Integrand size = 19, antiderivative size = 163 \[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{7 b x^{4/3}}+\frac {10 a \sqrt {b \sqrt [3]{x}+a x}}{7 b^2 x^{2/3}}+\frac {5 a^{7/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{7 b^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \]
-6/7*(b*x^(1/3)+a*x)^(1/2)/b/x^(4/3)+10/7*a*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(2 /3)+5/7*a^(7/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/c os(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)/b^(9/4)/(b*x^(1/3)+a*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36 \[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=-\frac {6 \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {1}{2},-\frac {3}{4},-\frac {a x^{2/3}}{b}\right )}{7 x \sqrt {b \sqrt [3]{x}+a x}} \]
(-6*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[-7/4, 1/2, -3/4, -((a*x^(2/3 ))/b)])/(7*x*Sqrt[b*x^(1/3) + a*x])
Time = 0.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1924, 1931, 1931, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a x+b \sqrt [3]{x}}} \, dx\) |
\(\Big \downarrow \) 1924 |
\(\displaystyle 3 \int \frac {1}{x^{4/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle 3 \left (-\frac {5 a \int \frac {1}{x^{2/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle 3 \left (-\frac {5 a \left (-\frac {a \int \frac {1}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{3 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle 3 \left (-\frac {5 a \left (-\frac {a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {x^{2/3} a+b} \sqrt [6]{x}}d\sqrt [3]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )\) |
\(\Big \downarrow \) 266 |
\(\displaystyle 3 \left (-\frac {5 a \left (-\frac {2 a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle 3 \left (-\frac {5 a \left (-\frac {a^{3/4} \sqrt [6]{x} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {a x^{2/3}+b} \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b \sqrt [3]{x}} \sqrt {a x^{4/3}+b}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )\) |
3*((-2*Sqrt[b*x^(1/3) + a*x])/(7*b*x^(4/3)) - (5*a*((-2*Sqrt[b*x^(1/3) + a *x])/(3*b*x^(2/3)) - (a^(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*b^(5/4)*Sqrt[b*x^(1/3) + a*x ]*Sqrt[b + a*x^(4/3)])))/(7*b))
3.2.55.3.1 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^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*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]) && LtQ[ m + j*p + 1, 0]
Time = 2.02 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {5 a \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) x^{\frac {4}{3}}+4 a b x +10 x^{\frac {5}{3}} a^{2}-6 b^{2} x^{\frac {1}{3}}}{7 b^{2} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, x^{\frac {4}{3}}}\) | \(142\) |
derivativedivides | \(-\frac {6 \sqrt {b \,x^{\frac {1}{3}}+a x}}{7 b \,x^{\frac {4}{3}}}+\frac {10 a \sqrt {b \,x^{\frac {1}{3}}+a x}}{7 b^{2} x^{\frac {2}{3}}}+\frac {5 a \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{7 b^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\) | \(158\) |
1/7*(5*a*(-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))*x^(4 /3)+4*a*b*x+10*x^(5/3)*a^2-6*b^2*x^(1/3))/b^2/(x^(1/3)*(b+a*x^(2/3)))^(1/2 )/x^(4/3)
\[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}} x^{2}} \,d x } \]
integral((a^2*x^2 - a*b*x^(4/3) + b^2*x^(2/3))*sqrt(a*x + b*x^(1/3))/(a^3* x^5 + b^3*x^3), x)
\[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {1}{x^{2} \sqrt {a x + b \sqrt [3]{x}}}\, dx \]
\[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}} x^{2}} \,d x } \]
\[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{x^2 \sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {1}{x^2\,\sqrt {a\,x+b\,x^{1/3}}} \,d x \]