Integrand size = 29, antiderivative size = 140 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=-\frac {\left (2 a c^2-d^2\right ) x^3}{3 b^2 c^3}+\frac {2 d \left (2 a c^2-d^2\right ) \sqrt {a+b x^3}}{3 b^3 c^4}-\frac {2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\frac {\left (a+b x^3\right )^2}{6 b^3 c}+\frac {2 \left (a c^2-d^2\right )^2 \log \left (d+c \sqrt {a+b x^3}\right )}{3 b^3 c^5} \]
-1/3*(2*a*c^2-d^2)*x^3/b^2/c^3-2/9*d*(b*x^3+a)^(3/2)/b^3/c^2+1/6*(b*x^3+a) ^2/b^3/c+2/3*(a*c^2-d^2)^2*ln(d+c*(b*x^3+a)^(1/2))/b^3/c^5+2/3*d*(2*a*c^2- d^2)*(b*x^3+a)^(1/2)/b^3/c^4
Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {3 c^2 \left (a+b x^3\right ) \left (-3 a c^2+2 d^2+b c^2 x^3\right )-4 c d \sqrt {a+b x^3} \left (-5 a c^2+3 d^2+b c^2 x^3\right )+12 \left (-a c^2+d^2\right )^2 \log \left (d+c \sqrt {a+b x^3}\right )}{18 b^3 c^5} \]
(3*c^2*(a + b*x^3)*(-3*a*c^2 + 2*d^2 + b*c^2*x^3) - 4*c*d*Sqrt[a + b*x^3]* (-5*a*c^2 + 3*d^2 + b*c^2*x^3) + 12*(-(a*c^2) + d^2)^2*Log[d + c*Sqrt[a + b*x^3]])/(18*b^3*c^5)
Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2586, 7267, 476, 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.
\(\displaystyle \int \frac {x^8}{d \sqrt {a+b x^3}+a c+b c x^3} \, dx\) |
\(\Big \downarrow \) 2586 |
\(\displaystyle \frac {1}{3} \int \frac {x^6}{b c x^3+a c+d \sqrt {b x^3+a}}dx^3\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {2 \int \frac {\left (a-x^6\right )^2}{\sqrt {b x^3+a} c+d}d\sqrt {b x^3+a}}{3 b^3}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {2 \int \left (\frac {x^9}{c}-\frac {d x^6}{c^2}+\frac {2 a c^2 d-d^3}{c^4}-\frac {\left (2 a c^2-d^2\right ) \sqrt {b x^3+a}}{c^3}+\frac {\left (a c^2-d^2\right )^2}{c^4 \left (\sqrt {b x^3+a} c+d\right )}\right )d\sqrt {b x^3+a}}{3 b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (\frac {\left (a c^2-d^2\right )^2 \log \left (c \sqrt {a+b x^3}+d\right )}{c^5}+\frac {d \sqrt {a+b x^3} \left (2 a c^2-d^2\right )}{c^4}-\frac {x^6 \left (2 a c^2-d^2\right )}{2 c^3}-\frac {d x^9}{3 c^2}+\frac {x^{12}}{4 c}\right )}{3 b^3}\) |
(2*(-1/2*((2*a*c^2 - d^2)*x^6)/c^3 - (d*x^9)/(3*c^2) + x^12/(4*c) + (d*(2* a*c^2 - d^2)*Sqrt[a + b*x^3])/c^4 + ((a*c^2 - d^2)^2*Log[d + c*Sqrt[a + b* x^3]])/c^5))/(3*b^3)
3.6.52.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[1/n Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* d, 0] && IntegerQ[(m + 1)/n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(124)=248\).
Time = 0.39 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.37
method | result | size |
default | \(d \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}}}{9 b^{3} c^{2}}+\frac {2 a^{2} \sqrt {b \,x^{3}+a}}{3 b^{3} d^{2}}-\frac {\left (a^{2} c^{4}-2 a \,c^{2} d^{2}+d^{4}\right ) \left (d \ln \left (c \sqrt {b \,x^{3}+a}-d \right )-d \ln \left (d +c \sqrt {b \,x^{3}+a}\right )+2 c \sqrt {b \,x^{3}+a}\right )}{3 d^{2} c^{5} b^{3}}\right )-a c \left (-\frac {x^{3}}{3 b^{2} c^{2}}+\frac {\left (-a^{2} c^{4}+2 a \,c^{2} d^{2}-d^{4}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 d^{2} c^{4} b^{3}}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3} d^{2}}\right )-b c \left (-\frac {\frac {1}{2} b \,c^{2} x^{6}-2 c^{2} x^{3} a +d^{2} x^{3}}{3 c^{4} b^{3}}+\frac {\left (a^{3} c^{6}-3 a^{2} c^{4} d^{2}+3 a \,c^{2} d^{4}-d^{6}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 c^{6} b^{4} d^{2}}-\frac {a^{3} \ln \left (b \,x^{3}+a \right )}{3 b^{4} d^{2}}\right )\) | \(332\) |
elliptic | \(\frac {\sqrt {b \,x^{3}+a}\, \left (d +c \sqrt {b \,x^{3}+a}\right ) \left (c \left (\frac {\frac {1}{2} b \,c^{2} x^{6}-c^{2} x^{3} a +d^{2} x^{3}}{3 b^{2} c^{4}}+\frac {\left (a^{2} c^{4}-2 a \,c^{2} d^{2}+d^{4}\right ) \ln \left (b \,c^{2} x^{3}+a \,c^{2}-d^{2}\right )}{3 b^{3} c^{6}}\right )-\frac {2 d \,x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2} c^{2}}+\frac {2 \left (\frac {\left (a \,c^{2}-d^{2}\right ) d}{b^{2} c^{4}}+\frac {2 d a}{3 b^{2} c^{2}}\right ) \sqrt {b \,x^{3}+a}}{3 b}-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,c^{2} \textit {\_Z}^{3}+a \,c^{2}-d^{2}\right )}{\sum }\frac {\left (a^{2} c^{4}-2 a \,c^{2} d^{2}+d^{4}\right ) \left (-b^{2} a \right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i b \left (2 x +\frac {\left (-b^{2} a \right )^{\frac {1}{3}}-i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}{b}\right )}{\left (-b^{2} a \right )^{\frac {1}{3}}}}\, \sqrt {\frac {b \left (x -\frac {\left (-b^{2} a \right )^{\frac {1}{3}}}{b}\right )}{-3 \left (-b^{2} a \right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i b \left (2 x +\frac {\left (-b^{2} a \right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}{b}\right )}{2 \left (-b^{2} a \right )^{\frac {1}{3}}}}\, \left (i \left (-b^{2} a \right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha b -i \left (-b^{2} a \right )^{\frac {2}{3}} \sqrt {3}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} b^{2}-\left (-b^{2} a \right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b -\left (-b^{2} a \right )^{\frac {2}{3}}\right ) \Pi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-b^{2} a \right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-b^{2} a \right )^{\frac {1}{3}}}}}{3}, -\frac {c^{2} \left (2 i \left (-b^{2} a \right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} b -i \left (-b^{2} a \right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, a b -3 \left (-b^{2} a \right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 a b \right )}{2 b \,d^{2}}, \sqrt {\frac {i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-b^{2} a \right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-b^{2} a \right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{2 \sqrt {b \,x^{3}+a}}\right )}{3 d \,b^{5} c^{4}}\right )}{a c +b c \,x^{3}+d \sqrt {b \,x^{3}+a}}\) | \(646\) |
d*(-2/9/b^3/c^2*(b*x^3+a)^(3/2)+2/3*a^2/b^3/d^2*(b*x^3+a)^(1/2)-1/3*(a^2*c ^4-2*a*c^2*d^2+d^4)/d^2/c^5/b^3*(d*ln(c*(b*x^3+a)^(1/2)-d)-d*ln(d+c*(b*x^3 +a)^(1/2))+2*c*(b*x^3+a)^(1/2)))-a*c*(-1/3/b^2/c^2*x^3+1/3*(-a^2*c^4+2*a*c ^2*d^2-d^4)/d^2/c^4/b^3*ln(b*c^2*x^3+a*c^2-d^2)+1/3*a^2/b^3/d^2*ln(b*x^3+a ))-b*c*(-1/3/c^4/b^3*(1/2*b*c^2*x^6-2*c^2*x^3*a+d^2*x^3)+1/3/c^6/b^4*(a^3* c^6-3*a^2*c^4*d^2+3*a*c^2*d^4-d^6)/d^2*ln(b*c^2*x^3+a*c^2-d^2)-1/3/b^4*a^3 /d^2*ln(b*x^3+a))
Time = 0.30 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.36 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {3 \, b^{2} c^{4} x^{6} - 6 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{3} + 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right ) - 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c - d\right ) - 4 \, {\left (b c^{3} d x^{3} - 5 \, a c^{3} d + 3 \, c d^{3}\right )} \sqrt {b x^{3} + a}}{18 \, b^{3} c^{5}} \]
1/18*(3*b^2*c^4*x^6 - 6*(a*b*c^4 - b*c^2*d^2)*x^3 + 6*(a^2*c^4 - 2*a*c^2*d ^2 + d^4)*log(b*c^2*x^3 + a*c^2 - d^2) + 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*l og(sqrt(b*x^3 + a)*c + d) - 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 + a)*c - d) - 4*(b*c^3*d*x^3 - 5*a*c^3*d + 3*c*d^3)*sqrt(b*x^3 + a))/(b^3 *c^5)
Time = 2.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\begin {cases} \frac {2 \left (\frac {\left (a + b x^{3}\right )^{2}}{12 c} - \frac {d \left (a + b x^{3}\right )^{\frac {3}{2}}}{9 c^{2}} + \frac {\left (a + b x^{3}\right ) \left (- 2 a c^{2} + d^{2}\right )}{6 c^{3}} + \frac {\sqrt {a + b x^{3}} \cdot \left (2 a c^{2} d - d^{3}\right )}{3 c^{4}} + \frac {\left (a c^{2} - d^{2}\right )^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{3}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{3}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{3 c^{4}}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{9}}{3 \cdot \left (3 \sqrt {a} d + 3 a c\right )} & \text {otherwise} \end {cases} \]
Piecewise((2*((a + b*x**3)**2/(12*c) - d*(a + b*x**3)**(3/2)/(9*c**2) + (a + b*x**3)*(-2*a*c**2 + d**2)/(6*c**3) + sqrt(a + b*x**3)*(2*a*c**2*d - d* *3)/(3*c**4) + (a*c**2 - d**2)**2*Piecewise((sqrt(a + b*x**3)/d, Eq(c, 0)) , (log(c*sqrt(a + b*x**3) + d)/c, True))/(3*c**4))/b**3, Ne(b, 0)), (x**9/ (3*(3*sqrt(a)*d + 3*a*c)), True))
Time = 0.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {\frac {3 \, {\left (b x^{3} + a\right )}^{2} c^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} c^{2} d - 6 \, {\left (2 \, a c^{3} - c d^{2}\right )} {\left (b x^{3} + a\right )} + 12 \, {\left (2 \, a c^{2} d - d^{3}\right )} \sqrt {b x^{3} + a}}{c^{4}} + \frac {12 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right )}{c^{5}}}{18 \, b^{3}} \]
1/18*((3*(b*x^3 + a)^2*c^3 - 4*(b*x^3 + a)^(3/2)*c^2*d - 6*(2*a*c^3 - c*d^ 2)*(b*x^3 + a) + 12*(2*a*c^2*d - d^3)*sqrt(b*x^3 + a))/c^4 + 12*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 + a)*c + d)/c^5)/b^3
Time = 0.32 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {2 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{3 \, b^{3} c^{5}} + \frac {3 \, {\left (b x^{3} + a\right )}^{2} b^{9} c^{3} - 12 \, {\left (b x^{3} + a\right )} a b^{9} c^{3} - 4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{9} c^{2} d + 24 \, \sqrt {b x^{3} + a} a b^{9} c^{2} d + 6 \, {\left (b x^{3} + a\right )} b^{9} c d^{2} - 12 \, \sqrt {b x^{3} + a} b^{9} d^{3}}{18 \, b^{12} c^{4}} \]
2/3*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(abs(sqrt(b*x^3 + a)*c + d))/(b^3*c^5 ) + 1/18*(3*(b*x^3 + a)^2*b^9*c^3 - 12*(b*x^3 + a)*a*b^9*c^3 - 4*(b*x^3 + a)^(3/2)*b^9*c^2*d + 24*sqrt(b*x^3 + a)*a*b^9*c^2*d + 6*(b*x^3 + a)*b^9*c* d^2 - 12*sqrt(b*x^3 + a)*b^9*d^3)/(b^12*c^4)
Time = 17.83 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.43 \[ \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx=\frac {\left (\frac {2\,d\,\left (a\,c^2-d^2\right )}{b^2\,c^4}+\frac {4\,a\,d}{3\,b^2\,c^2}\right )\,\sqrt {b\,x^3+a}}{3\,b}+\frac {x^6}{6\,b\,c}-\frac {x^3\,\left (a\,c^2-d^2\right )}{3\,b^2\,c^3}+\frac {\ln \left (\frac {d+c\,\sqrt {b\,x^3+a}}{d-c\,\sqrt {b\,x^3+a}}\right )\,{\left (a\,c^2-d^2\right )}^2}{3\,b^3\,c^5}+\frac {\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )\,\left (a^2\,c^4-2\,a\,c^2\,d^2+d^4\right )}{3\,b^3\,c^5}-\frac {2\,d\,x^3\,\sqrt {b\,x^3+a}}{9\,b^2\,c^2} \]
(((2*d*(a*c^2 - d^2))/(b^2*c^4) + (4*a*d)/(3*b^2*c^2))*(a + b*x^3)^(1/2))/ (3*b) + x^6/(6*b*c) - (x^3*(a*c^2 - d^2))/(3*b^2*c^3) + (log((d + c*(a + b *x^3)^(1/2))/(d - c*(a + b*x^3)^(1/2)))*(a*c^2 - d^2)^2)/(3*b^3*c^5) + (lo g(a*c^2 - d^2 + b*c^2*x^3)*(d^4 + a^2*c^4 - 2*a*c^2*d^2))/(3*b^3*c^5) - (2 *d*x^3*(a + b*x^3)^(1/2))/(9*b^2*c^2)