Integrand size = 15, antiderivative size = 80 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=\frac {5}{2} a b \sqrt {a+b x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}-\frac {5}{2} a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
5/6*b*(b*x^2+a)^(3/2)-1/2*(b*x^2+a)^(5/2)/x^2-5/2*a^(3/2)*b*arctanh((b*x^2 +a)^(1/2)/a^(1/2))+5/2*a*b*(b*x^2+a)^(1/2)
Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {a+b x^2} \left (-3 a^2+14 a b x^2+2 b^2 x^4\right )}{6 x^2}-\frac {5}{2} a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
(Sqrt[a + b*x^2]*(-3*a^2 + 14*a*b*x^2 + 2*b^2*x^4))/(6*x^2) - (5*a^(3/2)*b *ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2
Time = 0.18 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {243, 51, 60, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2}}{x^4}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} b \int \frac {\left (b x^2+a\right )^{3/2}}{x^2}dx^2-\frac {\left (a+b x^2\right )^{5/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} b \left (a \int \frac {\sqrt {b x^2+a}}{x^2}dx^2+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} b \left (a \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} b \left (a \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{2} b \left (a \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x^2}\right )\) |
(-((a + b*x^2)^(5/2)/x^2) + (5*b*((2*(a + b*x^2)^(3/2))/3 + a*(2*Sqrt[a + b*x^2] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])))/2)/2
3.4.92.3.1 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/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Time = 1.89 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85
method | result | size |
pseudoelliptic | \(-\frac {5 \left (\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) a^{2} b \,x^{2}-\frac {2 \left (\sqrt {a}\, b^{2} x^{4}+7 a^{\frac {3}{2}} b \,x^{2}-\frac {3 a^{\frac {5}{2}}}{2}\right ) \sqrt {b \,x^{2}+a}}{15}\right )}{2 \sqrt {a}\, x^{2}}\) | \(68\) |
risch | \(-\frac {a^{2} \sqrt {b \,x^{2}+a}}{2 x^{2}}+\frac {b^{2} x^{2} \sqrt {b \,x^{2}+a}}{3}+\frac {7 a b \sqrt {b \,x^{2}+a}}{3}-\frac {5 b \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\) | \(78\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\) | \(91\) |
-5/2/a^(1/2)*(arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^2*b*x^2-2/15*(a^(1/2)*b^2 *x^4+7*a^(3/2)*b*x^2-3/2*a^(5/2))*(b*x^2+a)^(1/2))/x^2
Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.78 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{12 \, x^{2}}, \frac {15 \, \sqrt {-a} a b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{6 \, x^{2}}\right ] \]
[1/12*(15*a^(3/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2 ) + 2*(2*b^2*x^4 + 14*a*b*x^2 - 3*a^2)*sqrt(b*x^2 + a))/x^2, 1/6*(15*sqrt( -a)*a*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2*b^2*x^4 + 14*a*b*x^2 - 3 *a^2)*sqrt(b*x^2 + a))/x^2]
Time = 2.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=- \frac {a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{2 x^{2}} + \frac {7 a^{\frac {3}{2}} b \sqrt {1 + \frac {b x^{2}}{a}}}{3} + \frac {5 a^{\frac {3}{2}} b \log {\left (\frac {b x^{2}}{a} \right )}}{4} - \frac {5 a^{\frac {3}{2}} b \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2} + \frac {\sqrt {a} b^{2} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{3} \]
-a**(5/2)*sqrt(1 + b*x**2/a)/(2*x**2) + 7*a**(3/2)*b*sqrt(1 + b*x**2/a)/3 + 5*a**(3/2)*b*log(b*x**2/a)/4 - 5*a**(3/2)*b*log(sqrt(1 + b*x**2/a) + 1)/ 2 + sqrt(a)*b**2*x**2*sqrt(1 + b*x**2/a)/3
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=-\frac {5}{2} \, a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} a b - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{2 \, a x^{2}} \]
-5/2*a^(3/2)*b*arcsinh(a/(sqrt(a*b)*abs(x))) + 5/6*(b*x^2 + a)^(3/2)*b + 1 /2*(b*x^2 + a)^(5/2)*b/a + 5/2*sqrt(b*x^2 + a)*a*b - 1/2*(b*x^2 + a)^(7/2) /(a*x^2)
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=\frac {\frac {15 \, a^{2} b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, \sqrt {b x^{2} + a} a b^{2} - \frac {3 \, \sqrt {b x^{2} + a} a^{2} b}{x^{2}}}{6 \, b} \]
1/6*(15*a^2*b^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 2*(b*x^2 + a)^ (3/2)*b^2 + 12*sqrt(b*x^2 + a)*a*b^2 - 3*sqrt(b*x^2 + a)*a^2*b/x^2)/b
Time = 4.77 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^3} \, dx=\frac {b\,{\left (b\,x^2+a\right )}^{3/2}}{3}-\frac {a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+2\,a\,b\,\sqrt {b\,x^2+a}+\frac {a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \]