Integrand size = 15, antiderivative size = 72 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=a^2 \sqrt {a+b x^2}+\frac {1}{3} a \left (a+b x^2\right )^{3/2}+\frac {1}{5} \left (a+b x^2\right )^{5/2}-a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:
a^2*(b*x^2+a)^(1/2)+1/3*a*(b*x^2+a)^(3/2)+1/5*(b*x^2+a)^(5/2)-a^(5/2)*arct anh((b*x^2+a)^(1/2)/a^(1/2))
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {1}{15} \sqrt {a+b x^2} \left (23 a^2+11 a b x^2+3 b^2 x^4\right )-a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Input:
Integrate[(a + b*x^2)^(5/2)/x,x]
Output:
(Sqrt[a + b*x^2]*(23*a^2 + 11*a*b*x^2 + 3*b^2*x^4))/15 - a^(5/2)*ArcTanh[S qrt[a + b*x^2]/Sqrt[a]]
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {243, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2}}{x^2}dx^2\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (a \int \frac {\left (b x^2+a\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (a \left (a \int \frac {\sqrt {b x^2+a}}{x^2}dx^2+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (a \left (a \left (a \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+2 \sqrt {a+b x^2}\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (a \left (a \left (2 \sqrt {a+b x^2}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^2\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^2\right )^{5/2}\right )\) |
Input:
Int[(a + b*x^2)^(5/2)/x,x]
Output:
((2*(a + b*x^2)^(5/2))/5 + a*((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
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 = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.74
method | result | size |
pseudoelliptic | \(-a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\frac {\sqrt {b \,x^{2}+a}\, \left (3 b^{2} x^{4}+11 a b \,x^{2}+23 a^{2}\right )}{15}\) | \(53\) |
default | \(\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\) | \(67\) |
Input:
int((b*x^2+a)^(5/2)/x,x,method=_RETURNVERBOSE)
Output:
-a^(5/2)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+1/15*(b*x^2+a)^(1/2)*(3*b^2*x^4+ 11*a*b*x^2+23*a^2)
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\left [\frac {1}{2} \, a^{\frac {5}{2}} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + \frac {1}{15} \, {\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\right )} \sqrt {b x^{2} + a}, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + \frac {1}{15} \, {\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\right )} \sqrt {b x^{2} + a}\right ] \] Input:
integrate((b*x^2+a)^(5/2)/x,x, algorithm="fricas")
Output:
[1/2*a^(5/2)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 1/15*(3 *b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt(b*x^2 + a), sqrt(-a)*a^2*arctan(sqrt( b*x^2 + a)*sqrt(-a)/a) + 1/15*(3*b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt(b*x^2 + a)]
Time = 2.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {23 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{15} + \frac {a^{\frac {5}{2}} \log {\left (\frac {b x^{2}}{a} \right )}}{2} - a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )} + \frac {11 a^{\frac {3}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{15} + \frac {\sqrt {a} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{5} \] Input:
integrate((b*x**2+a)**(5/2)/x,x)
Output:
23*a**(5/2)*sqrt(1 + b*x**2/a)/15 + a**(5/2)*log(b*x**2/a)/2 - a**(5/2)*lo g(sqrt(1 + b*x**2/a) + 1) + 11*a**(3/2)*b*x**2*sqrt(1 + b*x**2/a)/15 + sqr t(a)*b**2*x**4*sqrt(1 + b*x**2/a)/5
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=-a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a + \sqrt {b x^{2} + a} a^{2} \] Input:
integrate((b*x^2+a)^(5/2)/x,x, algorithm="maxima")
Output:
-a^(5/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/5*(b*x^2 + a)^(5/2) + 1/3*(b*x^ 2 + a)^(3/2)*a + sqrt(b*x^2 + a)*a^2
Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {a^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a + \sqrt {b x^{2} + a} a^{2} \] Input:
integrate((b*x^2+a)^(5/2)/x,x, algorithm="giac")
Output:
a^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/5*(b*x^2 + a)^(5/2) + 1/ 3*(b*x^2 + a)^(3/2)*a + sqrt(b*x^2 + a)*a^2
Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {{\left (b\,x^2+a\right )}^{5/2}}{5}+a^2\,\sqrt {b\,x^2+a}+a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \] Input:
int((a + b*x^2)^(5/2)/x,x)
Output:
a^(5/2)*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i + (a*(a + b*x^2)^(3/2))/3 + (a + b*x^2)^(5/2)/5 + a^2*(a + b*x^2)^(1/2)
Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {23 \sqrt {b \,x^{2}+a}\, a^{2}}{15}+\frac {11 \sqrt {b \,x^{2}+a}\, a b \,x^{2}}{15}+\frac {\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}{5}+\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}-\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} \] Input:
int((b*x^2+a)^(5/2)/x,x)
Output:
(23*sqrt(a + b*x**2)*a**2 + 11*sqrt(a + b*x**2)*a*b*x**2 + 3*sqrt(a + b*x* *2)*b**2*x**4 + 15*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sq rt(a))*a**2 - 15*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt (a))*a**2)/15