\(\int (a+\frac {b}{x^4})^{5/2} x^2 \, dx\) [2078]

Optimal result

Integrand size = 15, antiderivative size = 146 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^2 \, dx=-\frac {20 a b \sqrt {a+\frac {b}{x^4}}}{21 x}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{21 x}+\frac {1}{3} \left (a+\frac {b}{x^4}\right )^{5/2} x^3-\frac {20 a^{7/4} b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 \sqrt {a+\frac {b}{x^4}}} \]

[Out]

-10/21*b*(a+b/x^4)^(3/2)/x+1/3*(a+b/x^4)^(5/2)*x^3-20/21*a*b*(a+b/x^4)^(1/2)/x-20/21*a^(7/4)*b^(3/4)*(cos(2*ar
ccot(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)/(a+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {342, 283, 201, 226} \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^2 \, dx=-\frac {20 a^{7/4} b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 \sqrt {a+\frac {b}{x^4}}}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{21 x}-\frac {20 a b \sqrt {a+\frac {b}{x^4}}}{21 x}+\frac {1}{3} x^3 \left (a+\frac {b}{x^4}\right )^{5/2} \]

[In]

Int[(a + b/x^4)^(5/2)*x^2,x]

[Out]

(-20*a*b*Sqrt[a + b/x^4])/(21*x) - (10*b*(a + b/x^4)^(3/2))/(21*x) + ((a + b/x^4)^(5/2)*x^3)/3 - (20*a^(7/4)*b
^(3/4)*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])/(21*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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 \frac {\left (a+b x^4\right )^{5/2}}{x^4} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \left (a+\frac {b}{x^4}\right )^{5/2} x^3-\frac {1}{3} (10 b) \text {Subst}\left (\int \left (a+b x^4\right )^{3/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{21 x}+\frac {1}{3} \left (a+\frac {b}{x^4}\right )^{5/2} x^3-\frac {1}{7} (20 a b) \text {Subst}\left (\int \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {20 a b \sqrt {a+\frac {b}{x^4}}}{21 x}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{21 x}+\frac {1}{3} \left (a+\frac {b}{x^4}\right )^{5/2} x^3-\frac {1}{21} \left (40 a^2 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {20 a b \sqrt {a+\frac {b}{x^4}}}{21 x}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{21 x}+\frac {1}{3} \left (a+\frac {b}{x^4}\right )^{5/2} x^3-\frac {20 a^{7/4} 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 ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{21 \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 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^2 \, dx=-\frac {b^2 \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {7}{4},-\frac {3}{4},-\frac {a x^4}{b}\right )}{7 x^5 \sqrt {1+\frac {a x^4}{b}}} \]

[In]

Integrate[(a + b/x^4)^(5/2)*x^2,x]

[Out]

-1/7*(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-5/2, -7/4, -3/4, -((a*x^4)/b)])/(x^5*Sqrt[1 + (a*x^4)/b])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\left (7 a^{2} x^{8}-16 a b \,x^{4}-3 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{21 x^{5}}+\frac {40 a^{2} b \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{21 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\)	\(131\)
default	\(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{3} \left (7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}+40 a^{2} b \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) x^{7}-9 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} b \,x^{8}-19 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,b^{2} x^{4}-3 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{3}\right )}{21 \left (a \,x^{4}+b \right )^{3} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(181\)

[In]

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

[Out]

1/21*(7*a^2*x^8-16*a*b*x^4-3*b^2)/x^5*((a*x^4+b)/x^4)^(1/2)+40/21*a^2*b/(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 [F]

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

[In]

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

[Out]

integral((a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt((a*x^4 + b)/x^4)/x^6, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.92 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.30 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x^2 \, dx=- \frac {a^{\frac {5}{2}} x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

-a**(5/2)*x**3*gamma(-3/4)*hyper((-5/2, -3/4), (1/4,), b*exp_polar(I*pi)/(a*x**4))/(4*gamma(1/4))

Maxima [F]

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

[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]

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

[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 [F(-1)]

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

[In]

int(x^2*(a + b/x^4)^(5/2),x)

[Out]

int(x^2*(a + b/x^4)^(5/2), x)