Integrand size = 15, antiderivative size = 69 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=-\frac {a \sqrt {a+b x^2}}{4 x^4}-\frac {5 b \sqrt {a+b x^2}}{8 x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \] Output:
-1/4*a*(b*x^2+a)^(1/2)/x^4-5/8*b*(b*x^2+a)^(1/2)/x^2-3/8*b^2*arctanh((b*x^ 2+a)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=\frac {\left (-2 a-5 b x^2\right ) \sqrt {a+b x^2}}{8 x^4}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \] Input:
Integrate[(a + b*x^2)^(3/2)/x^5,x]
Output:
((-2*a - 5*b*x^2)*Sqrt[a + b*x^2])/(8*x^4) - (3*b^2*ArcTanh[Sqrt[a + b*x^2 ]/Sqrt[a]])/(8*Sqrt[a])
Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {243, 51, 51, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2}}{x^6}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \int \frac {\sqrt {b x^2+a}}{x^4}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )\) |
Input:
Int[(a + b*x^2)^(3/2)/x^5,x]
Output:
(-1/2*(a + b*x^2)^(3/2)/x^4 + (3*b*(-(Sqrt[a + b*x^2]/x^2) - (b*ArcTanh[Sq rt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/4)/2
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] :> 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 = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (5 b \,x^{2}+2 a \right )}{8 x^{4}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 \sqrt {a}}\) | \(57\) |
pseudoelliptic | \(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) b^{2} x^{4}-5 b \,x^{2} \sqrt {a}\, \sqrt {b \,x^{2}+a}-2 a^{\frac {3}{2}} \sqrt {b \,x^{2}+a}}{8 x^{4} \sqrt {a}}\) | \(64\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\) | \(101\) |
Input:
int((b*x^2+a)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/8*(b*x^2+a)^(1/2)*(5*b*x^2+2*a)/x^4-3/8*b^2/a^(1/2)*ln((2*a+2*a^(1/2)*( b*x^2+a)^(1/2))/x)
Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a x^{4}}, \frac {3 \, \sqrt {-a} b^{2} x^{4} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a b x^{2} + 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a x^{4}}\right ] \] Input:
integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="fricas")
Output:
[1/16*(3*sqrt(a)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^ 2) - 2*(5*a*b*x^2 + 2*a^2)*sqrt(b*x^2 + a))/(a*x^4), 1/8*(3*sqrt(-a)*b^2*x ^4*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - (5*a*b*x^2 + 2*a^2)*sqrt(b*x^2 + a ))/(a*x^4)]
Time = 1.53 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=- \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{4 x^{3}} - \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{8 x} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 \sqrt {a}} \] Input:
integrate((b*x**2+a)**(3/2)/x**5,x)
Output:
-a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(4*x**3) - 5*b**(3/2)*sqrt(a/(b*x**2) + 1) /(8*x) - 3*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*sqrt(a))
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=-\frac {3 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {b x^{2} + a} b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{4 \, a x^{4}} \] Input:
integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="maxima")
Output:
-3/8*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/8*(b*x^2 + a)^(3/2)*b^2 /a^2 + 3/8*sqrt(b*x^2 + a)*b^2/a - 1/8*(b*x^2 + a)^(5/2)*b/(a^2*x^2) - 1/4 *(b*x^2 + a)^(5/2)/(a*x^4)
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x^{2} + a} a b^{3}}{b^{2} x^{4}}}{8 \, b} \] Input:
integrate((b*x^2+a)^(3/2)/x^5,x, algorithm="giac")
Output:
1/8*(3*b^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (5*(b*x^2 + a)^(3/2 )*b^3 - 3*sqrt(b*x^2 + a)*a*b^3)/(b^2*x^4))/b
Time = 0.60 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=\frac {3\,a\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {5\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4} \] Input:
int((a + b*x^2)^(3/2)/x^5,x)
Output:
(3*a*(a + b*x^2)^(1/2))/(8*x^4) - (3*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2))) /(8*a^(1/2)) - (5*(a + b*x^2)^(3/2))/(8*x^4)
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{2}-5 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}}{8 a \,x^{4}} \] Input:
int((b*x^2+a)^(3/2)/x^5,x)
Output:
( - 2*sqrt(a + b*x**2)*a**2 - 5*sqrt(a + b*x**2)*a*b*x**2 + 3*sqrt(a)*log( (sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4 - 3*sqrt(a)*lo g((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4)/(8*a*x**4)