Integrand size = 13, antiderivative size = 100 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {33}{8} b^2 \sqrt {a+\frac {b}{\sqrt [3]{x}}} \sqrt [3]{x}+\frac {13}{4} a b \sqrt {a+\frac {b}{\sqrt [3]{x}}} x^{2/3}+a^2 \sqrt {a+\frac {b}{\sqrt [3]{x}}} x+\frac {15 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{8 \sqrt {a}} \] Output:
33/8*b^2*(a+b/x^(1/3))^(1/2)*x^(1/3)+13/4*a*b*(a+b/x^(1/3))^(1/2)*x^(2/3)+ a^2*(a+b/x^(1/3))^(1/2)*x+15/8*b^3*arctanh((a+b/x^(1/3))^(1/2)/a^(1/2))/a^ (1/2)
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {1}{8} \left (\sqrt {a+\frac {b}{\sqrt [3]{x}}} \left (33 b^2+26 a b \sqrt [3]{x}+8 a^2 x^{2/3}\right ) \sqrt [3]{x}+\frac {15 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \] Input:
Integrate[(a + b/x^(1/3))^(5/2),x]
Output:
(Sqrt[a + b/x^(1/3)]*(33*b^2 + 26*a*b*x^(1/3) + 8*a^2*x^(2/3))*x^(1/3) + ( 15*b^3*ArcTanh[Sqrt[a + b/x^(1/3)]/Sqrt[a]])/Sqrt[a])/8
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {774, 798, 51, 51, 51, 73, 221}
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.
\(\displaystyle \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -3 \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{x^{4/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -3 \left (\frac {5}{6} b \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{3/2}}{x}d\frac {1}{\sqrt [3]{x}}-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -3 \left (\frac {5}{6} b \left (\frac {3}{4} b \int \frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{x^{2/3}}d\frac {1}{\sqrt [3]{x}}-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{3/2}}{2 x^{2/3}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -3 \left (\frac {5}{6} b \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{\sqrt {a+\frac {b}{\sqrt [3]{x}}} \sqrt [3]{x}}d\frac {1}{\sqrt [3]{x}}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt [3]{x}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{3/2}}{2 x^{2/3}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -3 \left (\frac {5}{6} b \left (\frac {3}{4} b \left (\int \frac {1}{\frac {x^{2/3}}{b}-\frac {a}{b}}d\sqrt {a+\frac {b}{\sqrt [3]{x}}}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt [3]{x}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{3/2}}{2 x^{2/3}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -3 \left (\frac {5}{6} b \left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt [3]{x}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{3/2}}{2 x^{2/3}}\right )-\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2}}{3 x}\right )\) |
Input:
Int[(a + b/x^(1/3))^(5/2),x]
Output:
-3*(-1/3*(a + b/x^(1/3))^(5/2)/x + (5*b*(-1/2*(a + b/x^(1/3))^(3/2)/x^(2/3 ) + (3*b*(-(Sqrt[a + b/x^(1/3)]/x^(1/3)) - (b*ArcTanh[Sqrt[a + b/x^(1/3)]/ Sqrt[a]])/Sqrt[a]))/4))/6)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.37 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {\sqrt {\frac {b +a \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}}\, x^{\frac {1}{3}} \left (16 \left (a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} a^{\frac {5}{2}}+36 a^{\frac {5}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b \,x^{\frac {1}{3}}+66 a^{\frac {3}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b^{2}+15 a \ln \left (\frac {2 \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, \sqrt {a}+2 a \,x^{\frac {1}{3}}+b}{2 \sqrt {a}}\right ) b^{3}\right )}{16 \sqrt {\left (b +a \,x^{\frac {1}{3}}\right ) x^{\frac {1}{3}}}\, a^{\frac {3}{2}}}\) | \(135\) |
derivativedivides | \(\frac {\left (\frac {b +a \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}\right )^{\frac {5}{2}} x \left (16 \left (a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} a^{\frac {5}{2}}+36 a^{\frac {5}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b \,x^{\frac {1}{3}}+66 a^{\frac {3}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b^{2}+15 a \ln \left (\frac {2 \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, \sqrt {a}+2 a \,x^{\frac {1}{3}}+b}{2 \sqrt {a}}\right ) b^{3}\right )}{16 \left (b +a \,x^{\frac {1}{3}}\right )^{2} \sqrt {\left (b +a \,x^{\frac {1}{3}}\right ) x^{\frac {1}{3}}}\, a^{\frac {3}{2}}}\) | \(142\) |
Input:
int((a+b/x^(1/3))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/16*((b+a*x^(1/3))/x^(1/3))^(1/2)*x^(1/3)*(16*(a*x^(2/3)+b*x^(1/3))^(3/2) *a^(5/2)+36*a^(5/2)*(a*x^(2/3)+b*x^(1/3))^(1/2)*b*x^(1/3)+66*a^(3/2)*(a*x^ (2/3)+b*x^(1/3))^(1/2)*b^2+15*a*ln(1/2*(2*(a*x^(2/3)+b*x^(1/3))^(1/2)*a^(1 /2)+2*a*x^(1/3)+b)/a^(1/2))*b^3)/((b+a*x^(1/3))*x^(1/3))^(1/2)/a^(3/2)
Timed out. \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a+b/x^(1/3))^(5/2),x, algorithm="fricas")
Output:
Timed out
Time = 6.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=a^{2} \sqrt {b} x^{\frac {5}{6}} \sqrt {\frac {a \sqrt [3]{x}}{b} + 1} + \frac {13 a b^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}}{4} + \frac {33 b^{\frac {5}{2}} \sqrt [6]{x} \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}}{8} + \frac {15 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt [6]{x}}{\sqrt {b}} \right )}}{8 \sqrt {a}} \] Input:
integrate((a+b/x**(1/3))**(5/2),x)
Output:
a**2*sqrt(b)*x**(5/6)*sqrt(a*x**(1/3)/b + 1) + 13*a*b**(3/2)*sqrt(x)*sqrt( a*x**(1/3)/b + 1)/4 + 33*b**(5/2)*x**(1/6)*sqrt(a*x**(1/3)/b + 1)/8 + 15*b **3*asinh(sqrt(a)*x**(1/6)/sqrt(b))/(8*sqrt(a))
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.31 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=-\frac {15 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{\frac {1}{3}}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{\frac {1}{3}}}} + \sqrt {a}}\right )}{16 \, \sqrt {a}} + \frac {33 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{\frac {3}{2}} a b^{3} + 15 \, \sqrt {a + \frac {b}{x^{\frac {1}{3}}}} a^{2} b^{3}}{8 \, {\left ({\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} - 3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} a + 3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} a^{2} - a^{3}\right )}} \] Input:
integrate((a+b/x^(1/3))^(5/2),x, algorithm="maxima")
Output:
-15/16*b^3*log((sqrt(a + b/x^(1/3)) - sqrt(a))/(sqrt(a + b/x^(1/3)) + sqrt (a)))/sqrt(a) + 1/8*(33*(a + b/x^(1/3))^(5/2)*b^3 - 40*(a + b/x^(1/3))^(3/ 2)*a*b^3 + 15*sqrt(a + b/x^(1/3))*a^2*b^3)/((a + b/x^(1/3))^3 - 3*(a + b/x ^(1/3))^2*a + 3*(a + b/x^(1/3))*a^2 - a^3)
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.22 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=-\frac {15 \, b^{3} \log \left ({\left | -\sqrt {a} x^{\frac {1}{6}} + \sqrt {a x^{\frac {1}{3}} + b} \right |}\right )}{8 \, \sqrt {a} \mathrm {sgn}\left (x\right )^{\frac {1}{3}}} + \frac {1}{8} \, {\left (\frac {33 \, b^{2}}{\mathrm {sgn}\left (x\right )^{\frac {1}{3}}} + 2 \, {\left (\frac {4 \, a^{2} x^{\frac {1}{3}}}{\mathrm {sgn}\left (x\right )^{\frac {1}{3}}} + \frac {13 \, a b}{\mathrm {sgn}\left (x\right )^{\frac {1}{3}}}\right )} x^{\frac {1}{3}}\right )} \sqrt {a x^{\frac {1}{3}} + b} x^{\frac {1}{6}} - \frac {15 \, {\left (-i \, \sqrt {3} b^{3} \log \left ({\left | b \right |}\right ) + b^{3} \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 i \, \sqrt {3} \sqrt {a} + 32 \, \sqrt {a}} \] Input:
integrate((a+b/x^(1/3))^(5/2),x, algorithm="giac")
Output:
-15/8*b^3*log(abs(-sqrt(a)*x^(1/6) + sqrt(a*x^(1/3) + b)))/(sqrt(a)*sgn(x) ^(1/3)) + 1/8*(33*b^2/sgn(x)^(1/3) + 2*(4*a^2*x^(1/3)/sgn(x)^(1/3) + 13*a* b/sgn(x)^(1/3))*x^(1/3))*sqrt(a*x^(1/3) + b)*x^(1/6) - 15*(-I*sqrt(3)*b^3* log(abs(b)) + b^3*log(abs(b)))*sgn(x)/(32*I*sqrt(3)*sqrt(a) + 32*sqrt(a))
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.38 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {6\,x\,{\left (a+\frac {b}{x^{1/3}}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {a\,x^{1/3}}{b}\right )}{{\left (\frac {a\,x^{1/3}}{b}+1\right )}^{5/2}} \] Input:
int((a + b/x^(1/3))^(5/2),x)
Output:
(6*x*(a + b/x^(1/3))^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(a*x^(1/3))/b))/(( a*x^(1/3))/b + 1)^(5/2)
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {8 x^{\frac {5}{6}} \sqrt {x^{\frac {1}{3}} a +b}\, a^{3}+33 x^{\frac {1}{6}} \sqrt {x^{\frac {1}{3}} a +b}\, a \,b^{2}+26 \sqrt {x}\, \sqrt {x^{\frac {1}{3}} a +b}\, a^{2} b +15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}+x^{\frac {1}{6}} \sqrt {a}}{\sqrt {b}}\right ) b^{3}}{8 a} \] Input:
int((a+b/x^(1/3))^(5/2),x)
Output:
(8*x**(5/6)*sqrt(x**(1/3)*a + b)*a**3 + 33*x**(1/6)*sqrt(x**(1/3)*a + b)*a *b**2 + 26*sqrt(x)*sqrt(x**(1/3)*a + b)*a**2*b + 15*sqrt(a)*log((sqrt(x**( 1/3)*a + b) + x**(1/6)*sqrt(a))/sqrt(b))*b**3)/(8*a)