\(\int \frac {(a+\frac {b}{x^4})^{5/2}}{x^2} \, dx\) [2080]

Optimal result

Integrand size = 15, antiderivative size = 147 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {20 a^2 \sqrt {a+\frac {b}{x^4}}}{77 x}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {20 a^{11/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}} \]

[Out]

-10/77*a*(a+b/x^4)^(3/2)/x-1/11*(a+b/x^4)^(5/2)/x-20/77*a^2*(a+b/x^4)^(1/2)/x-20/77*a^(11/4)*(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^(1/4)/(a+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {342, 201, 226} \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {20 a^{11/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {20 a^2 \sqrt {a+\frac {b}{x^4}}}{77 x}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x} \]

[In]

Int[(a + b/x^4)^(5/2)/x^2,x]

[Out]

(-20*a^2*Sqrt[a + b/x^4])/(77*x) - (10*a*(a + b/x^4)^(3/2))/(77*x) - (a + b/x^4)^(5/2)/(11*x) - (20*a^(11/4)*S
qrt[(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])/(77*b^(1/4)*Sqrt[a + b/x^4])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \left (a+b x^4\right )^{5/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {1}{11} (10 a) \text {Subst}\left (\int \left (a+b x^4\right )^{3/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {1}{77} \left (60 a^2\right ) \text {Subst}\left (\int \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {20 a^2 \sqrt {a+\frac {b}{x^4}}}{77 x}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {1}{77} \left (40 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {20 a^2 \sqrt {a+\frac {b}{x^4}}}{77 x}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {20 a^{11/4} \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 )}{77 \sqrt [4]{b} \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 = 54, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {b^2 \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {5}{2},-\frac {7}{4},-\frac {a x^4}{b}\right )}{11 x^9 \sqrt {1+\frac {a x^4}{b}}} \]

[In]

Integrate[(a + b/x^4)^(5/2)/x^2,x]

[Out]

-1/11*(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-11/4, -5/2, -7/4, -((a*x^4)/b)])/(x^9*Sqrt[1 + (a*x^4)/b])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\left (37 a^{2} x^{8}+24 a b \,x^{4}+7 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{77 x^{9}}+\frac {40 a^{3} \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{77 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\)	\(130\)
default	\(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-40 a^{3} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) x^{11}+37 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}+61 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} b \,x^{8}+31 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,b^{2} x^{4}+7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{3}\right )}{77 x \left (a \,x^{4}+b \right )^{3} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(177\)

[In]

int((a+b/x^4)^(5/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/77*(37*a^2*x^8+24*a*b*x^4+7*b^2)/x^9*((a*x^4+b)/x^4)^(1/2)+40/77*a^3/(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)*((a*x^4
+b)/x^4)^(1/2)*x^2

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {40 \, a^{2} \sqrt {b} x^{9} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (37 \, a^{2} x^{8} + 24 \, a b x^{4} + 7 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{77 \, x^{9}} \]

[In]

integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="fricas")

[Out]

-1/77*(40*a^2*sqrt(b)*x^9*(-a/b)^(3/4)*elliptic_f(arcsin(x*(-a/b)^(1/4)), -1) + (37*a^2*x^8 + 24*a*b*x^4 + 7*b
^2)*sqrt((a*x^4 + b)/x^4))/x^9

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=- \frac {a^{\frac {5}{2}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate((a+b/x**4)**(5/2)/x**2,x)

[Out]

-a**(5/2)*gamma(1/4)*hyper((-5/2, 1/4), (5/4,), b*exp_polar(I*pi)/(a*x**4))/(4*x*gamma(5/4))

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \]

[In]

integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="maxima")

[Out]

integrate((a + b/x^4)^(5/2)/x^2, x)

Giac [F]

\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \]

[In]

integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="giac")

[Out]

integrate((a + b/x^4)^(5/2)/x^2, x)

Mupad [B] (verification not implemented)

Time = 6.69 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {{\left (a\,x^4+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {b}{a\,x^4}\right )}{x\,{\left (\frac {b}{a}+x^4\right )}^{5/2}} \]

[In]

int((a + b/x^4)^(5/2)/x^2,x)

[Out]

-((b + a*x^4)^(5/2)*hypergeom([-5/2, 1/4], 5/4, -b/(a*x^4)))/(x*(b/a + x^4)^(5/2))