Integrand size = 21, antiderivative size = 118 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2}}{b \sqrt {b x^2+c x^4}}+\frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 b^{5/4} \sqrt [4]{c} \sqrt {b x^2+c x^4}} \] Output:
x^(3/2)/b/(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^(5/4)/c^(1/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.87 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.51 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (1+\sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{b \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^(5/2)/(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)]))/(b*Sqrt[x^2*(b + c*x^2)])
Time = 0.40 (sec) , antiderivative size = 118, 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 = {1428, 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.
\(\displaystyle \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1428 |
\(\displaystyle \frac {\int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{2 b}+\frac {x^{3/2}}{b \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{2 b \sqrt {b x^2+c x^4}}+\frac {x^{3/2}}{b \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{b \sqrt {b x^2+c x^4}}+\frac {x^{3/2}}{b \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 b^{5/4} \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {x^{3/2}}{b \sqrt {b x^2+c x^4}}\) |
Input:
Int[x^(5/2)/(b*x^2 + c*x^4)^(3/2),x]
Output:
x^(3/2)/(b*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^(5/4)*c^(1/4)*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)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] + Simp[d^2* ((m + 4*p + 3)/(2*b*(p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[p, -1]
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 = 1.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (\sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+2 c x \right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c b}\) | \(123\) |
Input:
int(x^(5/2)/(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/b
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (c x^{3} + b x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} c \sqrt {x}}{b c^{2} x^{3} + b^{2} c x} \] Input:
integrate(x^(5/2)/(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))/(b*c^2*x^3 + b^2*c*x)
\[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{\frac {5}{2}}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**(5/2)/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**(5/2)/(x**2*(b + c*x**2))**(3/2), x)
\[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^(5/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate(x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)
\[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^(5/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
integrate(x^(5/2)/(c*x^4 + b*x^2)^(3/2), x)
Timed out. \[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:
int(x^(5/2)/(b*x^2 + c*x^4)^(3/2),x)
Output:
int(x^(5/2)/(b*x^2 + c*x^4)^(3/2), x)
\[ \int \frac {x^{5/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \] Input:
int(x^(5/2)/(c*x^4+b*x^2)^(3/2),x)
Output:
int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)