Integrand size = 13, antiderivative size = 101 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=-\frac {3 b^2 \sqrt {a+\frac {b}{x^{2/3}}}}{2 \sqrt [3]{x}}+7 a b \sqrt {a+\frac {b}{x^{2/3}}} \sqrt [3]{x}+a^2 \sqrt {a+\frac {b}{x^{2/3}}} x-\frac {15}{2} a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^{2/3}}} \sqrt [3]{x}}\right ) \] Output:
-3/2*b^2*(a+b/x^(2/3))^(1/2)/x^(1/3)+7*a*b*(a+b/x^(2/3))^(1/2)*x^(1/3)+a^2 *(a+b/x^(2/3))^(1/2)*x-15/2*a*b^(3/2)*arctanh(b^(1/2)/(a+b/x^(2/3))^(1/2)/ x^(1/3))
Time = 5.95 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=\frac {\sqrt {a+\frac {b}{x^{2/3}}} \sqrt [3]{x} \left (\frac {\sqrt {b+a x^{2/3}} \left (-3 b^2+14 a b x^{2/3}+2 a^2 x^{4/3}\right )}{2 x^{2/3}}-\frac {15}{2} a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b+a x^{2/3}}}{\sqrt {b}}\right )\right )}{\sqrt {b+a x^{2/3}}} \] Input:
Integrate[(a + b/x^(2/3))^(5/2),x]
Output:
(Sqrt[a + b/x^(2/3)]*x^(1/3)*((Sqrt[b + a*x^(2/3)]*(-3*b^2 + 14*a*b*x^(2/3 ) + 2*a^2*x^(4/3)))/(2*x^(2/3)) - (15*a*b^(3/2)*ArcTanh[Sqrt[b + a*x^(2/3) ]/Sqrt[b]])/2))/Sqrt[b + a*x^(2/3)]
Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {774, 858, 247, 247, 211, 224, 219}
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.
\(\displaystyle \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -3 \int \frac {\left (a+b x^{2/3}\right )^{5/2}}{x^{4/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -3 \left (\frac {5}{3} b \int \frac {\left (a+b x^{2/3}\right )^{3/2}}{x^{2/3}}d\frac {1}{\sqrt [3]{x}}-\frac {\left (a+b x^{2/3}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -3 \left (\frac {5}{3} b \left (3 b \int \sqrt {a+b x^{2/3}}d\frac {1}{\sqrt [3]{x}}-\frac {\left (a+b x^{2/3}\right )^{3/2}}{\sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -3 \left (\frac {5}{3} b \left (3 b \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+b x^{2/3}}}d\frac {1}{\sqrt [3]{x}}+\frac {\sqrt {a+b x^{2/3}}}{2 \sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{3/2}}{\sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -3 \left (\frac {5}{3} b \left (3 b \left (\frac {1}{2} a \int \frac {1}{1-b x^{2/3}}d\frac {1}{\sqrt {a+b x^{2/3}} \sqrt [3]{x}}+\frac {\sqrt {a+b x^{2/3}}}{2 \sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{3/2}}{\sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{5/2}}{3 x}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -3 \left (\frac {5}{3} b \left (3 b \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt [3]{x} \sqrt {a+b x^{2/3}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+b x^{2/3}}}{2 \sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{3/2}}{\sqrt [3]{x}}\right )-\frac {\left (a+b x^{2/3}\right )^{5/2}}{3 x}\right )\) |
Input:
Int[(a + b/x^(2/3))^(5/2),x]
Output:
-3*(-1/3*(a + b*x^(2/3))^(5/2)/x + (5*b*(-((a + b*x^(2/3))^(3/2)/x^(1/3)) + 3*b*(Sqrt[a + b*x^(2/3)]/(2*x^(1/3)) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b*x^ (2/3)]*x^(1/3))])/(2*Sqrt[b]))))/3)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Time = 0.38 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.19
method | result | size |
derivativedivides | \(-\frac {\left (\frac {b +a \,x^{\frac {2}{3}}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}} x \left (-3 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {5}{2}} a \,x^{\frac {2}{3}}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {b +a \,x^{\frac {2}{3}}}}{x^{\frac {1}{3}}}\right ) b^{\frac {5}{2}} a \,x^{\frac {2}{3}}+3 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}-5 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} a b \,x^{\frac {2}{3}}-15 \sqrt {b +a \,x^{\frac {2}{3}}}\, a \,b^{2} x^{\frac {2}{3}}\right )}{2 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {5}{2}} b}\) | \(120\) |
default | \(-\frac {\left (\frac {b +a \,x^{\frac {2}{3}}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}} x \left (-3 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {5}{2}} a \,x^{\frac {2}{3}}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {b +a \,x^{\frac {2}{3}}}}{x^{\frac {1}{3}}}\right ) b^{\frac {5}{2}} a \,x^{\frac {2}{3}}+3 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {7}{2}}-5 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} a b \,x^{\frac {2}{3}}-15 \sqrt {b +a \,x^{\frac {2}{3}}}\, a \,b^{2} x^{\frac {2}{3}}\right )}{2 \left (b +a \,x^{\frac {2}{3}}\right )^{\frac {5}{2}} b}\) | \(120\) |
Input:
int((a+b/x^(2/3))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/2*((b+a*x^(2/3))/x^(2/3))^(5/2)*x*(-3*(b+a*x^(2/3))^(5/2)*a*x^(2/3)+15* ln(2*(b^(1/2)*(b+a*x^(2/3))^(1/2)+b)/x^(1/3))*b^(5/2)*a*x^(2/3)+3*(b+a*x^( 2/3))^(7/2)-5*(b+a*x^(2/3))^(3/2)*a*b*x^(2/3)-15*(b+a*x^(2/3))^(1/2)*a*b^2 *x^(2/3))/(b+a*x^(2/3))^(5/2)/b
Timed out. \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a+b/x^(2/3))^(5/2),x, algorithm="fricas")
Output:
Timed out
Time = 7.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.21 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=a^{2} \sqrt {b} x^{\frac {2}{3}} \sqrt {\frac {a x^{\frac {2}{3}}}{b} + 1} + 7 a b^{\frac {3}{2}} \sqrt {\frac {a x^{\frac {2}{3}}}{b} + 1} + \frac {15 a b^{\frac {3}{2}} \log {\left (\frac {a x^{\frac {2}{3}}}{b} \right )}}{4} - \frac {15 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a x^{\frac {2}{3}}}{b} + 1} + 1 \right )}}{2} - \frac {3 b^{\frac {5}{2}} \sqrt {\frac {a x^{\frac {2}{3}}}{b} + 1}}{2 x^{\frac {2}{3}}} \] Input:
integrate((a+b/x**(2/3))**(5/2),x)
Output:
a**2*sqrt(b)*x**(2/3)*sqrt(a*x**(2/3)/b + 1) + 7*a*b**(3/2)*sqrt(a*x**(2/3 )/b + 1) + 15*a*b**(3/2)*log(a*x**(2/3)/b)/4 - 15*a*b**(3/2)*log(sqrt(a*x* *(2/3)/b + 1) + 1)/2 - 3*b**(5/2)*sqrt(a*x**(2/3)/b + 1)/(2*x**(2/3))
Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx={\left (a + \frac {b}{x^{\frac {2}{3}}}\right )}^{\frac {3}{2}} a x + \frac {15}{4} \, a b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{\frac {2}{3}}}} x^{\frac {1}{3}} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{\frac {2}{3}}}} x^{\frac {1}{3}} + \sqrt {b}}\right ) + 6 \, \sqrt {a + \frac {b}{x^{\frac {2}{3}}}} a b x^{\frac {1}{3}} - \frac {3 \, \sqrt {a + \frac {b}{x^{\frac {2}{3}}}} a b^{2} x^{\frac {1}{3}}}{2 \, {\left ({\left (a + \frac {b}{x^{\frac {2}{3}}}\right )} x^{\frac {2}{3}} - b\right )}} \] Input:
integrate((a+b/x^(2/3))^(5/2),x, algorithm="maxima")
Output:
(a + b/x^(2/3))^(3/2)*a*x + 15/4*a*b^(3/2)*log((sqrt(a + b/x^(2/3))*x^(1/3 ) - sqrt(b))/(sqrt(a + b/x^(2/3))*x^(1/3) + sqrt(b))) + 6*sqrt(a + b/x^(2/ 3))*a*b*x^(1/3) - 3/2*sqrt(a + b/x^(2/3))*a*b^2*x^(1/3)/((a + b/x^(2/3))*x ^(2/3) - b)
Exception generated. \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b/x^(2/3))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:simplify: Polynomials do not have t he same dimension Error: Bad Argument Value
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.37 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=\frac {x\,{\left (a+\frac {b}{x^{2/3}}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {3}{2};\ -\frac {1}{2};\ -\frac {b}{a\,x^{2/3}}\right )}{{\left (\frac {b}{a\,x^{2/3}}+1\right )}^{5/2}} \] Input:
int((a + b/x^(2/3))^(5/2),x)
Output:
(x*(a + b/x^(2/3))^(5/2)*hypergeom([-5/2, -3/2], -1/2, -b/(a*x^(2/3))))/(b /(a*x^(2/3)) + 1)^(5/2)
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.15 \[ \int \left (a+\frac {b}{x^{2/3}}\right )^{5/2} \, dx=\frac {14 x^{\frac {2}{3}} \sqrt {x^{\frac {2}{3}} a +b}\, a b +2 x^{\frac {4}{3}} \sqrt {x^{\frac {2}{3}} a +b}\, a^{2}-3 \sqrt {x^{\frac {2}{3}} a +b}\, b^{2}+15 x^{\frac {2}{3}} \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {x^{\frac {2}{3}} a +b}+x^{\frac {1}{3}} \sqrt {a}-\sqrt {b}}{\sqrt {b}}\right ) a b -15 x^{\frac {2}{3}} \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {x^{\frac {2}{3}} a +b}+x^{\frac {1}{3}} \sqrt {a}+\sqrt {b}}{\sqrt {b}}\right ) a b}{2 x^{\frac {2}{3}}} \] Input:
int((a+b/x^(2/3))^(5/2),x)
Output:
(14*x**(2/3)*sqrt(x**(2/3)*a + b)*a*b + 2*x**(1/3)*sqrt(x**(2/3)*a + b)*a* *2*x - 3*sqrt(x**(2/3)*a + b)*b**2 + 15*x**(2/3)*sqrt(b)*log((sqrt(x**(2/3 )*a + b) + x**(1/3)*sqrt(a) - sqrt(b))/sqrt(b))*a*b - 15*x**(2/3)*sqrt(b)* log((sqrt(x**(2/3)*a + b) + x**(1/3)*sqrt(a) + sqrt(b))/sqrt(b))*a*b)/(2*x **(2/3))