Integrand size = 11, antiderivative size = 250 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {12 a \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{5 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\left (a+\frac {b}{x^4}\right )^{3/2} x+\frac {12 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}}-\frac {6 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}} \]
(a+b/x^4)^(3/2)*x-6/5*b*(a+b/x^4)^(1/2)/x^3-12/5*a*b^(1/2)*(a+b/x^4)^(1/2) /x/(a^(1/2)+b^(1/2)/x^2)+12/5*a^(5/4)*b^(1/4)*(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)/(a+b/x^4)^(1/2)-6/5*a^(5/4)*b^(1/4)*(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)/(a+b/x^4)^(1/2)
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 = 0.21 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {a x^4}{b}\right )}{5 x^3 \sqrt {1+\frac {a x^4}{b}}} \]
-1/5*(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-3/2, -5/4, -1/4, -((a*x^4)/b)]) /(x^3*Sqrt[1 + (a*x^4)/b])
Time = 0.35 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {773, 809, 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.
\(\displaystyle \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 773 |
\(\displaystyle -\int \left (a+\frac {b}{x^4}\right )^{3/2} x^2d\frac {1}{x}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \left (\frac {2}{5} a \int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )\) |
\(\Big \downarrow \) 834 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle x \left (a+\frac {b}{x^4}\right )^{3/2}-6 b \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}\right )+\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\right )\) |
(a + b/x^4)^(3/2)*x - 6*b*(Sqrt[a + b/x^4]/(5*x^3) + (2*a*(-((-(Sqrt[a + b /x^4]/((Sqrt[a] + Sqrt[b]/x^2)*x)) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[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)
3.21.71.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && !IntegerQ[p]
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] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(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] && IntB inomialQ[a, b, c, n, m, p, x]
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[((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 = 0.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.56
method | result | size |
risch | \(-\frac {\left (7 a \,x^{4}+b \right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{5 x^{3}}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\) | \(140\) |
default | \(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x \left (12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}-8 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a b \,x^{4}-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2}\right )}{5 \left (a \,x^{4}+b \right )^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(229\) |
-1/5*(7*a*x^4+b)/x^3*((a*x^4+b)/x^4)^(1/2)+12/5*I*a^(3/2)*b^(1/2)/(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
\[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=\int { {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.17 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=- \frac {a^{\frac {3}{2}} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]
-a**(3/2)*x*gamma(-1/4)*hyper((-3/2, -1/4), (3/4,), b*exp_polar(I*pi)/(a*x **4))/(4*gamma(3/4))
\[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=\int { {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} \,d x } \]
\[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=\int { {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} \,d x } \]
Time = 6.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.15 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {x\,{\left (a+\frac {b}{x^4}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ -\frac {a\,x^4}{b}\right )}{5\,{\left (\frac {a\,x^4}{b}+1\right )}^{3/2}} \]