Integrand size = 22, antiderivative size = 76 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=-a c \sqrt {c+\frac {d}{x^2}}-\frac {1}{3} a \left (c+\frac {d}{x^2}\right )^{3/2}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}+a c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right ) \]
-1/3*a*(c+d/x^2)^(3/2)-1/5*b*(c+d/x^2)^(5/2)/d+a*c^(3/2)*arctanh((c+d/x^2) ^(1/2)/c^(1/2))-a*c*(c+d/x^2)^(1/2)
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (-\frac {3 b \left (d+c x^2\right )^2}{d}-5 a x^2 \left (d+4 c x^2\right )-\frac {15 a c^{3/2} x^5 \log \left (-\sqrt {c} x+\sqrt {d+c x^2}\right )}{\sqrt {d+c x^2}}\right )}{15 x^4} \]
(Sqrt[c + d/x^2]*((-3*b*(d + c*x^2)^2)/d - 5*a*x^2*(d + 4*c*x^2) - (15*a*c ^(3/2)*x^5*Log[-(Sqrt[c]*x) + Sqrt[d + c*x^2]])/Sqrt[d + c*x^2]))/(15*x^4)
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 90, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^2d\frac {1}{x^2}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (-a \int \left (c+\frac {d}{x^2}\right )^{3/2} x^2d\frac {1}{x^2}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (-a \left (c \int \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x^2}+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (-a \left (c \left (c \int \frac {x^2}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}+2 \sqrt {c+\frac {d}{x^2}}\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-a \left (c \left (\frac {2 c \int \frac {1}{\frac {1}{d x^4}-\frac {c}{d}}d\sqrt {c+\frac {d}{x^2}}}{d}+2 \sqrt {c+\frac {d}{x^2}}\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-a \left (c \left (2 \sqrt {c+\frac {d}{x^2}}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}\right )\) |
((-2*b*(c + d/x^2)^(5/2))/(5*d) - a*((2*(c + d/x^2)^(3/2))/3 + c*(2*Sqrt[c + d/x^2] - 2*Sqrt[c]*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])))/2
3.10.48.3.1 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 + 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.10 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.42
method | result | size |
risch | \(-\frac {\left (20 a c d \,x^{4}+3 b \,c^{2} x^{4}+5 a \,d^{2} x^{2}+6 b c d \,x^{2}+3 b \,d^{2}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{15 x^{4} d}+\frac {c^{\frac {3}{2}} a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}{\sqrt {c \,x^{2}+d}}\) | \(108\) |
default | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-10 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {5}{2}} a \,x^{6}+10 \left (c \,x^{2}+d \right )^{\frac {5}{2}} c^{\frac {3}{2}} a \,x^{4}-15 \sqrt {c \,x^{2}+d}\, c^{\frac {5}{2}} a d \,x^{6}-15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) a \,c^{2} d^{2} x^{5}+5 \left (c \,x^{2}+d \right )^{\frac {5}{2}} \sqrt {c}\, a d \,x^{2}+3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} \sqrt {c}\, b d \right )}{15 x^{2} \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{2} \sqrt {c}}\) | \(153\) |
-1/15*(20*a*c*d*x^4+3*b*c^2*x^4+5*a*d^2*x^2+6*b*c*d*x^2+3*b*d^2)/x^4/d*((c *x^2+d)/x^2)^(1/2)+c^(3/2)*a*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*((c*x^2+d)/x^2) ^(1/2)*x/(c*x^2+d)^(1/2)
Time = 0.31 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.80 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=\left [\frac {15 \, a c^{\frac {3}{2}} d x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} + {\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{30 \, d x^{4}}, -\frac {15 \, a \sqrt {-c} c d x^{4} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} + {\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, d x^{4}}\right ] \]
[1/30*(15*a*c^(3/2)*d*x^4*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^ 2) - d) - 2*((3*b*c^2 + 20*a*c*d)*x^4 + 3*b*d^2 + (6*b*c*d + 5*a*d^2)*x^2) *sqrt((c*x^2 + d)/x^2))/(d*x^4), -1/15*(15*a*sqrt(-c)*c*d*x^4*arctan(sqrt( -c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + ((3*b*c^2 + 20*a*c*d)*x^4 + 3 *b*d^2 + (6*b*c*d + 5*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d*x^4)]
Time = 9.84 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=\frac {\begin {cases} - \frac {2 a c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x^{2}}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - 2 a c \sqrt {c + \frac {d}{x^{2}}} - \frac {2 a \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {2 b \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5 d} & \text {for}\: d \neq 0 \\- a c^{\frac {3}{2}} \log {\left (- \frac {b c^{\frac {3}{2}}}{x^{2}} \right )} - \frac {b c^{\frac {3}{2}}}{x^{2}} & \text {otherwise} \end {cases}}{2} \]
Piecewise((-2*a*c**2*atan(sqrt(c + d/x**2)/sqrt(-c))/sqrt(-c) - 2*a*c*sqrt (c + d/x**2) - 2*a*(c + d/x**2)**(3/2)/3 - 2*b*(c + d/x**2)**(5/2)/(5*d), Ne(d, 0)), (-a*c**(3/2)*log(-b*c**(3/2)/x**2) - b*c**(3/2)/x**2, True))/2
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=-\frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) + 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} + 6 \, \sqrt {c + \frac {d}{x^{2}}} c\right )} a \]
-1/5*b*(c + d/x^2)^(5/2)/d - 1/6*(3*c^(3/2)*log((sqrt(c + d/x^2) - sqrt(c) )/(sqrt(c + d/x^2) + sqrt(c))) + 2*(c + d/x^2)^(3/2) + 6*sqrt(c + d/x^2)*c )*a
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (60) = 120\).
Time = 0.84 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.34 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=-\frac {1}{2} \, a c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {3}{2}} d \mathrm {sgn}\left (x\right ) - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {3}{2}} d^{3} \mathrm {sgn}\left (x\right ) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {3}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 3 \, b c^{\frac {5}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 20 \, a c^{\frac {3}{2}} d^{5} \mathrm {sgn}\left (x\right )\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{5}} \]
-1/2*a*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + d))^2)*sgn(x) + 2/15*(15*(sqr t(c)*x - sqrt(c*x^2 + d))^8*b*c^(5/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(3/2)*d*sgn(x) - 90*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(3/2)* d^2*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(5/2)*d^2*sgn(x) + 110 *(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(3/2)*d^3*sgn(x) - 70*(sqrt(c)*x - sq rt(c*x^2 + d))^2*a*c^(3/2)*d^4*sgn(x) + 3*b*c^(5/2)*d^4*sgn(x) + 20*a*c^(3 /2)*d^5*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5
Time = 10.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x} \, dx=a\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {a\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{3}-a\,c\,\sqrt {c+\frac {d}{x^2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}\,{\left (c\,x^2+d\right )}^2}{5\,d\,x^4} \]