Integrand size = 7, antiderivative size = 78 \[ \int \frac {1}{5+x^3} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{5}-2 x}{\sqrt {3} \sqrt [3]{5}}\right )}{\sqrt {3} 5^{2/3}}+\frac {\log \left (\sqrt [3]{5}+x\right )}{3\ 5^{2/3}}-\frac {\log \left (5^{2/3}-\sqrt [3]{5} x+x^2\right )}{6\ 5^{2/3}} \] Output:
1/15*ln(5^(1/3)+x)*5^(1/3)-1/30*ln(5^(2/3)-5^(1/3)*x+x^2)*5^(1/3)-1/15*arc tan(1/15*(5^(1/3)-2*x)*5^(2/3)*3^(1/2))*5^(1/3)*3^(1/2)
Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5+x^3} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {-5+2\ 5^{2/3} x}{5 \sqrt {3}}\right )+2 \log \left (5+5^{2/3} x\right )-\log \left (5-5^{2/3} x+\sqrt [3]{5} x^2\right )}{6\ 5^{2/3}} \] Input:
Integrate[(5 + x^3)^(-1),x]
Output:
(2*Sqrt[3]*ArcTan[(-5 + 2*5^(2/3)*x)/(5*Sqrt[3])] + 2*Log[5 + 5^(2/3)*x] - Log[5 - 5^(2/3)*x + 5^(1/3)*x^2])/(6*5^(2/3))
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {750, 16, 1142, 25, 1082, 217, 1103}
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.
\(\displaystyle \int \frac {1}{x^3+5} \, dx\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {\int \frac {2 \sqrt [3]{5}-x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx}{3\ 5^{2/3}}+\frac {\int \frac {1}{x+\sqrt [3]{5}}dx}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\int \frac {2 \sqrt [3]{5}-x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\frac {3}{2} \sqrt [3]{5} \int \frac {1}{x^2-\sqrt [3]{5} x+5^{2/3}}dx-\frac {1}{2} \int -\frac {\sqrt [3]{5}-2 x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {3}{2} \sqrt [3]{5} \int \frac {1}{x^2-\sqrt [3]{5} x+5^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{5}-2 x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt [3]{5}-2 x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx+3 \int \frac {1}{-\left (1-\frac {2 x}{\sqrt [3]{5}}\right )^2-3}d\left (1-\frac {2 x}{\sqrt [3]{5}}\right )}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt [3]{5}-2 x}{x^2-\sqrt [3]{5} x+5^{2/3}}dx-\sqrt {3} \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{5}}}{\sqrt {3}}\right )}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {-\sqrt {3} \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{5}}}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^2-\sqrt [3]{5} x+5^{2/3}\right )}{3\ 5^{2/3}}+\frac {\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}\) |
Input:
Int[(5 + x^3)^(-1),x]
Output:
Log[5^(1/3) + x]/(3*5^(2/3)) + (-(Sqrt[3]*ArcTan[(1 - (2*x)/5^(1/3))/Sqrt[ 3]]) - Log[5^(2/3) - 5^(1/3)*x + x^2]/2)/(3*5^(2/3))
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
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_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+5\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3}\) | \(22\) |
default | \(\frac {\ln \left (5^{\frac {1}{3}}+x \right ) 5^{\frac {1}{3}}}{15}-\frac {\ln \left (5^{\frac {2}{3}}-5^{\frac {1}{3}} x +x^{2}\right ) 5^{\frac {1}{3}}}{30}+\frac {5^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,5^{\frac {2}{3}} x}{5}-1\right )}{3}\right )}{15}\) | \(54\) |
meijerg | \(\frac {5^{\frac {1}{3}} \left (\frac {x \ln \left (1+\frac {5^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{5}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \ln \left (1-\frac {5^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{5}+\frac {5^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{5}\right )}{2 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, 5^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{10-5^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{15}\) | \(96\) |
Input:
int(1/(x^3+5),x,method=_RETURNVERBOSE)
Output:
1/3*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3+5))
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5+x^3} \, dx=\frac {1}{5} \cdot 25^{\frac {1}{6}} \sqrt {\frac {1}{3}} \arctan \left (\frac {1}{25} \cdot 25^{\frac {1}{6}} \sqrt {\frac {1}{3}} {\left (2 \cdot 25^{\frac {2}{3}} x - 5 \cdot 25^{\frac {1}{3}}\right )}\right ) - \frac {1}{150} \cdot 25^{\frac {2}{3}} \log \left (5 \, x^{2} - 25^{\frac {2}{3}} x + 5 \cdot 25^{\frac {1}{3}}\right ) + \frac {1}{75} \cdot 25^{\frac {2}{3}} \log \left (5 \, x + 25^{\frac {2}{3}}\right ) \] Input:
integrate(1/(x^3+5),x, algorithm="fricas")
Output:
1/5*25^(1/6)*sqrt(1/3)*arctan(1/25*25^(1/6)*sqrt(1/3)*(2*25^(2/3)*x - 5*25 ^(1/3))) - 1/150*25^(2/3)*log(5*x^2 - 25^(2/3)*x + 5*25^(1/3)) + 1/75*25^( 2/3)*log(5*x + 25^(2/3))
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {1}{5+x^3} \, dx=\frac {\sqrt [3]{5} \log {\left (x + \sqrt [3]{5} \right )}}{15} - \frac {\sqrt [3]{5} \log {\left (x^{2} - \sqrt [3]{5} x + 5^{\frac {2}{3}} \right )}}{30} + \frac {\sqrt {3} \cdot \sqrt [3]{5} \operatorname {atan}{\left (\frac {2 \sqrt {3} \cdot 5^{\frac {2}{3}} x}{15} - \frac {\sqrt {3}}{3} \right )}}{15} \] Input:
integrate(1/(x**3+5),x)
Output:
5**(1/3)*log(x + 5**(1/3))/15 - 5**(1/3)*log(x**2 - 5**(1/3)*x + 5**(2/3)) /30 + sqrt(3)*5**(1/3)*atan(2*sqrt(3)*5**(2/3)*x/15 - sqrt(3)/3)/15
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.73 \[ \int \frac {1}{5+x^3} \, dx=\frac {1}{15} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {1}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (2 \, x - 5^{\frac {1}{3}}\right )}\right ) - \frac {1}{30} \cdot 5^{\frac {1}{3}} \log \left (x^{2} - 5^{\frac {1}{3}} x + 5^{\frac {2}{3}}\right ) + \frac {1}{15} \cdot 5^{\frac {1}{3}} \log \left (x + 5^{\frac {1}{3}}\right ) \] Input:
integrate(1/(x^3+5),x, algorithm="maxima")
Output:
1/15*5^(1/3)*sqrt(3)*arctan(1/15*5^(2/3)*sqrt(3)*(2*x - 5^(1/3))) - 1/30*5 ^(1/3)*log(x^2 - 5^(1/3)*x + 5^(2/3)) + 1/15*5^(1/3)*log(x + 5^(1/3))
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {1}{5+x^3} \, dx=\frac {1}{15} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {1}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (2 \, x - 5^{\frac {1}{3}}\right )}\right ) - \frac {1}{30} \cdot 5^{\frac {1}{3}} \log \left (x^{2} - 5^{\frac {1}{3}} x + 5^{\frac {2}{3}}\right ) + \frac {1}{15} \cdot 5^{\frac {1}{3}} \log \left ({\left | x + 5^{\frac {1}{3}} \right |}\right ) \] Input:
integrate(1/(x^3+5),x, algorithm="giac")
Output:
1/15*5^(1/3)*sqrt(3)*arctan(1/15*5^(2/3)*sqrt(3)*(2*x - 5^(1/3))) - 1/30*5 ^(1/3)*log(x^2 - 5^(1/3)*x + 5^(2/3)) + 1/15*5^(1/3)*log(abs(x + 5^(1/3)))
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {1}{5+x^3} \, dx=\frac {5^{1/3}\,\ln \left (x+5^{1/3}\right )}{15}+\frac {5^{1/3}\,\ln \left (x+\frac {5^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{30}-\frac {5^{1/3}\,\ln \left (x-\frac {5^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{30} \] Input:
int(1/(x^3 + 5),x)
Output:
(5^(1/3)*log(x + 5^(1/3)))/15 + (5^(1/3)*log(x + (5^(1/3)*(3^(1/2)*1i - 1) )/2)*(3^(1/2)*1i - 1))/30 - (5^(1/3)*log(x - (5^(1/3)*(3^(1/2)*1i + 1))/2) *(3^(1/2)*1i + 1))/30
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {1}{5+x^3} \, dx=\frac {5^{\frac {1}{3}} \left (-2 \sqrt {3}\, \mathit {atan} \left (\frac {\left (5^{\frac {1}{3}}-2 x \right ) 5^{\frac {2}{3}}}{5 \sqrt {3}}\right )-\mathrm {log}\left (5^{\frac {2}{3}}-5^{\frac {1}{3}} x +x^{2}\right )+2 \,\mathrm {log}\left (5^{\frac {1}{3}}+x \right )\right )}{30} \] Input:
int(1/(x^3+5),x)
Output:
(5**(1/3)*( - 2*sqrt(3)*atan((5**(1/3) - 2*x)/(sqrt(3)*5**(1/3))) - log(5* *(2/3) - 5**(1/3)*x + x**2) + 2*log(5**(1/3) + x)))/30