [3]{x^2+1}\right )}{20 \sqrt [3]{10}}+\frac {3 \log \left (\sqrt [3]{10}-\sqrt [3]{x^2+1}\right )}{10 \sqrt [3]{10}}\right )}{\sqrt [3]{(x+3)^3 \left (x^2+1\right )}}\) |
((3 + x)*(1 + x^2)^(1/3)*((2*x*AppellF1[1/2, 1, 1/3, 3/2, x^2/9, -x^2])/15 + (x*AppellF1[1/2, 1, 1/3, 3/2, 4*x^2, -x^2])/5 + (Sqrt[3]*ArcTan[(5^(1/3 ) + 2^(2/3)*(1 + x^2)^(1/3))/(Sqrt[3]*5^(1/3))])/(5*10^(1/3)) + (Sqrt[3]*A rcTan[(5^(1/3) + 2*2^(2/3)*(1 + x^2)^(1/3))/(Sqrt[3]*5^(1/3))])/(10*10^(1/ 3)) - Log[1 - 4*x^2]/(20*10^(1/3)) - Log[9 - x^2]/(10*10^(1/3)) + (3*Log[1 0^(1/3) - 2*(1 + x^2)^(1/3)])/(20*10^(1/3)) + (3*Log[10^(1/3) - (1 + x^2)^ (1/3)])/(10*10^(1/3))))/((3 + x)^3*(1 + x^2))^(1/3)
3.29.75.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 10.52 (sec) , antiderivative size = 3732, normalized size of antiderivative = 12.12
-1/50*ln(-(640746115263450*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3 -100)+2500*_Z^2)+21887249427900*(x^5+9*x^4+28*x^3+36*x^2+27*x+27)^(2/3)-17 78460897600*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2 )*x^5-72330055488*RootOf(_Z^3-100)*x^5+9561129209820*RootOf(_Z^3-100)*x+56 966325626250*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^ 2)*x^4+368474867221500*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100 )+2500*_Z^2)*x^3+551878647286500*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootO f(_Z^3-100)+2500*_Z^2)*x^2+235090299901500*RootOf(81*RootOf(_Z^3-100)^2+45 0*_Z*RootOf(_Z^3-100)+2500*_Z^2)*x+2316822089850*RootOf(_Z^3-100)*x^4+1498 5883371420*RootOf(_Z^3-100)*x^3+22444920343620*RootOf(_Z^3-100)*x^2-177752 9944800*(x^5+9*x^4+28*x^3+36*x^2+27*x+27)^(2/3)*RootOf(81*RootOf(_Z^3-100) ^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2)*RootOf(_Z^3-100)^2*x+1235042290000*R ootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z ^3-100)^2*x^4+50229205200*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3- 100)+2500*_Z^2)*RootOf(_Z^3-100)^3*x^4+13894225762500*RootOf(81*RootOf(_Z^ 3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100)^2*x^2+56507 8558500*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2)*Ro otOf(_Z^3-100)^3*x^2+26059162803786*RootOf(_Z^3-100)+154380286250*RootOf(8 1*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+2500*_Z^2)^2*RootOf(_Z^3-100) ^2*x^5+6278650650*RootOf(81*RootOf(_Z^3-100)^2+450*_Z*RootOf(_Z^3-100)+...
Exception generated. \[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\int \frac {x + 1}{\sqrt [3]{\left (x + 3\right )^{3} \left (x^{2} + 1\right )} \left (2 x + 1\right )}\, dx \]
\[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\int { \frac {x + 1}{{\left (x^{5} + 9 \, x^{4} + 28 \, x^{3} + 36 \, x^{2} + 27 \, x + 27\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
\[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\int { \frac {x + 1}{{\left (x^{5} + 9 \, x^{4} + 28 \, x^{3} + 36 \, x^{2} + 27 \, x + 27\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1+x}{(1+2 x) \sqrt [3]{27+27 x+36 x^2+28 x^3+9 x^4+x^5}} \, dx=\int \frac {x+1}{\left (2\,x+1\right )\,{\left (x^5+9\,x^4+28\,x^3+36\,x^2+27\,x+27\right )}^{1/3}} \,d x \]