Integrand size = 15, antiderivative size = 131 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {1}{2 a x^3 \sqrt {a+b x^4}}-\frac {5 \sqrt {a+b x^4}}{6 a^2 x^3}-\frac {5 b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}} \] Output:
1/2/a/x^3/(b*x^4+a)^(1/2)-5/6*(b*x^4+a)^(1/2)/a^2/x^3-5/12*b^(3/4)*(a^(1/2 )+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2 *arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(9/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {b x^4}{a}\right )}{3 a x^3 \sqrt {a+b x^4}} \] Input:
Integrate[1/(x^4*(a + b*x^4)^(3/2)),x]
Output:
-1/3*(Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-3/4, 3/2, 1/4, -((b*x^4)/a)]) /(a*x^3*Sqrt[a + b*x^4])
Time = 0.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {5 \int \frac {1}{x^4 \sqrt {b x^4+a}}dx}{2 a}+\frac {1}{2 a x^3 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {5 \left (-\frac {b \int \frac {1}{\sqrt {b x^4+a}}dx}{3 a}-\frac {\sqrt {a+b x^4}}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {5 \left (-\frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt {a+b x^4}}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \sqrt {a+b x^4}}\) |
Input:
Int[1/(x^4*(a + b*x^4)^(3/2)),x]
Output:
1/(2*a*x^3*Sqrt[a + b*x^4]) + (5*(-1/3*Sqrt[a + b*x^4]/(a*x^3) - (b^(3/4)* (Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellipt icF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 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]
Result contains complex when optimal does not.
Time = 1.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {b x}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(113\) |
elliptic | \(-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {b x}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(113\) |
risch | \(-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {b \left (b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{3 a^{2}}\) | \(217\) |
Input:
int(1/x^4/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/3*(b*x^4+a)^(1/2)/a^2/x^3-1/2*b/a^2*x/((x^4+a/b)*b)^(1/2)-5/6*b/a^2/(I/ a^(1/2)*b^(1/2))^(1/2)*(1-I*b^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*b^(1/2)*x^2/a^ (1/2))^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (b x^{7} + a x^{3}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, b x^{4} + 2 \, a\right )} \sqrt {b x^{4} + a}}{6 \, {\left (a^{2} b x^{7} + a^{3} x^{3}\right )}} \] Input:
integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/6*(5*(b*x^7 + a*x^3)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/ 4)), -1) - (5*b*x^4 + 2*a)*sqrt(b*x^4 + a))/(a^2*b*x^7 + a^3*x^3)
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {1}{4}\right )} \] Input:
integrate(1/x**4/(b*x**4+a)**(3/2),x)
Output:
gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/ 2)*x**3*gamma(1/4))
\[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^4 + a)^(3/2)*x^4), x)
\[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(3/2)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int(1/(x^4*(a + b*x^4)^(3/2)),x)
Output:
int(1/(x^4*(a + b*x^4)^(3/2)), x)
\[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {b \,x^{4}+a}}{b^{2} x^{12}+2 a b \,x^{8}+a^{2} x^{4}}d x \] Input:
int(1/x^4/(b*x^4+a)^(3/2),x)
Output:
int(sqrt(a + b*x**4)/(a**2*x**4 + 2*a*b*x**8 + b**2*x**12),x)