Integrand size = 21, antiderivative size = 149 \[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=-\frac {10 b \sqrt {b x^2+c x^4}}{21 c^2 \sqrt {x}}+\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}+\frac {5 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 c^{9/4} \sqrt {b x^2+c x^4}} \] Output:
-10/21*b*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/2)+2/7*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c +5/21*b^(7/4)*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))/c^(9/4)/( 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 = 86, normalized size of antiderivative = 0.58 \[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (-5 b^2-2 b c x^2+3 c^2 x^4+5 b^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{21 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^(9/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*x^(3/2)*(-5*b^2 - 2*b*c*x^2 + 3*c^2*x^4 + 5*b^2*Sqrt[1 + (c*x^2)/b]*Hyp ergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(21*c^2*Sqrt[x^2*(b + c*x^2) ])
Time = 0.48 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1429, 1429, 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.
\(\displaystyle \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\) |
Input:
Int[x^(9/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) - (5*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c *Sqrt[x]) - (b^(3/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/ 4)*Sqrt[b*x^2 + c*x^4])))/(7*c)
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)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[m + 2*p - 1, 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]
Time = 0.71 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\sqrt {x}\, \left (5 b^{2} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+6 c^{3} x^{5}-4 b \,c^{2} x^{3}-10 b^{2} c x \right )}{21 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{3}}\) | \(137\) |
risch | \(-\frac {2 \left (-3 c \,x^{2}+5 b \right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{21 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {5 b^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 c^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(178\) |
Input:
int(x^(9/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/21/(c*x^4+b*x^2)^(1/2)*x^(1/2)*(5*b^2*(-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))+6*c^3*x^5-4*b*c^2*x^3-10*b^2*c*x)/c^3
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.38 \[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \, {\left (5 \, b^{2} \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} {\left (3 \, c^{2} x^{2} - 5 \, b c\right )} \sqrt {x}\right )}}{21 \, c^{3} x} \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
2/21*(5*b^2*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c*x^4 + b*x ^2)*(3*c^2*x^2 - 5*b*c)*sqrt(x))/(c^3*x)
\[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{\frac {9}{2}}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**(9/2)/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**(9/2)/sqrt(x**2*(b + c*x**2)), x)
\[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {9}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(x^(9/2)/sqrt(c*x^4 + b*x^2), x)
\[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {9}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
integrate(x^(9/2)/sqrt(c*x^4 + b*x^2), x)
Timed out. \[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{9/2}}{\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int(x^(9/2)/(b*x^2 + c*x^4)^(1/2),x)
Output:
int(x^(9/2)/(b*x^2 + c*x^4)^(1/2), x)
\[ \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {-10 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b +6 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c \,x^{2}+5 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b^{2}}{21 c^{2}} \] Input:
int(x^(9/2)/(c*x^4+b*x^2)^(1/2),x)
Output:
( - 10*sqrt(x)*sqrt(b + c*x**2)*b + 6*sqrt(x)*sqrt(b + c*x**2)*c*x**2 + 5* int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3),x)*b**2)/(21*c**2)