Integrand size = 21, antiderivative size = 119 \[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^{3/2}}{c \sqrt {b x^2+c x^4}}+\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{b} c^{5/4} \sqrt {b x^2+c x^4}} \] Output:
-x^(3/2)/c/(c*x^4+b*x^2)^(1/2)+1/2*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/ 2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1 /2*2^(1/2))/b^(1/4)/c^(5/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.50 \[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (-1+\sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^(9/2)/(b*x^2 + c*x^4)^(3/2),x]
Output:
(x^(3/2)*(-1 + Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x ^2)/b)]))/(c*Sqrt[x^2*(b + c*x^2)])
Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1427, 1431, 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.
\(\displaystyle \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1427 |
\(\displaystyle \frac {\int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{2 c}-\frac {x^{3/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{2 c \sqrt {b x^2+c x^4}}-\frac {x^{3/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{c \sqrt {b x^2+c x^4}}-\frac {x^{3/2}}{c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{b} c^{5/4} \sqrt {b x^2+c x^4}}-\frac {x^{3/2}}{c \sqrt {b x^2+c x^4}}\) |
Input:
Int[x^(9/2)/(b*x^2 + c*x^4)^(3/2),x]
Output:
-(x^(3/2)/(c*Sqrt[b*x^2 + c*x^4])) + (x*(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*b^(1/4)*c^(5/4)*Sqrt[b*x^2 + c*x^4])
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[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] - Simp[d^4*( (m + 2*p - 1)/(2*c*(p + 1))) Int[(d*x)^(m - 4)*(b*x^2 + c*x^4)^(p + 1), x ], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[p, -1] && GtQ[m + 2*p + 1, 2]
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]
Time = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (\sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-2 c x \right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{2}}\) | \(120\) |
Input:
int(x^(9/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*((-b*c)^(1/2)*((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)*(-c /(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1 /2*2^(1/2))-2*c*x)/c^2
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (c x^{3} + b x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} c \sqrt {x}}{c^{3} x^{3} + b c^{2} x} \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
((c*x^3 + b*x)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x^4 + b* x^2)*c*sqrt(x))/(c^3*x^3 + b*c^2*x)
\[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{\frac {9}{2}}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**(9/2)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**(9/2)/(x**2*(b + c*x**2))**(3/2), x)
\[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {9}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate(x^(9/2)/(c*x^4 + b*x^2)^(3/2), x)
\[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {9}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
integrate(x^(9/2)/(c*x^4 + b*x^2)^(3/2), x)
Timed out. \[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{9/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:
int(x^(9/2)/(b*x^2 + c*x^4)^(3/2),x)
Output:
int(x^(9/2)/(b*x^2 + c*x^4)^(3/2), x)
\[ \int \frac {x^{9/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-2 \sqrt {x}\, \sqrt {c \,x^{2}+b}+\left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{2}+\left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b c \,x^{2}}{c \left (c \,x^{2}+b \right )} \] Input:
int(x^(9/2)/(c*x^4+b*x^2)^(3/2),x)
Output:
( - 2*sqrt(x)*sqrt(b + c*x**2) + int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b**2 + int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b*c*x**2)/(c*(b + c*x**2))