Integrand size = 19, antiderivative size = 216 \[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=-\frac {78 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^4}+\frac {234 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^3}-\frac {26 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^2}+\frac {2 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a}+\frac {39 b^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 a^{17/4} \sqrt {b \sqrt [3]{x}+a x}} \]
-78/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^4+234/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^( 1/2)/a^3-26/55*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)/a^2+2/5*x^2*(b*x^(1/3)+a*x) ^(1/2)/a+39/77*b^(15/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2) ^(1/2)/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticF(sin(2*arctan(a^(1/ 4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3)) /(x^(1/3)*a^(1/2)+b^(1/2))^2)^(1/2)/a^(17/4)/(b*x^(1/3)+a*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.57 \[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x} \left (-195 b^4-78 a b^3 x^{2/3}+26 a^2 b^2 x^{4/3}-14 a^3 b x^2+77 a^4 x^{8/3}+195 b^4 \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^{2/3}}{b}\right )\right )}{385 a^4 \left (b+a x^{2/3}\right )} \]
(2*Sqrt[b*x^(1/3) + a*x]*(-195*b^4 - 78*a*b^3*x^(2/3) + 26*a^2*b^2*x^(4/3) - 14*a^3*b*x^2 + 77*a^4*x^(8/3) + 195*b^4*Sqrt[1 + (a*x^(2/3))/b]*Hyperge ometric2F1[1/4, 1/2, 5/4, -((a*x^(2/3))/b)]))/(385*a^4*(b + a*x^(2/3)))
Time = 0.44 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1924, 1930, 1930, 1930, 1930, 1917, 266, 761}
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.
| \(\displaystyle \int \frac {x^2}{\sqrt {a x+b \sqrt [3]{x}}} \, dx\) |
| \(\Big \downarrow \) 1924 |
| \(\displaystyle 3 \int \frac {x^{8/3}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \int \frac {x^2}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{15 a}\right )\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \int \frac {x^{4/3}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{11 a}\right )}{15 a}\right )\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \int \frac {x^{2/3}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{7 a}\right )}{11 a}\right )}{15 a}\right )\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b \int \frac {1}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{3 a}\right )}{7 a}\right )}{11 a}\right )}{15 a}\right )\) |
| \(\Big \downarrow \) 1917 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {x^{2/3} a+b} \sqrt [6]{x}}d\sqrt [3]{x}}{3 a \sqrt {a x+b \sqrt [3]{x}}}\right )}{7 a}\right )}{11 a}\right )}{15 a}\right )\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {2 b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{3 a \sqrt {a x+b \sqrt [3]{x}}}\right )}{7 a}\right )}{11 a}\right )}{15 a}\right )\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle 3 \left (\frac {2 x^2 \sqrt {a x+b \sqrt [3]{x}}}{15 a}-\frac {13 b \left (\frac {2 x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{11 a}-\frac {9 b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b^{3/4} \sqrt [6]{x} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {a x^{2/3}+b} \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 a^{5/4} \sqrt {a x+b \sqrt [3]{x}} \sqrt {a x^{4/3}+b}}\right )}{7 a}\right )}{11 a}\right )}{15 a}\right )\) |
3*((2*x^2*Sqrt[b*x^(1/3) + a*x])/(15*a) - (13*b*((2*x^(4/3)*Sqrt[b*x^(1/3) + a*x])/(11*a) - (9*b*((2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(7*a) - (5*b*((2 *Sqrt[b*x^(1/3) + a*x])/(3*a) - (b^(3/4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[ b + a*x^(2/3)]*x^(1/6)*Sqrt[(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^(2/3))^2] *EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(3*a^(5/4)*Sqrt[b*x^ (1/3) + a*x]*Sqrt[b + a*x^(4/3)])))/(7*a)))/(11*a)))/(15*a))
3.2.51.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp [1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x ], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j ] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1 ]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))) I nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
Time = 2.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(-\frac {-52 a^{3} b^{2} x^{\frac {5}{3}}+28 a^{4} b \,x^{\frac {7}{3}}-195 b^{4} \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+156 a^{2} b^{3} x -154 a^{5} x^{3}+390 a \,b^{4} x^{\frac {1}{3}}}{385 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{5}}\) | \(163\) |
| derivativedivides | \(\frac {2 x^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{5 a}-\frac {26 b \,x^{\frac {4}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{55 a^{2}}+\frac {234 b^{2} x^{\frac {2}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 a^{3}}-\frac {78 b^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 a^{4}}+\frac {39 b^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a^{5} \sqrt {b \,x^{\frac {1}{3}}+a x}}\) | \(201\) |
-1/385*(-52*a^3*b^2*x^(5/3)+28*a^4*b*x^(7/3)-195*b^4*(-a*b)^(1/2)*((a*x^(1 /3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^ (1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^ (1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+156*a^2*b^3*x-154*a^5*x^3+390*a*b^ 4*x^(1/3))/(x^(1/3)*(b+a*x^(2/3)))^(1/2)/a^5
\[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {x^{2}}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]
\[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {x^{2}}{\sqrt {a x + b \sqrt [3]{x}}}\, dx \]
\[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {x^{2}}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]
\[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {x^{2}}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {x^2}{\sqrt {a\,x+b\,x^{1/3}}} \,d x \]