Integrand size = 17, antiderivative size = 186 \[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=-\frac {8 a^3 \sqrt {a x+b x^3}}{231 b^2}+\frac {8 a^2 x^2 \sqrt {a x+b x^3}}{385 b}+\frac {4}{55} a x^4 \sqrt {a x+b x^3}+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac {4 a^{15/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{231 b^{9/4} \sqrt {a x+b x^3}} \]
2/15*x^3*(b*x^3+a*x)^(3/2)-8/231*a^3*(b*x^3+a*x)^(1/2)/b^2+8/385*a^2*x^2*( b*x^3+a*x)^(1/2)/b+4/55*a*x^4*(b*x^3+a*x)^(1/2)+4/231*a^(15/4)*(cos(2*arct an(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4) ))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+ x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(9/4)/(b*x^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 = 94, normalized size of antiderivative = 0.51 \[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\frac {2 \sqrt {x \left (a+b x^2\right )} \left (-\left (\left (5 a-11 b x^2\right ) \left (a+b x^2\right )^2 \sqrt {1+\frac {b x^2}{a}}\right )+5 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{165 b^2 \sqrt {1+\frac {b x^2}{a}}} \]
(2*Sqrt[x*(a + b*x^2)]*(-((5*a - 11*b*x^2)*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/ a]) + 5*a^3*Hypergeometric2F1[-3/2, 1/4, 5/4, -((b*x^2)/a)]))/(165*b^2*Sqr t[1 + (b*x^2)/a])
Time = 0.37 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1927, 1927, 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.
| \(\displaystyle \int x^2 \left (a x+b x^3\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1927 |
| \(\displaystyle \frac {2}{5} a \int x^3 \sqrt {b x^3+a x}dx+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 1927 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \int \frac {x^4}{\sqrt {b x^3+a x}}dx+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \int \frac {x^2}{\sqrt {b x^3+a x}}dx}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 1930 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^3+a x}}dx}{3 b}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 1917 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{3 b \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2}{5} a \left (\frac {2}{11} a \left (\frac {2 x^2 \sqrt {a x+b x^3}}{7 b}-\frac {5 a \left (\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}}\right )}{7 b}\right )+\frac {2}{11} x^4 \sqrt {a x+b x^3}\right )+\frac {2}{15} x^3 \left (a x+b x^3\right )^{3/2}\) |
(2*x^3*(a*x + b*x^3)^(3/2))/15 + (2*a*((2*x^4*Sqrt[a*x + b*x^3])/11 + (2*a *((2*x^2*Sqrt[a*x + b*x^3])/(7*b) - (5*a*((2*Sqrt[a*x + b*x^3])/(3*b) - (a ^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x )^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(3*b^(5/4)*Sqrt[ a*x + b*x^3])))/(7*b)))/11))/5
3.1.46.3.1 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[((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 + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
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 = 169, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(-\frac {2 \left (-77 b^{3} x^{6}-119 a \,b^{2} x^{4}-12 a^{2} b \,x^{2}+20 a^{3}\right ) x \left (b \,x^{2}+a \right )}{1155 b^{2} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {4 a^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\) | \(169\) |
| default | \(\frac {2 b \,x^{6} \sqrt {b \,x^{3}+a x}}{15}+\frac {34 a \,x^{4} \sqrt {b \,x^{3}+a x}}{165}+\frac {8 a^{2} x^{2} \sqrt {b \,x^{3}+a x}}{385 b}-\frac {8 a^{3} \sqrt {b \,x^{3}+a x}}{231 b^{2}}+\frac {4 a^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\) | \(188\) |
| elliptic | \(\frac {2 b \,x^{6} \sqrt {b \,x^{3}+a x}}{15}+\frac {34 a \,x^{4} \sqrt {b \,x^{3}+a x}}{165}+\frac {8 a^{2} x^{2} \sqrt {b \,x^{3}+a x}}{385 b}-\frac {8 a^{3} \sqrt {b \,x^{3}+a x}}{231 b^{2}}+\frac {4 a^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{231 b^{3} \sqrt {b \,x^{3}+a x}}\) | \(188\) |
-2/1155*(-77*b^3*x^6-119*a*b^2*x^4-12*a^2*b*x^2+20*a^3)/b^2*x*(b*x^2+a)/(x *(b*x^2+a))^(1/2)+4/231*a^4/b^3*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^(1 /2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2) *b)^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^ (1/2),1/2*2^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (20 \, a^{4} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (77 \, b^{4} x^{6} + 119 \, a b^{3} x^{4} + 12 \, a^{2} b^{2} x^{2} - 20 \, a^{3} b\right )} \sqrt {b x^{3} + a x}\right )}}{1155 \, b^{3}} \]
2/1155*(20*a^4*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) + (77*b^4*x^6 + 1 19*a*b^3*x^4 + 12*a^2*b^2*x^2 - 20*a^3*b)*sqrt(b*x^3 + a*x))/b^3
\[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\int x^{2} \left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}\, dx \]
\[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a x\right )}^{\frac {3}{2}} x^{2} \,d x } \]
\[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\int { {\left (b x^{3} + a x\right )}^{\frac {3}{2}} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (a x+b x^3\right )^{3/2} \, dx=\int x^2\,{\left (b\,x^3+a\,x\right )}^{3/2} \,d x \]