Integrand size = 48, antiderivative size = 496 \[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {5 \sqrt {3} \sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}}{12 \sqrt [3]{10}+22 \sqrt [3]{10} x-6 \sqrt [3]{10} x^2-4 \sqrt [3]{10} x^3+5 \sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (3+7 x+2 x^2\right )}{5 \sqrt [3]{10}}+\frac {\log \left (9+42 x+61 x^2+28 x^3+4 x^4\right )}{10 \sqrt [3]{10}}+\frac {\log \left (-6 \sqrt [3]{10}-11 \sqrt [3]{10} x+3 \sqrt [3]{10} x^2+2 \sqrt [3]{10} x^3+5 \sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}\right )}{5 \sqrt [3]{10}}-\frac {\log \left (36\ 10^{2/3}+132\ 10^{2/3} x+85\ 10^{2/3} x^2-90\ 10^{2/3} x^3-35\ 10^{2/3} x^4+12\ 10^{2/3} x^5+4\ 10^{2/3} x^6+\left (30 \sqrt [3]{10}+55 \sqrt [3]{10} x-15 \sqrt [3]{10} x^2-10 \sqrt [3]{10} x^3\right ) \sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}+25 \left (27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8\right )^{2/3}\right )}{10 \sqrt [3]{10}} \]
1/50*3^(1/2)*arctan(5*3^(1/2)*(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^ 3+522*x^2+189*x+27)^(1/3)/(12*10^(1/3)+22*10^(1/3)*x-6*10^(1/3)*x^2-4*10^( 1/3)*x^3+5*(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189*x+27) ^(1/3)))*10^(2/3)-1/50*ln(2*x^2+7*x+3)*10^(2/3)+1/100*ln(4*x^4+28*x^3+61*x ^2+42*x+9)*10^(2/3)+1/50*ln(-6*10^(1/3)-11*10^(1/3)*x+3*10^(1/3)*x^2+2*10^ (1/3)*x^3+5*(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189*x+27 )^(1/3))*10^(2/3)-1/100*ln(36*10^(2/3)+132*10^(2/3)*x+85*10^(2/3)*x^2-90*1 0^(2/3)*x^3-35*10^(2/3)*x^4+12*10^(2/3)*x^5+4*10^(2/3)*x^6+(30*10^(1/3)+55 *10^(1/3)*x-15*10^(1/3)*x^2-10*10^(1/3)*x^3)*(8*x^8+84*x^7+338*x^6+679*x^5 +825*x^4+784*x^3+522*x^2+189*x+27)^(1/3)+25*(8*x^8+84*x^7+338*x^6+679*x^5+ 825*x^4+784*x^3+522*x^2+189*x+27)^(2/3))*10^(2/3)
Time = 0.25 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.38 \[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=-\frac {\sqrt [3]{1+x^2} \left (3+7 x+2 x^2\right ) \left (2 \sqrt {3} \arctan \left (\frac {4 \sqrt [3]{10}-2 \sqrt [3]{10} x+5 \sqrt [3]{1+x^2}}{5 \sqrt {3} \sqrt [3]{1+x^2}}\right )-2 \log \left (-2 \sqrt [3]{10}+\sqrt [3]{10} x+5 \sqrt [3]{1+x^2}\right )+\log \left (4\ 10^{2/3}-4\ 10^{2/3} x+10^{2/3} x^2-5 \sqrt [3]{10} (-2+x) \sqrt [3]{1+x^2}+25 \left (1+x^2\right )^{2/3}\right )\right )}{10 \sqrt [3]{10} \sqrt [3]{\left (1+x^2\right ) \left (3+7 x+2 x^2\right )^3}} \]
Integrate[(1 + x)/(27 + 189*x + 522*x^2 + 784*x^3 + 825*x^4 + 679*x^5 + 33 8*x^6 + 84*x^7 + 8*x^8)^(1/3),x]
-1/10*((1 + x^2)^(1/3)*(3 + 7*x + 2*x^2)*(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*1 0^(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)*((1 + x^2)*(3 + 7*x + 2*x^2)^3)^(1/3))
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.
\(\displaystyle \int \frac {x+1}{\sqrt [3]{8 x^8+84 x^7+338 x^6+679 x^5+825 x^4+784 x^3+522 x^2+189 x+27}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x+1}{\sqrt [3]{\left (x^2+1\right ) \left (2 x^2+7 x+3\right )^3}}dx\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle \frac {\sqrt [3]{x^2+1} \left (2 x^2+7 x+3\right ) \int \frac {x+1}{\sqrt [3]{x^2+1} \left (2 x^2+7 x+3\right )}dx}{\sqrt [3]{\left (x^2+1\right ) \left (2 x^2+7 x+3\right )^3}}\) |
\(\Big \downarrow \) 1376 |
\(\displaystyle \frac {\sqrt [3]{x^2+1} \left (2 x^2+7 x+3\right ) \int \frac {x+1}{\sqrt [3]{x^2+1} \left (2 x^2+7 x+3\right )}dx}{\sqrt [3]{\left (x^2+1\right ) \left (2 x^2+7 x+3\right )^3}}\) |
Int[(1 + x)/(27 + 189*x + 522*x^2 + 784*x^3 + 825*x^4 + 679*x^5 + 338*x^6 + 84*x^7 + 8*x^8)^(1/3),x]
3.31.70.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, c, d, e, f, g, h, p, q}, x]
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 = 11.60 (sec) , antiderivative size = 5455, normalized size of antiderivative = 11.00
int((1+x)/(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189*x+27)^ (1/3),x,method=_RETURNVERBOSE)
Exception generated. \[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\text {Exception raised: TypeError} \]
integrate((1+x)/(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189* x+27)^(1/3),x, algorithm="fricas")
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\int \frac {x + 1}{\sqrt [3]{\left (x + 3\right )^{3} \left (2 x + 1\right )^{3} \left (x^{2} + 1\right )}}\, dx \]
integrate((1+x)/(8*x**8+84*x**7+338*x**6+679*x**5+825*x**4+784*x**3+522*x* *2+189*x+27)**(1/3),x)
\[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\int { \frac {x + 1}{{\left (8 \, x^{8} + 84 \, x^{7} + 338 \, x^{6} + 679 \, x^{5} + 825 \, x^{4} + 784 \, x^{3} + 522 \, x^{2} + 189 \, x + 27\right )}^{\frac {1}{3}}} \,d x } \]
integrate((1+x)/(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189* x+27)^(1/3),x, algorithm="maxima")
integrate((x + 1)/(8*x^8 + 84*x^7 + 338*x^6 + 679*x^5 + 825*x^4 + 784*x^3 + 522*x^2 + 189*x + 27)^(1/3), x)
\[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\int { \frac {x + 1}{{\left (8 \, x^{8} + 84 \, x^{7} + 338 \, x^{6} + 679 \, x^{5} + 825 \, x^{4} + 784 \, x^{3} + 522 \, x^{2} + 189 \, x + 27\right )}^{\frac {1}{3}}} \,d x } \]
integrate((1+x)/(8*x^8+84*x^7+338*x^6+679*x^5+825*x^4+784*x^3+522*x^2+189* x+27)^(1/3),x, algorithm="giac")
integrate((x + 1)/(8*x^8 + 84*x^7 + 338*x^6 + 679*x^5 + 825*x^4 + 784*x^3 + 522*x^2 + 189*x + 27)^(1/3), x)
Timed out. \[ \int \frac {1+x}{\sqrt [3]{27+189 x+522 x^2+784 x^3+825 x^4+679 x^5+338 x^6+84 x^7+8 x^8}} \, dx=\int \frac {x+1}{{\left (8\,x^8+84\,x^7+338\,x^6+679\,x^5+825\,x^4+784\,x^3+522\,x^2+189\,x+27\right )}^{1/3}} \,d x \]
int((x + 1)/(189*x + 522*x^2 + 784*x^3 + 825*x^4 + 679*x^5 + 338*x^6 + 84* x^7 + 8*x^8 + 27)^(1/3),x)