Integrand size = 26, antiderivative size = 137 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {3 a^2 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 b^3}-\frac {6 a \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{5 b^3}+\frac {\left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{2 b^3} \] Output:
3/4*a^2*(a+b*x^(1/3))^3*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3-6/5*a*(a +b*x^(1/3))^4*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3+1/2*(a+b*x^(1/3))^ 5*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.49 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^{3/2} \left (20 a^3 x+45 a^2 b x^{4/3}+36 a b^2 x^{5/3}+10 b^3 x^2\right )}{20 \left (a+b \sqrt [3]{x}\right )^3} \] Input:
Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(3/2),x]
Output:
(((a + b*x^(1/3))^2)^(3/2)*(20*a^3*x + 45*a^2*b*x^(4/3) + 36*a*b^2*x^(5/3) + 10*b^3*x^2))/(20*(a + b*x^(1/3))^3)
Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.68, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 774, 27, 49, 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.
\(\displaystyle \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \int \left (\sqrt [3]{x} b^2+a b\right )^3dx}{a b^3+b^4 \sqrt [3]{x}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \int b^3 \left (a+b \sqrt [3]{x}\right )^3 x^{2/3}d\sqrt [3]{x}}{a b^3+b^4 \sqrt [3]{x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 b^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \int \left (a+b \sqrt [3]{x}\right )^3 x^{2/3}d\sqrt [3]{x}}{a b^3+b^4 \sqrt [3]{x}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {3 b^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \int \left (x^{2/3} a^3+3 b x a^2+3 b^2 x^{4/3} a+b^3 x^{5/3}\right )d\sqrt [3]{x}}{a b^3+b^4 \sqrt [3]{x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 b^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (\frac {a^3 x}{3}+\frac {3}{4} a^2 b x^{4/3}+\frac {3}{5} a b^2 x^{5/3}+\frac {b^3 x^2}{6}\right )}{a b^3+b^4 \sqrt [3]{x}}\) |
Input:
Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(3/2),x]
Output:
(3*b^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]*((a^3*x)/3 + (3*a^2*b*x^(4/ 3))/4 + (3*a*b^2*x^(5/3))/5 + (b^3*x^2)/6))/(a*b^3 + b^4*x^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39
method | result | size |
derivativedivides | \(\frac {{\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )}^{\frac {3}{2}} x \left (10 b^{3} x +36 a \,b^{2} x^{\frac {2}{3}}+45 a^{2} b \,x^{\frac {1}{3}}+20 a^{3}\right )}{20 \left (a +b \,x^{\frac {1}{3}}\right )^{3}}\) | \(54\) |
default | \(\frac {\left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{\frac {3}{2}} \left (36 b^{2} a \,x^{\frac {5}{3}}+45 a^{2} b \,x^{\frac {4}{3}}+10 b^{3} x^{2}+20 a^{3} x \right )}{20 \left (a +b \,x^{\frac {1}{3}}\right )^{3}}\) | \(65\) |
Input:
int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/20*((a+b*x^(1/3))^2)^(3/2)*x*(10*b^3*x+36*a*b^2*x^(2/3)+45*a^2*b*x^(1/3) +20*a^3)/(a+b*x^(1/3))^3
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.23 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {1}{2} \, b^{3} x^{2} + \frac {9}{5} \, a b^{2} x^{\frac {5}{3}} + \frac {9}{4} \, a^{2} b x^{\frac {4}{3}} + a^{3} x \] Input:
integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(3/2),x, algorithm="fricas")
Output:
1/2*b^3*x^2 + 9/5*a*b^2*x^(5/3) + 9/4*a^2*b*x^(4/3) + a^3*x
Time = 0.62 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.28 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=3 \left (\begin {cases} \sqrt {a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}} \left (\frac {a^{5}}{60 b^{3}} - \frac {a^{4} \sqrt [3]{x}}{60 b^{2}} + \frac {a^{3} x^{\frac {2}{3}}}{60 b} + \frac {19 a^{2} x}{60} + \frac {13 a b x^{\frac {4}{3}}}{30} + \frac {b^{2} x^{\frac {5}{3}}}{6}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {5}{2}}}{5} - \frac {2 a^{2} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {9}{2}}}{9}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**(3/2),x)
Output:
3*Piecewise((sqrt(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))*(a**5/(60*b**3) - a**4*x**(1/3)/(60*b**2) + a**3*x**(2/3)/(60*b) + 19*a**2*x/60 + 13*a*b*x* *(4/3)/30 + b**2*x**(5/3)/6), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x**(1/3)) **(5/2)/5 - 2*a**2*(a**2 + 2*a*b*x**(1/3))**(7/2)/7 + (a**2 + 2*a*b*x**(1/ 3))**(9/2)/9)/(4*a**3*b**3), Ne(a*b, 0)), (x*(a**2)**(3/2)/3, True))
Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{\frac {1}{3}}}{4 \, b^{2}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} x^{\frac {1}{3}}}{2 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} a}{10 \, b^{3}} \] Input:
integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(3/2),x, algorithm="maxima")
Output:
3/4*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(3/2)*a^2*x^(1/3)/b^2 + 3/4*(b^2*x ^(2/3) + 2*a*b*x^(1/3) + a^2)^(3/2)*a^3/b^3 + 1/2*(b^2*x^(2/3) + 2*a*b*x^( 1/3) + a^2)^(5/2)*x^(1/3)/b^2 - 7/10*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^( 5/2)*a/b^3
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {1}{2} \, b^{3} x^{2} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {9}{5} \, a b^{2} x^{\frac {5}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {9}{4} \, a^{2} b x^{\frac {4}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) \] Input:
integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(3/2),x, algorithm="giac")
Output:
1/2*b^3*x^2*sgn(b*x^(1/3) + a) + 9/5*a*b^2*x^(5/3)*sgn(b*x^(1/3) + a) + 9/ 4*a^2*b*x^(4/3)*sgn(b*x^(1/3) + a) + a^3*x*sgn(b*x^(1/3) + a)
Timed out. \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\int {\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^{3/2} \,d x \] Input:
int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(3/2),x)
Output:
int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.24 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx=\frac {x \left (36 x^{\frac {2}{3}} a \,b^{2}+45 x^{\frac {1}{3}} a^{2} b +20 a^{3}+10 b^{3} x \right )}{20} \] Input:
int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(3/2),x)
Output:
(x*(36*x**(2/3)*a*b**2 + 45*x**(1/3)*a**2*b + 20*a**3 + 10*b**3*x))/20