Integrand size = 15, antiderivative size = 69 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 b}{4 a^2 \sqrt {a+b x^4}}-\frac {1}{4 a x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \]
3/4*b*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(5/2)-3/4*b/a^2/(b*x^4+a)^(1/2)-1 /4/a/x^4/(b*x^4+a)^(1/2)
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-a-3 b x^4}{4 a^2 x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \]
(-a - 3*b*x^4)/(4*a^2*x^4*Sqrt[a + b*x^4]) + (3*b*ArcTanh[Sqrt[a + b*x^4]/ Sqrt[a]])/(4*a^(5/2))
Time = 0.19 (sec) , antiderivative size = 74, 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 = {798, 52, 61, 73, 221}
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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (b x^4+a\right )^{3/2}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/2}}dx^4}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {\int \frac {1}{x^4 \sqrt {b x^4+a}}dx^4}{a}+\frac {2}{a \sqrt {a+b x^4}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{a b}+\frac {2}{a \sqrt {a+b x^4}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x^4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
(-(1/(a*x^4*Sqrt[a + b*x^4])) - (3*b*(2/(a*Sqrt[a + b*x^4]) - (2*ArcTanh[S qrt[a + b*x^4]/Sqrt[a]])/a^(3/2)))/(2*a))/4
3.9.55.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.41 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{4}+a}}{\sqrt {a}}\right ) \sqrt {b \,x^{4}+a}\, b \,x^{4}-3 b \,x^{4} \sqrt {a}-a^{\frac {3}{2}}}{4 x^{4} a^{\frac {5}{2}} \sqrt {b \,x^{4}+a}}\) | \(62\) |
default | \(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
risch | \(-\frac {\sqrt {b \,x^{4}+a}}{4 a^{2} x^{4}}-\frac {b}{2 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
elliptic | \(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
1/4*(3*arctanh((b*x^4+a)^(1/2)/a^(1/2))*(b*x^4+a)^(1/2)*b*x^4-3*b*x^4*a^(1 /2)-a^(3/2))/x^4/a^(5/2)/(b*x^4+a)^(1/2)
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.51 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}, -\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{4} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}\right ] \]
[1/8*(3*(b^2*x^8 + a*b*x^4)*sqrt(a)*log((b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2*(3*a*b*x^4 + a^2)*sqrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4), -1/4*(3*(b^2*x^8 + a*b*x^4)*sqrt(-a)*arctan(sqrt(b*x^4 + a)*sqrt(-a)/a) + (3*a*b*x^4 + a^2)*sqrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4)]
Time = 1.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=- \frac {1}{4 a \sqrt {b} x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {3 \sqrt {b}}{4 a^{2} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {5}{2}}} \]
-1/(4*a*sqrt(b)*x**6*sqrt(a/(b*x**4) + 1)) - 3*sqrt(b)/(4*a**2*x**2*sqrt(a /(b*x**4) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x**2))/(4*a**(5/2))
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{4} + a} a^{3}\right )}} - \frac {3 \, b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \]
-1/4*(3*(b*x^4 + a)*b - 2*a*b)/((b*x^4 + a)^(3/2)*a^2 - sqrt(b*x^4 + a)*a^ 3) - 3/8*b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/a^ (5/2)
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x^{4} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{4} + a} a\right )} a^{2}} \]
-3/4*b*arctan(sqrt(b*x^4 + a)/sqrt(-a))/(sqrt(-a)*a^2) - 1/4*(3*(b*x^4 + a )*b - 2*a*b)/(((b*x^4 + a)^(3/2) - sqrt(b*x^4 + a)*a)*a^2)
Time = 5.85 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{4\,a^{5/2}}-\frac {1}{4\,a\,x^4\,\sqrt {b\,x^4+a}}-\frac {3\,b}{4\,a^2\,\sqrt {b\,x^4+a}} \]