Integrand size = 19, antiderivative size = 213 \[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=-\frac {12 a \sqrt {b \sqrt [3]{x}+a x}}{55 x^2}-\frac {24 a^2 \sqrt {b \sqrt [3]{x}+a x}}{385 b x^{4/3}}+\frac {8 a^3 \sqrt {b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac {4 a^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 b^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \]
-2/5*(b*x^(1/3)+a*x)^(3/2)/x^3-12/55*a*(b*x^(1/3)+a*x)^(1/2)/x^2-24/385*a^ 2*(b*x^(1/3)+a*x)^(1/2)/b/x^(4/3)+8/77*a^3*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(2/ 3)+4/77*a^(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)/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 = 10.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.29 \[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=-\frac {2 b \sqrt {b \sqrt [3]{x}+a x} \operatorname {Hypergeometric2F1}\left (-\frac {15}{4},-\frac {3}{2},-\frac {11}{4},-\frac {a x^{2/3}}{b}\right )}{5 \sqrt {1+\frac {a x^{2/3}}{b}} x^{8/3}} \]
(-2*b*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-15/4, -3/2, -11/4, -((a*x^( 2/3))/b)])/(5*Sqrt[1 + (a*x^(2/3))/b]*x^(8/3))
Time = 0.43 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1924, 1926, 1926, 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.
\(\displaystyle \int \frac {\left (a x+b \sqrt [3]{x}\right )^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 1924 |
\(\displaystyle 3 \int \frac {\left (\sqrt [3]{x} b+a x\right )^{3/2}}{x^{10/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle 3 \left (\frac {2}{5} a \int \frac {\sqrt {\sqrt [3]{x} b+a x}}{x^{7/3}}d\sqrt [3]{x}-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \int \frac {1}{x^{4/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \left (-\frac {5 a \int \frac {1}{x^{2/3} \sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \left (-\frac {5 a \left (-\frac {a \int \frac {1}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{3 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \left (-\frac {5 a \left (-\frac {a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {x^{2/3} a+b} \sqrt [6]{x}}d\sqrt [3]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 266 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \left (-\frac {5 a \left (-\frac {2 a \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{3 b \sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {2}{11} a \left (-\frac {5 a \left (-\frac {a^{3/4} \sqrt [6]{x} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {a x^{2/3}+b} \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b \sqrt [3]{x}} \sqrt {a x^{4/3}+b}}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 b x^{2/3}}\right )}{7 b}-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{7 b x^{4/3}}\right )-\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{11 x^2}\right )-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{15 x^3}\right )\) |
3*((-2*(b*x^(1/3) + a*x)^(3/2))/(15*x^3) + (2*a*((-2*Sqrt[b*x^(1/3) + a*x] )/(11*x^2) + (2*a*((-2*Sqrt[b*x^(1/3) + a*x])/(7*b*x^(4/3)) - (5*a*((-2*Sq rt[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)))/11))/5)
3.2.46.3.1 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*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p *((n - j)/(c^n*(m + j*p + 1))) Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Integer sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
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 = 3.00 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {\frac {4 a^{3} \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) x^{\frac {14}{3}}}{77}-\frac {262 x^{\frac {11}{3}} a^{2} b^{2}}{385}+\frac {16 x^{\frac {13}{3}} a^{3} b}{385}-\frac {56 a \,b^{3} x^{3}}{55}+\frac {8 a^{4} x^{5}}{77}-\frac {2 x^{\frac {7}{3}} b^{4}}{5}}{b^{2} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, x^{\frac {14}{3}}}\) | \(168\) |
derivativedivides | \(-\frac {2 b \sqrt {b \,x^{\frac {1}{3}}+a x}}{5 x^{\frac {8}{3}}}-\frac {34 a \sqrt {b \,x^{\frac {1}{3}}+a x}}{55 x^{2}}-\frac {24 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 b \,x^{\frac {4}{3}}}+\frac {8 a^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 b^{2} x^{\frac {2}{3}}}+\frac {4 a^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 b^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\) | \(199\) |
2/385*(10*a^3*(-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^(14/3)-131*x^(11/3)*a^2*b^2+8*x^(13/3)*a^3*b-196*a*b^3*x^3+20*a^4*x^5-7 7*x^(7/3)*b^4)/b^2/(x^(1/3)*(b+a*x^(2/3)))^(1/2)/x^(14/3)
\[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]
\[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=\int \frac {\left (a x + b \sqrt [3]{x}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]
\[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]
\[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a\,x+b\,x^{1/3}\right )}^{3/2}}{x^4} \,d x \]