Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=-\frac {5 a^2 \sqrt {a+\frac {b}{x^4}}}{32 x^2}-\frac {5 a \left (a+\frac {b}{x^4}\right )^{3/2}}{48 x^2}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{12 x^2}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^4}} x^2}\right )}{32 \sqrt {b}} \]
-5/48*a*(a+b/x^4)^(3/2)/x^2-1/12*(a+b/x^4)^(5/2)/x^2-5/32*a^3*arctanh(b^(1 /2)/x^2/(a+b/x^4)^(1/2))/b^(1/2)-5/32*a^2*(a+b/x^4)^(1/2)/x^2
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {a+\frac {b}{x^4}} \left (-8 b^2-26 a b x^4-33 a^2 x^8-\frac {15 a^3 x^{12} \text {arctanh}\left (\frac {\sqrt {b+a x^4}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {b+a x^4}}\right )}{96 x^{10}} \]
(Sqrt[a + b/x^4]*(-8*b^2 - 26*a*b*x^4 - 33*a^2*x^8 - (15*a^3*x^12*ArcTanh[ Sqrt[b + a*x^4]/Sqrt[b]])/(Sqrt[b]*Sqrt[b + a*x^4])))/(96*x^10)
Time = 0.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {858, 807, 211, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x}d\frac {1}{x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right )^{5/2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (-\frac {5}{6} a \int \left (a+\frac {b}{x^2}\right )^{3/2}d\frac {1}{x^2}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x^2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (-\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {a+\frac {b}{x^2}}d\frac {1}{x^2}+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x^2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {1}{2} \left (-\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x^2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} \left (-\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{\sqrt {a+\frac {b}{x^2}} x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^2}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x^2}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x^2}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x^2}\right )\) |
(-1/6*(a + b/x^2)^(5/2)/x^2 - (5*a*((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))/6)/2
3.21.77.3.1 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[(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 = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05
method | result | size |
risch | \(-\frac {\left (33 a^{2} x^{8}+26 a b \,x^{4}+8 b^{2}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{96 x^{10}}-\frac {5 a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{32 \sqrt {b}\, \sqrt {a \,x^{4}+b}}\) | \(97\) |
default | \(-\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (15 a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{4}+b}}{x^{2}}\right ) x^{12}+33 a^{2} \sqrt {a \,x^{4}+b}\, x^{8} \sqrt {b}+26 b^{\frac {3}{2}} a \sqrt {a \,x^{4}+b}\, x^{4}+8 b^{\frac {5}{2}} \sqrt {a \,x^{4}+b}\right )}{96 x^{2} \left (a \,x^{4}+b \right )^{\frac {5}{2}} \sqrt {b}}\) | \(113\) |
-1/96*(33*a^2*x^8+26*a*b*x^4+8*b^2)/x^10*((a*x^4+b)/x^4)^(1/2)-5/32*a^3/b^ (1/2)*ln((2*b+2*b^(1/2)*(a*x^4+b)^(1/2))/x^2)*((a*x^4+b)/x^4)^(1/2)*x^2/(a *x^4+b)^(1/2)
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x^{10} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) - 2 \, {\left (33 \, a^{2} b x^{8} + 26 \, a b^{2} x^{4} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{192 \, b x^{10}}, \frac {15 \, a^{3} \sqrt {-b} x^{10} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) - {\left (33 \, a^{2} b x^{8} + 26 \, a b^{2} x^{4} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{96 \, b x^{10}}\right ] \]
[1/192*(15*a^3*sqrt(b)*x^10*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^ 4) + 2*b)/x^4) - 2*(33*a^2*b*x^8 + 26*a*b^2*x^4 + 8*b^3)*sqrt((a*x^4 + b)/ x^4))/(b*x^10), 1/96*(15*a^3*sqrt(-b)*x^10*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b) - (33*a^2*b*x^8 + 26*a*b^2*x^4 + 8*b^3)*sqrt((a*x^4 + b)/x^4 ))/(b*x^10)]
Time = 2.76 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=- \frac {11 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{4}}}}{32 x^{2}} - \frac {13 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{4}}}}{48 x^{6}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{4}}}}{12 x^{10}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{32 \sqrt {b}} \]
-11*a**(5/2)*sqrt(1 + b/(a*x**4))/(32*x**2) - 13*a**(3/2)*b*sqrt(1 + b/(a* x**4))/(48*x**6) - sqrt(a)*b**2*sqrt(1 + b/(a*x**4))/(12*x**10) - 5*a**3*a sinh(sqrt(b)/(sqrt(a)*x**2))/(32*sqrt(b))
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (72) = 144\).
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {5 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right )}{64 \, \sqrt {b}} - \frac {33 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} a^{3} x^{10} - 40 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{3} b x^{6} + 15 \, \sqrt {a + \frac {b}{x^{4}}} a^{3} b^{2} x^{2}}{96 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{3} x^{12} - 3 \, {\left (a + \frac {b}{x^{4}}\right )}^{2} b x^{8} + 3 \, {\left (a + \frac {b}{x^{4}}\right )} b^{2} x^{4} - b^{3}\right )}} \]
5/64*a^3*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x^2 + sqrt(b )))/sqrt(b) - 1/96*(33*(a + b/x^4)^(5/2)*a^3*x^10 - 40*(a + b/x^4)^(3/2)*a ^3*b*x^6 + 15*sqrt(a + b/x^4)*a^3*b^2*x^2)/((a + b/x^4)^3*x^12 - 3*(a + b/ x^4)^2*b*x^8 + 3*(a + b/x^4)*b^2*x^4 - b^3)
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\frac {\frac {15 \, a^{4} \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {33 \, {\left (a x^{4} + b\right )}^{\frac {5}{2}} a^{4} - 40 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{4} b + 15 \, \sqrt {a x^{4} + b} a^{4} b^{2}}{a^{3} x^{12}}}{96 \, a} \]
1/96*(15*a^4*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) - (33*(a*x^4 + b)^( 5/2)*a^4 - 40*(a*x^4 + b)^(3/2)*a^4*b + 15*sqrt(a*x^4 + b)*a^4*b^2)/(a^3*x ^12))/a
Timed out. \[ \int \frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (a+\frac {b}{x^4}\right )}^{5/2}}{x^3} \,d x \]