Integrand size = 21, antiderivative size = 290 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\frac {8 b^2 x^{3/2} \left (b+c x^2\right )}{15 \sqrt {c} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {4}{15} b \sqrt {x} \sqrt {b x^2+c x^4}+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}-\frac {8 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {b x^2+c x^4}}+\frac {4 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {b x^2+c x^4}} \]
2/9*(c*x^4+b*x^2)^(3/2)/x^(3/2)+8/15*b^2*x^(3/2)*(c*x^2+b)/c^(1/2)/(b^(1/2 )+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)+4/15*b*x^(1/2)*(c*x^4+b*x^2)^(1/2)-8/15*b ^(9/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^( 1/4)*x^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/ 2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^( 3/4)/(c*x^4+b*x^2)^(1/2)+4/15*b^(9/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1 /4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arct an(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/( b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.20 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\frac {2 b \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )}{3 \sqrt {1+\frac {c x^2}{b}}} \]
(2*b*Sqrt[x]*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-3/2, 3/4, 7/4, -((c* x^2)/b)])/(3*Sqrt[1 + (c*x^2)/b])
Time = 0.39 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1426, 1426, 1431, 266, 834, 27, 761, 1510}
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.
| \(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {2}{3} b \int \frac {\sqrt {c x^4+b x^2}}{\sqrt {x}}dx+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {2}{3} b \left (\frac {2}{5} b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {2}{3} b \left (\frac {2 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\) |
(2*(b*x^2 + c*x^4)^(3/2))/(9*x^(3/2)) + (2*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^ 4])/5 + (4*b*x*Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S qrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4) *Sqrt[b + c*x^2]))/Sqrt[c]) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x ^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)] , 1/2])/(2*c^(3/4)*Sqrt[b + c*x^2])))/(5*Sqrt[b*x^2 + c*x^4])))/3
3.4.68.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 0.18 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.78
| method | result | size |
| default | \(\frac {2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (5 c^{3} x^{6}+12 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+16 b \,c^{2} x^{4}+11 b^{2} c \,x^{2}\right )}{45 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c}\) | \(226\) |
| risch | \(\frac {2 \sqrt {x}\, \left (5 c \,x^{2}+11 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{45}+\frac {4 b^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(227\) |
2/45*(c*x^4+b*x^2)^(3/2)*(5*c^3*x^6+12*b^3*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2 ))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/ 2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-6 *b^3*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/ (-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2) )/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+16*b*c^2*x^4+11*b^2*c*x^2)/x^(7/2)/(c*x ^2+b)^2/c
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.21 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=-\frac {2 \, {\left (12 \, b^{2} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - \sqrt {c x^{4} + b x^{2}} {\left (5 \, c^{2} x^{2} + 11 \, b c\right )} \sqrt {x}\right )}}{45 \, c} \]
-2/45*(12*b^2*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/ c, 0, x)) - sqrt(c*x^4 + b*x^2)*(5*c^2*x^2 + 11*b*c)*sqrt(x))/c
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{5/2}} \,d x \]