Integrand size = 15, antiderivative size = 257 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{15 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{15 \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\frac {4 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {2 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}} \] Output:
-2/15*a*(a+b/x^4)^(1/2)/x^3-1/9*(a+b/x^4)^(3/2)/x^3-4/15*a^2*(a+b/x^4)^(1/ 2)/b^(1/2)/(a^(1/2)+b^(1/2)/x^2)/x+4/15*a^(9/4)*((a+b/x^4)/(a^(1/2)+b^(1/2 )/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*EllipticE(sin(2*arccot(a^(1/4)*x/b^( 1/4))),1/2*2^(1/2))/b^(3/4)/(a+b/x^4)^(1/2)-2/15*a^(9/4)*((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))/b^(3/4)/(a+b/x^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{2},-\frac {5}{4},-\frac {a x^4}{b}\right )}{9 x^7 \sqrt {1+\frac {a x^4}{b}}} \] Input:
Integrate[(a + b/x^4)^(3/2)/x^4,x]
Output:
-1/9*(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((a*x^4)/b)]) /(x^7*Sqrt[1 + (a*x^4)/b])
Time = 0.59 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {858, 811, 811, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle -\frac {2}{3} a \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^2}d\frac {1}{x}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle -\frac {2}{3} a \left (\frac {2}{5} a \int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}\) |
Input:
Int[(a + b/x^4)^(3/2)/x^4,x]
Output:
-1/9*(a + b/x^4)^(3/2)/x^3 - (2*a*(Sqrt[a + b/x^4]/(5*x^3) + (2*a*(-((-(Sq rt[a + b/x^4]/((Sqrt[a] + Sqrt[b]/x^2)*x)) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sq rt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*ArcTan[b^(1/4) /(a^(1/4)*x)], 1/2])/(b^(1/4)*Sqrt[a + b/x^4]))/Sqrt[b]) + (a^(1/4)*Sqrt[( a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2* ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(2*b^(3/4)*Sqrt[a + b/x^4])))/5))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
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]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.61
method | result | size |
risch | \(-\frac {\left (12 a^{2} x^{8}+11 a b \,x^{4}+5 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{45 x^{7} b}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{15 \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\) | \(156\) |
default | \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (-12 i a^{\frac {5}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{9} b \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+12 i a^{\frac {5}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{9} b \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+12 \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}+23 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}+16 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{4}+5 b^{\frac {7}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\right )}{45 x^{3} \left (a \,x^{4}+b \right )^{2} b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(251\) |
Input:
int((a+b/x^4)^(3/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/45*(12*a^2*x^8+11*a*b*x^4+5*b^2)/x^7/b*((a*x^4+b)/x^4)^(1/2)+4/15*I/b^( 1/2)*a^(5/2)/(I*a^(1/2)/b^(1/2))^(1/2)*(1-I*a^(1/2)/b^(1/2)*x^2)^(1/2)*(1+ I*a^(1/2)/b^(1/2)*x^2)^(1/2)/(a*x^4+b)*(EllipticF(x*(I*a^(1/2)/b^(1/2))^(1 /2),I)-EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I))*((a*x^4+b)/x^4)^(1/2)*x^2
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {12 \, a^{2} \sqrt {b} x^{7} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - 12 \, a^{2} \sqrt {b} x^{7} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (12 \, a^{2} x^{8} + 11 \, a b x^{4} + 5 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{45 \, b x^{7}} \] Input:
integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="fricas")
Output:
-1/45*(12*a^2*sqrt(b)*x^7*(-a/b)^(3/4)*elliptic_e(arcsin(x*(-a/b)^(1/4)), -1) - 12*a^2*sqrt(b)*x^7*(-a/b)^(3/4)*elliptic_f(arcsin(x*(-a/b)^(1/4)), - 1) + (12*a^2*x^8 + 11*a*b*x^4 + 5*b^2)*sqrt((a*x^4 + b)/x^4))/(b*x^7)
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.16 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=- \frac {a^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((a+b/x**4)**(3/2)/x**4,x)
Output:
-a**(3/2)*gamma(3/4)*hyper((-3/2, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**4) )/(4*x**3*gamma(7/4))
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="maxima")
Output:
integrate((a + b/x^4)^(3/2)/x^4, x)
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="giac")
Output:
integrate((a + b/x^4)^(3/2)/x^4, x)
Timed out. \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{x^4}\right )}^{3/2}}{x^4} \,d x \] Input:
int((a + b/x^4)^(3/2)/x^4,x)
Output:
int((a + b/x^4)^(3/2)/x^4, x)
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\frac {-7 \sqrt {a \,x^{4}+b}\, a \,x^{4}-\sqrt {a \,x^{4}+b}\, b +12 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a \,x^{14}+b \,x^{10}}d x \right ) b^{2} x^{9}}{21 x^{9}} \] Input:
int((a+b/x^4)^(3/2)/x^4,x)
Output:
( - 7*sqrt(a*x**4 + b)*a*x**4 - sqrt(a*x**4 + b)*b + 12*int(sqrt(a*x**4 + b)/(a*x**14 + b*x**10),x)*b**2*x**9)/(21*x**9)