Integrand size = 13, antiderivative size = 91 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=-\frac {15 a b \sqrt {a+\frac {b}{x^4}}}{16 x^2}-\frac {5 b \left (a+\frac {b}{x^4}\right )^{3/2}}{8 x^2}+\frac {1}{2} \left (a+\frac {b}{x^4}\right )^{5/2} x^2-\frac {15}{16} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^4}} x^2}\right ) \]
-5/8*b*(a+b/x^4)^(3/2)/x^2+1/2*(a+b/x^4)^(5/2)*x^2-15/16*a^2*arctanh(b^(1/ 2)/x^2/(a+b/x^4)^(1/2))*b^(1/2)-15/16*a*b*(a+b/x^4)^(1/2)/x^2
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=-\frac {\sqrt {a+\frac {b}{x^4}} \left (\sqrt {b+a x^4} \left (2 b^2+9 a b x^4-8 a^2 x^8\right )+15 a^2 \sqrt {b} x^8 \text {arctanh}\left (\frac {\sqrt {b+a x^4}}{\sqrt {b}}\right )\right )}{16 x^6 \sqrt {b+a x^4}} \]
-1/16*(Sqrt[a + b/x^4]*(Sqrt[b + a*x^4]*(2*b^2 + 9*a*b*x^4 - 8*a^2*x^8) + 15*a^2*Sqrt[b]*x^8*ArcTanh[Sqrt[b + a*x^4]/Sqrt[b]]))/(x^6*Sqrt[b + a*x^4] )
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {858, 807, 247, 211, 211, 224, 219}
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.
\(\displaystyle \int x \left (a+\frac {b}{x^4}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \left (a+\frac {b}{x^4}\right )^{5/2} x^3d\frac {1}{x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right )^{5/2} x^2d\frac {1}{x^2}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \int \left (a+\frac {b}{x^2}\right )^{3/2}d\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \int \sqrt {a+\frac {b}{x^2}}d\frac {1}{x^2}+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{\sqrt {a+\frac {b}{x^2}} x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^2}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )\right )\) |
((a + b/x^2)^(5/2)*x - 5*b*((a + b/x^2)^(3/2)/(4*x^2) + (3*a*(Sqrt[a + b/x ^2]/(2*x^2) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x^2)])/(2*Sqrt[b])))/4)) /2
3.21.75.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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]
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13
method | result | size |
risch | \(-\frac {b \left (9 a \,x^{4}+2 b \right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{16 x^{6}}+\frac {\left (\frac {\sqrt {a \,x^{4}+b}\, a^{2}}{2}-\frac {15 \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) a^{2}}{16}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{\sqrt {a \,x^{4}+b}}\) | \(103\) |
default | \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{2} \left (15 \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) a^{2} x^{8}-8 a^{2} x^{8} \sqrt {a \,x^{4}+b}+9 a b \sqrt {a \,x^{4}+b}\, x^{4}+2 b^{2} \sqrt {a \,x^{4}+b}\right )}{16 \left (a \,x^{4}+b \right )^{\frac {5}{2}}}\) | \(108\) |
-1/16*b*(9*a*x^4+2*b)/x^6*((a*x^4+b)/x^4)^(1/2)+(1/2*(a*x^4+b)^(1/2)*a^2-1 5/16*b^(1/2)*ln((2*b+2*b^(1/2)*(a*x^4+b)^(1/2))/x^2)*a^2)*((a*x^4+b)/x^4)^ (1/2)*x^2/(a*x^4+b)^(1/2)
Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.86 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x^{6} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) + 2 \, {\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{32 \, x^{6}}, \frac {15 \, a^{2} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) + {\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{16 \, x^{6}}\right ] \]
[1/32*(15*a^2*sqrt(b)*x^6*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4) + 2*(8*a^2*x^8 - 9*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^ 6, 1/16*(15*a^2*sqrt(-b)*x^6*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b) + (8*a^2*x^8 - 9*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^6]
Time = 2.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.36 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=\frac {a^{\frac {5}{2}} x^{2}}{2 \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {a^{\frac {3}{2}} b}{16 x^{2} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {11 \sqrt {a} b^{2}}{16 x^{6} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{16} - \frac {b^{3}}{8 \sqrt {a} x^{10} \sqrt {1 + \frac {b}{a x^{4}}}} \]
a**(5/2)*x**2/(2*sqrt(1 + b/(a*x**4))) - a**(3/2)*b/(16*x**2*sqrt(1 + b/(a *x**4))) - 11*sqrt(a)*b**2/(16*x**6*sqrt(1 + b/(a*x**4))) - 15*a**2*sqrt(b )*asinh(sqrt(b)/(sqrt(a)*x**2))/16 - b**3/(8*sqrt(a)*x**10*sqrt(1 + b/(a*x **4)))
Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.53 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=\frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} a^{2} x^{2} + \frac {15}{32} \, a^{2} \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right ) - \frac {9 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} b x^{6} - 7 \, \sqrt {a + \frac {b}{x^{4}}} a^{2} b^{2} x^{2}}{16 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{2} x^{8} - 2 \, {\left (a + \frac {b}{x^{4}}\right )} b x^{4} + b^{2}\right )}} \]
1/2*sqrt(a + b/x^4)*a^2*x^2 + 15/32*a^2*sqrt(b)*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x^2 + sqrt(b))) - 1/16*(9*(a + b/x^4)^(3/2)*a^2 *b*x^6 - 7*sqrt(a + b/x^4)*a^2*b^2*x^2)/((a + b/x^4)^2*x^8 - 2*(a + b/x^4) *b*x^4 + b^2)
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=\frac {\frac {15 \, a^{3} b \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 8 \, \sqrt {a x^{4} + b} a^{3} - \frac {9 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{3} b - 7 \, \sqrt {a x^{4} + b} a^{3} b^{2}}{a^{2} x^{8}}}{16 \, a} \]
1/16*(15*a^3*b*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) + 8*sqrt(a*x^4 + b)*a^3 - (9*(a*x^4 + b)^(3/2)*a^3*b - 7*sqrt(a*x^4 + b)*a^3*b^2)/(a^2*x^8) )/a
Timed out. \[ \int \left (a+\frac {b}{x^4}\right )^{5/2} x \, dx=\int x\,{\left (a+\frac {b}{x^4}\right )}^{5/2} \,d x \]