Integrand size = 11, antiderivative size = 112 \[ \int \left (a+b x^3\right )^{5/3} \, dx=\frac {5}{18} a x \left (a+b x^3\right )^{2/3}+\frac {1}{6} x \left (a+b x^3\right )^{5/3}+\frac {5 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} \sqrt [3]{b}}-\frac {5 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 \sqrt [3]{b}} \] Output:
5/18*a*x*(b*x^3+a)^(2/3)+1/6*x*(b*x^3+a)^(5/3)+5/27*a^2*arctan(1/3*(1+2*b^ (1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(1/3)-5/18*a^2*ln(-b^(1/3)*x+( b*x^3+a)^(1/3))/b^(1/3)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.17 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.43 \[ \int \left (a+b x^3\right )^{5/3} \, dx=\frac {3 \left (\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}+x\right ) \left (a+b x^3\right )^{5/3} \operatorname {AppellF1}\left (\frac {8}{3},-\frac {5}{3},-\frac {5}{3},\frac {11}{3},-\frac {\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}+x}{-\frac {\sqrt [3]{-1} \sqrt [3]{a}}{\sqrt [3]{b}}-\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}},-\frac {\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}+x}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}-\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}}\right )}{8 \left (1+\frac {\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}+x}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}-\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}}\right )^{5/3} \left (1+\frac {\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}+x}{-\frac {\sqrt [3]{-1} \sqrt [3]{a}}{\sqrt [3]{b}}-\frac {(-1)^{2/3} \sqrt [3]{a}}{\sqrt [3]{b}}}\right )^{5/3}} \] Input:
Integrate[(a + b*x^3)^(5/3),x]
Output:
(3*(((-1)^(2/3)*a^(1/3))/b^(1/3) + x)*(a + b*x^3)^(5/3)*AppellF1[8/3, -5/3 , -5/3, 11/3, -((((-1)^(2/3)*a^(1/3))/b^(1/3) + x)/(-(((-1)^(1/3)*a^(1/3)) /b^(1/3)) - ((-1)^(2/3)*a^(1/3))/b^(1/3))), -((((-1)^(2/3)*a^(1/3))/b^(1/3 ) + x)/(a^(1/3)/b^(1/3) - ((-1)^(2/3)*a^(1/3))/b^(1/3)))])/(8*(1 + (((-1)^ (2/3)*a^(1/3))/b^(1/3) + x)/(a^(1/3)/b^(1/3) - ((-1)^(2/3)*a^(1/3))/b^(1/3 )))^(5/3)*(1 + (((-1)^(2/3)*a^(1/3))/b^(1/3) + x)/(-(((-1)^(1/3)*a^(1/3))/ b^(1/3)) - ((-1)^(2/3)*a^(1/3))/b^(1/3)))^(5/3))
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {748, 748, 769}
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.
\(\displaystyle \int \left (a+b x^3\right )^{5/3} \, dx\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {5}{6} a \int \left (b x^3+a\right )^{2/3}dx+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {5}{6} a \left (\frac {2}{3} a \int \frac {1}{\sqrt [3]{b x^3+a}}dx+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\) |
\(\Big \downarrow \) 769 |
\(\displaystyle \frac {5}{6} a \left (\frac {2}{3} a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\) |
Input:
Int[(a + b*x^3)^(5/3),x]
Output:
(x*(a + b*x^3)^(5/3))/6 + (5*a*((x*(a + b*x^3)^(2/3))/3 + (2*a*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1 /3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))/3))/6
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[p]])
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* (x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
Time = 2.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.29
method | result | size |
pseudoelliptic | \(\frac {9 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {4}{3}} x^{4}+24 a x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {1}{3}}-10 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a^{2}-10 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2}+5 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2}}{54 b^{\frac {1}{3}}}\) | \(144\) |
Input:
int((b*x^3+a)^(5/3),x,method=_RETURNVERBOSE)
Output:
1/54*(9*(b*x^3+a)^(2/3)*b^(4/3)*x^4+24*a*x*(b*x^3+a)^(2/3)*b^(1/3)-10*3^(1 /2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)*a^2-10*ln( (-b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2+5*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1 /3)*x+(b*x^3+a)^(2/3))/x^2)*a^2)/b^(1/3)
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (85) = 170\).
Time = 0.09 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.56 \[ \int \left (a+b x^3\right )^{5/3} \, dx=\left [\frac {15 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} b x^{3} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} x\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} + 2 \, a\right ) - 10 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 5 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (3 \, b^{2} x^{4} + 8 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b}, -\frac {30 \, \sqrt {\frac {1}{3}} a^{2} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) + 10 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 5 \, a^{2} \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b^{2} x^{4} + 8 \, a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54 \, b}\right ] \] Input:
integrate((b*x^3+a)^(5/3),x, algorithm="fricas")
Output:
[1/54*(15*sqrt(1/3)*a^2*b*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^( 1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b* x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 10*a^2 *(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 5*a^2*(-b)^(2/3)*l og(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x ^2) + 3*(3*b^2*x^4 + 8*a*b*x)*(b*x^3 + a)^(2/3))/b, -1/54*(30*sqrt(1/3)*a^ 2*b*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1 /3))*sqrt(-(-b)^(1/3)/b)/x) + 10*a^2*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 5*a^2*(-b)^(2/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3) *(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 3*(3*b^2*x^4 + 8*a*b*x)*(b*x^3 + a)^(2/3))/b]
Result contains complex when optimal does not.
Time = 0.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.33 \[ \int \left (a+b x^3\right )^{5/3} \, dx=\frac {a^{\frac {5}{3}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \] Input:
integrate((b*x**3+a)**(5/3),x)
Output:
a**(5/3)*x*gamma(1/3)*hyper((-5/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a) /(3*gamma(4/3))
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (85) = 170\).
Time = 0.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.60 \[ \int \left (a+b x^3\right )^{5/3} \, dx=-\frac {5 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {1}{3}}} + \frac {5 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{54 \, b^{\frac {1}{3}}} - \frac {5 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {1}{3}}} - \frac {\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b}{x^{2}} - \frac {8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{x^{5}}}{18 \, {\left (b^{2} - \frac {2 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )}} \] Input:
integrate((b*x^3+a)^(5/3),x, algorithm="maxima")
Output:
-5/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^( 1/3))/b^(1/3) + 5/54*a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^ 3 + a)^(2/3)/x^2)/b^(1/3) - 5/27*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b ^(1/3) - 1/18*(5*(b*x^3 + a)^(2/3)*a^2*b/x^2 - 8*(b*x^3 + a)^(5/3)*a^2/x^5 )/(b^2 - 2*(b*x^3 + a)*b/x^3 + (b*x^3 + a)^2/x^6)
\[ \int \left (a+b x^3\right )^{5/3} \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {5}{3}} \,d x } \] Input:
integrate((b*x^3+a)^(5/3),x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(5/3), x)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.33 \[ \int \left (a+b x^3\right )^{5/3} \, dx=\frac {x\,{\left (b\,x^3+a\right )}^{5/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{3},\frac {1}{3};\ \frac {4}{3};\ -\frac {b\,x^3}{a}\right )}{{\left (\frac {b\,x^3}{a}+1\right )}^{5/3}} \] Input:
int((a + b*x^3)^(5/3),x)
Output:
(x*(a + b*x^3)^(5/3)*hypergeom([-5/3, 1/3], 4/3, -(b*x^3)/a))/((b*x^3)/a + 1)^(5/3)
\[ \int \left (a+b x^3\right )^{5/3} \, dx=\frac {4 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a x}{9}+\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{4}}{6}+\frac {5 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{2}}{9} \] Input:
int((b*x^3+a)^(5/3),x)
Output:
(8*(a + b*x**3)**(2/3)*a*x + 3*(a + b*x**3)**(2/3)*b*x**4 + 10*int((a + b* x**3)**(2/3)/(a + b*x**3),x)*a**2)/18