Integrand size = 15, antiderivative size = 131 \[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {5 \sqrt {a+\frac {b}{x^4}} x^3}{6 a^2}+\frac {5 b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+\frac {b}{x^4}}} \] Output:
-1/2*x^3/a/(a+b/x^4)^(1/2)+5/6*(a+b/x^4)^(1/2)*x^3/a^2+5/12*b^(3/4)*((a+b/ x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*InverseJacobiAM( 2*arccot(a^(1/4)*x/b^(1/4)),1/2*2^(1/2))/a^(9/4)/(a+b/x^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {5 b+2 a x^4-5 b \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^4}{b}\right )}{6 a^2 \sqrt {a+\frac {b}{x^4}} x} \] Input:
Integrate[x^2/(a + b/x^4)^(3/2),x]
Output:
(5*b + 2*a*x^4 - 5*b*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^4)/b)])/(6*a^2*Sqrt[a + b/x^4]*x)
Time = 0.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {858, 819, 847, 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.
\(\displaystyle \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {x^4}{\left (a+\frac {b}{x^4}\right )^{3/2}}d\frac {1}{x}\) |
\(\Big \downarrow \) 819 |
\(\displaystyle -\frac {5 \int \frac {x^4}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{2 a}-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle -\frac {5 \left (-\frac {b \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{3 a}-\frac {x^3 \sqrt {a+\frac {b}{x^4}}}{3 a}\right )}{2 a}-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {5 \left (-\frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}}-\frac {x^3 \sqrt {a+\frac {b}{x^4}}}{3 a}\right )}{2 a}-\frac {x^3}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
Input:
Int[x^2/(a + b/x^4)^(3/2),x]
Output:
-1/2*x^3/(a*Sqrt[a + b/x^4]) - (5*(-1/3*(Sqrt[a + b/x^4]*x^3)/a - (b^(3/4) *Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ellip ticF[2*ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(6*a^(5/4)*Sqrt[a + b/x^4])))/(2 *a)
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Result contains complex when optimal does not.
Time = 1.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {\left (a \,x^{4}+b \right ) \left (2 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{5}-5 b \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b x \right )}{6 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} a^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(133\) |
risch | \(\frac {a \,x^{4}+b}{3 a^{2} x \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}-\frac {b \left (b \left (\frac {x}{2 b \sqrt {\left (x^{4}+\frac {b}{a}\right ) a}}+\frac {\sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a \,x^{4}+b}}\right )+4 a \left (-\frac {x}{2 a \sqrt {\left (x^{4}+\frac {b}{a}\right ) a}}+\frac {\sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a \,x^{4}+b}}\right )\right ) \sqrt {a \,x^{4}+b}}{3 a^{2} \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}\) | \(253\) |
Input:
int(x^2/(a+b/x^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/6*(a*x^4+b)*(2*(I*a^(1/2)/b^(1/2))^(1/2)*a*x^5-5*b*(-(I*a^(1/2)*x^2-b^(1 /2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I *a^(1/2)/b^(1/2))^(1/2),I)+5*(I*a^(1/2)/b^(1/2))^(1/2)*b*x)/((a*x^4+b)/x^4 )^(3/2)/x^6/a^2/(I*a^(1/2)/b^(1/2))^(1/2)
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=-\frac {5 \, {\left (a x^{4} + b\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (2 \, a x^{7} + 5 \, b x^{3}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{6 \, {\left (a^{3} x^{4} + a^{2} b\right )}} \] Input:
integrate(x^2/(a+b/x^4)^(3/2),x, algorithm="fricas")
Output:
-1/6*(5*(a*x^4 + b)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin((-b/a)^(1/4)/x) , -1) - (2*a*x^7 + 5*b*x^3)*sqrt((a*x^4 + b)/x^4))/(a^3*x^4 + a^2*b)
Result contains complex when optimal does not.
Time = 0.79 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.32 \[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=- \frac {x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \] Input:
integrate(x**2/(a+b/x**4)**(3/2),x)
Output:
-x**3*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*exp_polar(I*pi)/(a*x**4))/( 4*a**(3/2)*gamma(1/4))
\[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2/(a+b/x^4)^(3/2),x, algorithm="maxima")
Output:
integrate(x^2/(a + b/x^4)^(3/2), x)
\[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2/(a+b/x^4)^(3/2),x, algorithm="giac")
Output:
integrate(x^2/(a + b/x^4)^(3/2), x)
Timed out. \[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\int \frac {x^2}{{\left (a+\frac {b}{x^4}\right )}^{3/2}} \,d x \] Input:
int(x^2/(a + b/x^4)^(3/2),x)
Output:
int(x^2/(a + b/x^4)^(3/2), x)
\[ \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {\sqrt {a \,x^{4}+b}\, a \,x^{5}+5 \sqrt {a \,x^{4}+b}\, b x -5 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a^{2} x^{8}+2 a b \,x^{4}+b^{2}}d x \right ) a \,b^{2} x^{4}-5 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a^{2} x^{8}+2 a b \,x^{4}+b^{2}}d x \right ) b^{3}}{3 a^{2} \left (a \,x^{4}+b \right )} \] Input:
int(x^2/(a+b/x^4)^(3/2),x)
Output:
(sqrt(a*x**4 + b)*a*x**5 + 5*sqrt(a*x**4 + b)*b*x - 5*int(sqrt(a*x**4 + b) /(a**2*x**8 + 2*a*b*x**4 + b**2),x)*a*b**2*x**4 - 5*int(sqrt(a*x**4 + b)/( a**2*x**8 + 2*a*b*x**4 + b**2),x)*b**3)/(3*a**2*(a*x**4 + b))