Integrand size = 15, antiderivative size = 147 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {20 a^2 \sqrt {a+\frac {b}{x^4}}}{77 x}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{77 x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}-\frac {20 a^{11/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}} \] Output:
-20/77*a^2*(a+b/x^4)^(1/2)/x-10/77*a*(a+b/x^4)^(3/2)/x-1/11*(a+b/x^4)^(5/2 )/x-20/77*a^(11/4)*((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^(1/4)/ (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 = 54, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {b^2 \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {5}{2},-\frac {7}{4},-\frac {a x^4}{b}\right )}{11 x^9 \sqrt {1+\frac {a x^4}{b}}} \] Input:
Integrate[(a + b/x^4)^(5/2)/x^2,x]
Output:
-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])
Time = 0.39 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {858, 748, 748, 748, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \left (a+\frac {b}{x^4}\right )^{5/2}d\frac {1}{x}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle -\frac {10}{11} a \int \left (a+\frac {b}{x^4}\right )^{3/2}d\frac {1}{x}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle -\frac {10}{11} a \left (\frac {6}{7} a \int \sqrt {a+\frac {b}{x^4}}d\frac {1}{x}+\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle -\frac {10}{11} a \left (\frac {6}{7} a \left (\frac {2}{3} a \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^4}}}{3 x}\right )+\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {10}{11} a \left (\frac {6}{7} a \left (\frac {a^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {a+\frac {b}{x^4}}}{3 x}\right )+\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}\right )-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{11 x}\) |
Input:
Int[(a + b/x^4)^(5/2)/x^2,x]
Output:
-1/11*(a + b/x^4)^(5/2)/x - (10*a*((a + b/x^4)^(3/2)/(7*x) + (6*a*(Sqrt[a + b/x^4]/(3*x) + (a^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqr t[a] + Sqrt[b]/x^2)*EllipticF[2*ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(3*b^(1 /4)*Sqrt[a + b/x^4])))/7))/11
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[p]])
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[(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]
Result contains complex when optimal does not.
Time = 0.89 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {\left (37 a^{2} x^{8}+24 a b \,x^{4}+7 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{77 x^{9}}+\frac {40 a^{3} \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{77 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\) | \(130\) |
default | \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-40 a^{3} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) x^{11}+37 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}+61 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} b \,x^{8}+31 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,b^{2} x^{4}+7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{3}\right )}{77 x \left (a \,x^{4}+b \right )^{3} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(177\) |
Input:
int((a+b/x^4)^(5/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-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
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {40 \, a^{2} \sqrt {b} x^{9} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (37 \, a^{2} x^{8} + 24 \, a b x^{4} + 7 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{77 \, x^{9}} \] Input:
integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="fricas")
Output:
-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
Result contains complex when optimal does not.
Time = 1.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=- \frac {a^{\frac {5}{2}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((a+b/x**4)**(5/2)/x**2,x)
Output:
-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))
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \] Input:
integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="maxima")
Output:
integrate((a + b/x^4)^(5/2)/x^2, x)
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=\int { \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \] Input:
integrate((a+b/x^4)^(5/2)/x^2,x, algorithm="giac")
Output:
integrate((a + b/x^4)^(5/2)/x^2, x)
Time = 1.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=-\frac {{\left (a\,x^4+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {b}{a\,x^4}\right )}{x\,{\left (\frac {b}{a}+x^4\right )}^{5/2}} \] Input:
int((a + b/x^4)^(5/2)/x^2,x)
Output:
-((b + a*x^4)^(5/2)*hypergeom([-5/2, 1/4], 5/4, -b/(a*x^4)))/(x*(b/a + x^4 )^(5/2))
\[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^2} \, dx=\frac {-3 \sqrt {a \,x^{4}+b}\, a^{2} x^{8}-\sqrt {a \,x^{4}+b}\, b^{2}-8 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a \,x^{16}+b \,x^{12}}d x \right ) b^{3} x^{11}}{3 x^{11}} \] Input:
int((a+b/x^4)^(5/2)/x^2,x)
Output:
( - 3*sqrt(a*x**4 + b)*a**2*x**8 - sqrt(a*x**4 + b)*b**2 - 8*int(sqrt(a*x* *4 + b)/(a*x**16 + b*x**12),x)*b**3*x**11)/(3*x**11)