Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=-\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}} \]
-1/6*(b*x^2+a)^(3/2)/x^6+1/16*b^3*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(3/2) -1/8*b*(b*x^2+a)^(1/2)/x^4-1/16*b^2*(b*x^2+a)^(1/2)/a/x^2
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a^2-14 a b x^2-3 b^2 x^4\right )}{48 a x^6}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}} \]
(Sqrt[a + b*x^2]*(-8*a^2 - 14*a*b*x^2 - 3*b^2*x^4))/(48*a*x^6) + (b^3*ArcT anh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*a^(3/2))
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {243, 51, 51, 52, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \int \frac {\sqrt {b x^2+a}}{x^6}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )-\frac {\sqrt {a+b x^2}}{2 x^4}\right )-\frac {\left (a+b x^2\right )^{3/2}}{3 x^6}\right )\) |
(-1/3*(a + b*x^2)^(3/2)/x^6 + (b*(-1/2*Sqrt[a + b*x^2]/x^4 + (b*(-(Sqrt[a + b*x^2]/(a*x^2)) + (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)))/4))/2)/ 2
3.4.76.3.1 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 + 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] :> 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 = 1.88 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (3 b^{2} x^{4}+14 a b \,x^{2}+8 a^{2}\right )}{48 x^{6} a}+\frac {b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{16 a^{\frac {3}{2}}}\) | \(71\) |
pseudoelliptic | \(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right ) b^{3} x^{6}-3 b^{2} x^{4} \sqrt {b \,x^{2}+a}\, \sqrt {a}-14 a^{\frac {3}{2}} b \,x^{2} \sqrt {b \,x^{2}+a}-8 \sqrt {b \,x^{2}+a}\, a^{\frac {5}{2}}}{48 a^{\frac {3}{2}} x^{6}}\) | \(84\) |
default | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\) | \(125\) |
-1/48*(b*x^2+a)^(1/2)*(3*b^2*x^4+14*a*b*x^2+8*a^2)/x^6/a+1/16*b^3/a^(3/2)* ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{6} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, a b^{2} x^{4} + 14 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{96 \, a^{2} x^{6}}, -\frac {3 \, \sqrt {-a} b^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, a b^{2} x^{4} + 14 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{48 \, a^{2} x^{6}}\right ] \]
[1/96*(3*sqrt(a)*b^3*x^6*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^ 2) - 2*(3*a*b^2*x^4 + 14*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a^2*x^6), -1 /48*(3*sqrt(-a)*b^3*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*a*b^2*x^4 + 14*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a^2*x^6)]
Time = 3.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=- \frac {a^{2}}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {11 a \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {17 b^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {b^{\frac {5}{2}}}{16 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {3}{2}}} \]
-a**2/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 11*a*sqrt(b)/(24*x**5*sqrt(a /(b*x**2) + 1)) - 17*b**(3/2)/(48*x**3*sqrt(a/(b*x**2) + 1)) - b**(5/2)/(1 6*a*x*sqrt(a/(b*x**2) + 1)) + b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(3/2) )
Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=\frac {b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}}{48 \, a^{3}} - \frac {\sqrt {b x^{2} + a} b^{3}}{16 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{48 \, a^{3} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{6 \, a x^{6}} \]
1/16*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 1/48*(b*x^2 + a)^(3/2)*b^ 3/a^3 - 1/16*sqrt(b*x^2 + a)*b^3/a^2 + 1/48*(b*x^2 + a)^(5/2)*b^2/(a^3*x^2 ) + 1/24*(b*x^2 + a)^(5/2)*b/(a^2*x^4) - 1/6*(b*x^2 + a)^(5/2)/(a*x^6)
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=-\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4} + 8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x^{2} + a} a^{2} b^{4}}{a b^{3} x^{6}}}{48 \, b} \]
-1/48*(3*b^4*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) + (3*(b*x^2 + a )^(5/2)*b^4 + 8*(b*x^2 + a)^(3/2)*a*b^4 - 3*sqrt(b*x^2 + a)*a^2*b^4)/(a*b^ 3*x^6))/b
Time = 5.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^7} \, dx=\frac {a\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {{\left (b\,x^2+a\right )}^{5/2}}{16\,a\,x^6}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \]