Integrand size = 24, antiderivative size = 48 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=-\frac {2 C \arctan \left (\frac {1-\sqrt [3]{b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {C \log \left (2+\sqrt [3]{b} x\right )}{\sqrt [3]{b}} \] Output:
-2/3*C*arctan(1/3*(1-b^(1/3)*x)*3^(1/2))*3^(1/2)/b^(1/3)+C*ln(2+b^(1/3)*x) /b^(1/3)
Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.58 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\frac {C \left (2 \sqrt {3} \arctan \left (\frac {-1+\sqrt [3]{b} x}{\sqrt {3}}\right )+2 \log \left (2+\sqrt [3]{b} x\right )-\log \left (4-2 \sqrt [3]{b} x+b^{2/3} x^2\right )+\log \left (8+b x^3\right )\right )}{3 \sqrt [3]{b}} \] Input:
Integrate[(8*C + b^(2/3)*C*x^2)/(8 + b*x^3),x]
Output:
(C*(2*Sqrt[3]*ArcTan[(-1 + b^(1/3)*x)/Sqrt[3]] + 2*Log[2 + b^(1/3)*x] - Lo g[4 - 2*b^(1/3)*x + b^(2/3)*x^2] + Log[8 + b*x^3]))/(3*b^(1/3))
Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2402, 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.
\(\displaystyle \int \frac {b^{2/3} C x^2+8 C}{b x^3+8} \, dx\) |
\(\Big \downarrow \) 2402 |
\(\displaystyle \frac {2 C \int \frac {1}{x^2-\frac {2 x}{\sqrt [3]{b}}+\frac {4}{b^{2/3}}}dx}{b^{2/3}}+\frac {C \int \frac {1}{x+\frac {2}{\sqrt [3]{b}}}dx}{\sqrt [3]{b}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {2 C \int \frac {1}{x^2-\frac {2 x}{\sqrt [3]{b}}+\frac {4}{b^{2/3}}}dx}{b^{2/3}}+\frac {C \log \left (\sqrt [3]{b} x+2\right )}{\sqrt [3]{b}}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 C \int \frac {1}{-\left (1-\sqrt [3]{b} x\right )^2-3}d\left (1-\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {C \log \left (\sqrt [3]{b} x+2\right )}{\sqrt [3]{b}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {C \log \left (\sqrt [3]{b} x+2\right )}{\sqrt [3]{b}}-\frac {2 C \arctan \left (\frac {1-\sqrt [3]{b} x}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}\) |
Input:
Int[(8*C + b^(2/3)*C*x^2)/(8 + b*x^3),x]
Output:
(-2*C*ArcTan[(1 - b^(1/3)*x)/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (C*Log[2 + b^(1 /3)*x])/b^(1/3)
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_) + (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[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x, 2]}, With[{q = a^(1/3)/b^(1/3)}, Simp[C /b Int[1/(q + x), x], x] + Simp[(B + C*q)/b Int[1/(q^2 - q*x + x^2), x] , x]] /; EqQ[A*b^(2/3) - a^(1/3)*b^(1/3)*B - 2*a^(2/3)*C, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]
Leaf count of result is larger than twice the leaf count of optimal. \(114\) vs. \(2(37)=74\).
Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.40
method | result | size |
default | \(C \left (\frac {8^{\frac {1}{3}} \ln \left (x +8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {1}{b}\right )^{\frac {2}{3}}}-\frac {8^{\frac {1}{3}} \ln \left (x^{2}-8^{\frac {1}{3}} \left (\frac {1}{b}\right )^{\frac {1}{3}} x +8^{\frac {2}{3}} \left (\frac {1}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {1}{b}\right )^{\frac {2}{3}}}+\frac {8^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {8^{\frac {2}{3}} x}{4 \left (\frac {1}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {1}{b}\right )^{\frac {2}{3}}}+\frac {\ln \left (b \,x^{3}+8\right )}{3 b^{\frac {1}{3}}}\right )\) | \(115\) |
meijerg | \(\frac {2 C \left (\frac {b^{\frac {1}{3}} x \ln \left (1+\frac {\left (b \,x^{3}\right )^{\frac {1}{3}}}{2}\right )}{\left (b \,x^{3}\right )^{\frac {1}{3}}}-\frac {b^{\frac {1}{3}} x \ln \left (1-\frac {\left (b \,x^{3}\right )^{\frac {1}{3}}}{2}+\frac {\left (b \,x^{3}\right )^{\frac {2}{3}}}{4}\right )}{2 \left (b \,x^{3}\right )^{\frac {1}{3}}}+\frac {b^{\frac {1}{3}} x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b \,x^{3}\right )^{\frac {1}{3}}}{4-\left (b \,x^{3}\right )^{\frac {1}{3}}}\right )}{\left (b \,x^{3}\right )^{\frac {1}{3}}}\right )}{3 b^{\frac {1}{3}}}+\frac {C \ln \left (1+\frac {b \,x^{3}}{8}\right )}{3 b^{\frac {1}{3}}}\) | \(123\) |
Input:
int((8*C+b^(2/3)*C*x^2)/(b*x^3+8),x,method=_RETURNVERBOSE)
Output:
C*(1/3/b*8^(1/3)/(1/b)^(2/3)*ln(x+8^(1/3)*(1/b)^(1/3))-1/6/b*8^(1/3)/(1/b) ^(2/3)*ln(x^2-8^(1/3)*(1/b)^(1/3)*x+8^(2/3)*(1/b)^(2/3))+1/3/b*8^(1/3)/(1/ b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(1/4*8^(2/3)/(1/b)^(1/3)*x-1))+1/3/b^( 1/3)*ln(b*x^3+8))
Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.79 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\left [\frac {\sqrt {\frac {1}{3}} C b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {b x^{3} + 6 \, \sqrt {\frac {1}{3}} {\left (b x^{2} + b^{\frac {2}{3}} x - 2 \, b^{\frac {1}{3}}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} - 6 \, b^{\frac {1}{3}} x - 4}{b x^{3} + 8}\right ) + C b^{\frac {2}{3}} \log \left (b x + 2 \, b^{\frac {2}{3}}\right )}{b}, \frac {2 \, \sqrt {\frac {1}{3}} C b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {2}{3}} x - b^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}}}\right ) + C b^{\frac {2}{3}} \log \left (b x + 2 \, b^{\frac {2}{3}}\right )}{b}\right ] \] Input:
integrate((8*C+b^(2/3)*C*x^2)/(b*x^3+8),x, algorithm="fricas")
Output:
[(sqrt(1/3)*C*b*sqrt(-1/b^(2/3))*log((b*x^3 + 6*sqrt(1/3)*(b*x^2 + b^(2/3) *x - 2*b^(1/3))*sqrt(-1/b^(2/3)) - 6*b^(1/3)*x - 4)/(b*x^3 + 8)) + C*b^(2/ 3)*log(b*x + 2*b^(2/3)))/b, (2*sqrt(1/3)*C*b^(2/3)*arctan(sqrt(1/3)*(b^(2/ 3)*x - b^(1/3))/b^(1/3)) + C*b^(2/3)*log(b*x + 2*b^(2/3)))/b]
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\operatorname {RootSum} {\left (3 t^{3} b^{\frac {5}{3}} - 3 t^{2} C b^{\frac {4}{3}} + t C^{2} b - C^{3} b^{\frac {2}{3}}, \left ( t \mapsto t \log {\left (x + \frac {3 t \sqrt [3]{b} - C}{C \sqrt [3]{b}} \right )} \right )\right )} \] Input:
integrate((8*C+b**(2/3)*C*x**2)/(b*x**3+8),x)
Output:
RootSum(3*_t**3*b**(5/3) - 3*_t**2*C*b**(4/3) + _t*C**2*b - C**3*b**(2/3), Lambda(_t, _t*log(x + (3*_t*b**(1/3) - C)/(C*b**(1/3)))))
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\frac {2 \, \sqrt {3} C \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {2}{3}} x - b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {1}{3}}} + \frac {C \log \left (\frac {b^{\frac {1}{3}} x + 2}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} \] Input:
integrate((8*C+b^(2/3)*C*x^2)/(b*x^3+8),x, algorithm="maxima")
Output:
2/3*sqrt(3)*C*arctan(1/3*sqrt(3)*(b^(2/3)*x - b^(1/3))/b^(1/3))/b^(1/3) + C*log((b^(1/3)*x + 2)/b^(1/3))/b^(1/3)
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (36) = 72\).
Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.40 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\frac {2}{3} \, \sqrt {3} C \left (-\frac {1}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (x + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {1}{b}\right )^{\frac {1}{3}}}\right ) - \frac {1}{3} \, {\left (C b^{\frac {2}{3}} \left (-\frac {1}{b}\right )^{\frac {2}{3}} + 2 \, C\right )} \left (-\frac {1}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - 2 \, \left (-\frac {1}{b}\right )^{\frac {1}{3}} \right |}\right ) + \frac {1}{3} \, {\left (C \left (-\frac {1}{b}\right )^{\frac {1}{3}} + \frac {C}{b^{\frac {1}{3}}}\right )} \log \left (x^{2} + 2 \, x \left (-\frac {1}{b}\right )^{\frac {1}{3}} + 4 \, \left (-\frac {1}{b}\right )^{\frac {2}{3}}\right ) \] Input:
integrate((8*C+b^(2/3)*C*x^2)/(b*x^3+8),x, algorithm="giac")
Output:
2/3*sqrt(3)*C*(-1/b)^(1/3)*arctan(1/3*sqrt(3)*(x + (-1/b)^(1/3))/(-1/b)^(1 /3)) - 1/3*(C*b^(2/3)*(-1/b)^(2/3) + 2*C)*(-1/b)^(1/3)*log(abs(x - 2*(-1/b )^(1/3))) + 1/3*(C*(-1/b)^(1/3) + C/b^(1/3))*log(x^2 + 2*x*(-1/b)^(1/3) + 4*(-1/b)^(2/3))
Time = 6.23 (sec) , antiderivative size = 147, normalized size of antiderivative = 3.06 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\sum _{k=1}^3\ln \left (-\frac {\left (C-\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,b^{1/3}\,3\right )\,\left (-C+\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right )\,b^{1/3}\,3+C\,b^{1/3}\,x\right )\,8}{b^{5/3}}\right )\,\mathrm {root}\left (27\,b^3\,z^3-27\,C\,b^{8/3}\,z^2+9\,C^2\,b^{7/3}\,z-9\,C^3\,b^2,z,k\right ) \] Input:
int((8*C + C*b^(2/3)*x^2)/(b*x^3 + 8),x)
Output:
symsum(log(-(8*(C - 3*root(27*b^3*z^3 - 27*C*b^(8/3)*z^2 + 9*C^2*b^(7/3)*z - 9*C^3*b^2, z, k)*b^(1/3))*(3*root(27*b^3*z^3 - 27*C*b^(8/3)*z^2 + 9*C^2 *b^(7/3)*z - 9*C^3*b^2, z, k)*b^(1/3) - C + C*b^(1/3)*x))/b^(5/3))*root(27 *b^3*z^3 - 27*C*b^(8/3)*z^2 + 9*C^2*b^(7/3)*z - 9*C^3*b^2, z, k), k, 1, 3)
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.71 \[ \int \frac {8 C+b^{2/3} C x^2}{8+b x^3} \, dx=\frac {c \left (2 \sqrt {3}\, \mathit {atan} \left (\frac {b^{\frac {1}{3}} x -1}{\sqrt {3}}\right )+3 \,\mathrm {log}\left (b^{\frac {1}{3}} x +2\right )\right )}{3 b^{\frac {1}{3}}} \] Input:
int((8*C+b^(2/3)*C*x^2)/(b*x^3+8),x)
Output:
(c*(2*sqrt(3)*atan((b**(1/3)*x - 1)/sqrt(3)) + 3*log(b**(1/3)*x + 2)))/(3* b**(1/3))