Integrand size = 15, antiderivative size = 69 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {3 b}{4 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {x^4}{4 a \sqrt {a+\frac {b}{x^4}}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4 a^{5/2}} \]
-3/4*b*arctanh((a+b/x^4)^(1/2)/a^(1/2))/a^(5/2)+3/4*b/a^2/(a+b/x^4)^(1/2)+ 1/4*x^4/a/(a+b/x^4)^(1/2)
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {\sqrt {a} x^2 \left (3 b+a x^4\right )-3 b \sqrt {b+a x^4} \log \left (\sqrt {a} x^2+\sqrt {b+a x^4}\right )}{4 a^{5/2} \sqrt {a+\frac {b}{x^4}} x^2} \]
(Sqrt[a]*x^2*(3*b + a*x^4) - 3*b*Sqrt[b + a*x^4]*Log[Sqrt[a]*x^2 + Sqrt[b + a*x^4]])/(4*a^(5/2)*Sqrt[a + b/x^4]*x^2)
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{4} \int \frac {x^8}{\left (a+\frac {b}{x^4}\right )^{3/2}}d\frac {1}{x^4}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (\frac {3 b \int \frac {x^4}{\left (a+\frac {b}{x^4}\right )^{3/2}}d\frac {1}{x^4}}{2 a}+\frac {x^4}{a \sqrt {a+\frac {b}{x^4}}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} \left (\frac {3 b \left (\frac {\int \frac {x^4}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x^4}}{a}+\frac {2}{a \sqrt {a+\frac {b}{x^4}}}\right )}{2 a}+\frac {x^4}{a \sqrt {a+\frac {b}{x^4}}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {3 b \left (\frac {2 \int \frac {1}{\frac {1}{b x^8}-\frac {a}{b}}d\sqrt {a+\frac {b}{x^4}}}{a b}+\frac {2}{a \sqrt {a+\frac {b}{x^4}}}\right )}{2 a}+\frac {x^4}{a \sqrt {a+\frac {b}{x^4}}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (\frac {3 b \left (\frac {2}{a \sqrt {a+\frac {b}{x^4}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}+\frac {x^4}{a \sqrt {a+\frac {b}{x^4}}}\right )\) |
(x^4/(a*Sqrt[a + b/x^4]) + (3*b*(2/(a*Sqrt[a + b/x^4]) - (2*ArcTanh[Sqrt[a + b/x^4]/Sqrt[a]])/a^(3/2)))/(2*a))/4
3.21.90.3.1 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 = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\left (a \,x^{4}+b \right ) \left (x^{6} a^{\frac {7}{2}}+3 a^{\frac {5}{2}} b \,x^{2}-3 b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right ) a^{2} \sqrt {a \,x^{4}+b}\right )}{4 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} a^{\frac {9}{2}}}\) | \(79\) |
risch | \(\frac {a \,x^{4}+b}{4 a^{2} \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}+\frac {\left (\frac {b \,x^{2}}{2 a^{2} \sqrt {a \,x^{4}+b}}-\frac {3 b \ln \left (x^{2} \sqrt {a}+\sqrt {a \,x^{4}+b}\right )}{4 a^{\frac {5}{2}}}\right ) \sqrt {a \,x^{4}+b}}{\sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}\) | \(96\) |
1/4*(a*x^4+b)*(x^6*a^(7/2)+3*a^(5/2)*b*x^2-3*b*ln(x^2*a^(1/2)+(a*x^4+b)^(1 /2))*a^2*(a*x^4+b)^(1/2))/((a*x^4+b)/x^4)^(3/2)/x^6/a^(9/2)
Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.78 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b x^{4} + b^{2}\right )} \sqrt {a} \log \left (-2 \, a x^{4} + 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + 2 \, {\left (a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{8 \, {\left (a^{4} x^{4} + a^{3} b\right )}}, \frac {3 \, {\left (a b x^{4} + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + {\left (a^{2} x^{8} + 3 \, a b x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{4 \, {\left (a^{4} x^{4} + a^{3} b\right )}}\right ] \]
[1/8*(3*(a*b*x^4 + b^2)*sqrt(a)*log(-2*a*x^4 + 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) + 2*(a^2*x^8 + 3*a*b*x^4)*sqrt((a*x^4 + b)/x^4))/(a^4*x^4 + a^3*b), 1/4*(3*(a*b*x^4 + b^2)*sqrt(-a)*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4 + b)) + (a^2*x^8 + 3*a*b*x^4)*sqrt((a*x^4 + b)/x^4))/(a^4*x ^4 + a^3*b)]
Time = 1.79 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {x^{6}}{4 a \sqrt {b} \sqrt {\frac {a x^{4}}{b} + 1}} + \frac {3 \sqrt {b} x^{2}}{4 a^{2} \sqrt {\frac {a x^{4}}{b} + 1}} - \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a} x^{2}}{\sqrt {b}} \right )}}{4 a^{\frac {5}{2}}} \]
x**6/(4*a*sqrt(b)*sqrt(a*x**4/b + 1)) + 3*sqrt(b)*x**2/(4*a**2*sqrt(a*x**4 /b + 1)) - 3*b*asinh(sqrt(a)*x**2/sqrt(b))/(4*a**(5/2))
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {3 \, {\left (a + \frac {b}{x^{4}}\right )} b - 2 \, a b}{4 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} - \sqrt {a + \frac {b}{x^{4}}} a^{3}\right )}} + \frac {3 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \]
1/4*(3*(a + b/x^4)*b - 2*a*b)/((a + b/x^4)^(3/2)*a^2 - sqrt(a + b/x^4)*a^3 ) + 3/8*b*log((sqrt(a + b/x^4) - sqrt(a))/(sqrt(a + b/x^4) + sqrt(a)))/a^( 5/2)
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {{\left (\frac {x^{4}}{a} + \frac {3 \, b}{a^{2}}\right )} x^{2}}{4 \, \sqrt {a x^{4} + b}} + \frac {3 \, b \log \left ({\left | -\sqrt {a} x^{2} + \sqrt {a x^{4} + b} \right |}\right )}{4 \, a^{\frac {5}{2}}} \]
1/4*(x^4/a + 3*b/a^2)*x^2/sqrt(a*x^4 + b) + 3/4*b*log(abs(-sqrt(a)*x^2 + s qrt(a*x^4 + b)))/a^(5/2)
Time = 6.44 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {3\,b}{4\,a^2\,\sqrt {a+\frac {b}{x^4}}}+\frac {x^4}{4\,a\,\sqrt {a+\frac {b}{x^4}}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4\,a^{5/2}} \]