\(\int \frac {(a+\frac {b}{x^4})^{3/2}}{x^4} \, dx\) [2073]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 257 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{15 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{15 \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\frac {4 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {2 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}} \]

[Out]

-1/9*(a+b/x^4)^(3/2)/x^3-2/15*a*(a+b/x^4)^(1/2)/x^3-4/15*a^2*(a+b/x^4)^(1/2)/x/b^(1/2)/(a^(1/2)+b^(1/2)/x^2)+4
/15*a^(9/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4)))*EllipticE(sin(2*arccot
(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)/b^(3/4)/(a+b
/x^4)^(1/2)-2/15*a^(9/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4)))*EllipticF
(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)
/b^(3/4)/(a+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {342, 285, 311, 226, 1210} \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {2 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {4 a^{9/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{15 \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{15 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3} \]

[In]

Int[(a + b/x^4)^(3/2)/x^4,x]

[Out]

(-2*a*Sqrt[a + b/x^4])/(15*x^3) - (a + b/x^4)^(3/2)/(9*x^3) - (4*a^2*Sqrt[a + b/x^4])/(15*Sqrt[b]*(Sqrt[a] + S
qrt[b]/x^2)*x) + (4*a^(9/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b/x^4]) - (2*a^(9/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]
/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b/x^4])

Rule 226

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]

Rule 285

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] + Dist[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]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
 x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 342

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] && IntegerQ[m]

Rule 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int x^2 \left (a+b x^4\right )^{3/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {1}{3} (2 a) \text {Subst}\left (\int x^2 \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right ) \\ & = -\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 {1}{15} \left (4 a^2\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\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 {\left (4 a^{5/2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{15 \sqrt {b}}+\frac {\left (4 a^{5/2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{15 \sqrt {b}} \\ & = -\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 ) F\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}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{2},-\frac {5}{4},-\frac {a x^4}{b}\right )}{9 x^7 \sqrt {1+\frac {a x^4}{b}}} \]

[In]

Integrate[(a + b/x^4)^(3/2)/x^4,x]

[Out]

-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])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.61

method result size
risch \(-\frac {\left (12 a^{2} x^{8}+11 a b \,x^{4}+5 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{45 x^{7} b}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{15 \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\) \(156\)
default \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (-12 i a^{\frac {5}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{9} b F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+12 i a^{\frac {5}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{9} b E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+12 \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}+23 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}+16 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{4}+5 b^{\frac {7}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\right )}{45 x^{3} \left (a \,x^{4}+b \right )^{2} b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) \(251\)

[In]

int((a+b/x^4)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=-\frac {12 \, a^{2} \sqrt {b} x^{7} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - 12 \, a^{2} \sqrt {b} x^{7} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (12 \, a^{2} x^{8} + 11 \, a b x^{4} + 5 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{45 \, b x^{7}} \]

[In]

integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="fricas")

[Out]

-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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.72 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.16 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=- \frac {a^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((a+b/x**4)**(3/2)/x**4,x)

[Out]

-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))

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]

[In]

integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((a + b/x^4)^(3/2)/x^4, x)

Giac [F]

\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \]

[In]

integrate((a+b/x^4)^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate((a + b/x^4)^(3/2)/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{x^4}\right )}^{3/2}}{x^4} \,d x \]

[In]

int((a + b/x^4)^(3/2)/x^4,x)

[Out]

int((a + b/x^4)^(3/2)/x^4, x)