Integrand size = 19, antiderivative size = 85 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=-\frac {(b c+2 a d) \sqrt {c+\frac {d}{x^2}}}{2 c x}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x}{c}-\frac {(b c+2 a d) \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{2 \sqrt {d}} \]
a*(c+d/x^2)^(3/2)*x/c-1/2*(2*a*d+b*c)*arctanh(d^(1/2)/x/(c+d/x^2)^(1/2))/d ^(1/2)-1/2*(2*a*d+b*c)*(c+d/x^2)^(1/2)/c/x
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (-b+2 a x^2-\frac {(b c+2 a d) x^2 \text {arctanh}\left (\frac {\sqrt {d+c x^2}}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {d+c x^2}}\right )}{2 x} \]
(Sqrt[c + d/x^2]*(-b + 2*a*x^2 - ((b*c + 2*a*d)*x^2*ArcTanh[Sqrt[d + c*x^2 ]/Sqrt[d]])/(Sqrt[d]*Sqrt[d + c*x^2])))/(2*x)
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {899, 359, 211, 224, 219}
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.
\(\displaystyle \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx\) |
\(\Big \downarrow \) 899 |
\(\displaystyle -\int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c}-\frac {(2 a d+b c) \int \sqrt {c+\frac {d}{x^2}}d\frac {1}{x}}{c}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c}-\frac {(2 a d+b c) \left (\frac {1}{2} c \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )}{c}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c}-\frac {(2 a d+b c) \left (\frac {1}{2} c \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c}-\frac {(2 a d+b c) \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 \sqrt {d}}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )}{c}\) |
(a*(c + d/x^2)^(3/2)*x)/c - ((b*c + 2*a*d)*(Sqrt[c + d/x^2]/(2*x) + (c*Arc Tanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(2*Sqrt[d])))/c
3.10.42.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {b \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{2 x}+\frac {\left (a \sqrt {c \,x^{2}+d}-\frac {\left (2 a d +b c \right ) \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right )}{2 \sqrt {d}}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}{\sqrt {c \,x^{2}+d}}\) | \(93\) |
default | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (2 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) a \,x^{2}+\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) b c \,x^{2}-2 \sqrt {c \,x^{2}+d}\, a d \,x^{2}-\sqrt {c \,x^{2}+d}\, b c \,x^{2}+\left (c \,x^{2}+d \right )^{\frac {3}{2}} b \right )}{2 x \sqrt {c \,x^{2}+d}\, d}\) | \(135\) |
-1/2*b/x*((c*x^2+d)/x^2)^(1/2)+(a*(c*x^2+d)^(1/2)-1/2*(2*a*d+b*c)/d^(1/2)* ln((2*d+2*d^(1/2)*(c*x^2+d)^(1/2))/x))*((c*x^2+d)/x^2)^(1/2)*x/(c*x^2+d)^( 1/2)
Time = 0.43 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=\left [\frac {{\left (b c + 2 \, a d\right )} \sqrt {d} x \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (2 \, a d x^{2} - b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, d x}, \frac {{\left (b c + 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a d x^{2} - b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, d x}\right ] \]
[1/4*((b*c + 2*a*d)*sqrt(d)*x*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x ^2) + 2*d)/x^2) + 2*(2*a*d*x^2 - b*d)*sqrt((c*x^2 + d)/x^2))/(d*x), 1/2*(( b*c + 2*a*d)*sqrt(-d)*x*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d )) + (2*a*d*x^2 - b*d)*sqrt((c*x^2 + d)/x^2))/(d*x)]
Time = 2.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=\frac {a \sqrt {c} x}{\sqrt {1 + \frac {d}{c x^{2}}}} - a \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {a d}{\sqrt {c} x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {b \sqrt {c} \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} - \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 \sqrt {d}} \]
a*sqrt(c)*x/sqrt(1 + d/(c*x**2)) - a*sqrt(d)*asinh(sqrt(d)/(sqrt(c)*x)) + a*d/(sqrt(c)*x*sqrt(1 + d/(c*x**2))) - b*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*x ) - b*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*sqrt(d))
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.56 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=\frac {1}{2} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )\right )} a - \frac {1}{4} \, {\left (\frac {2 \, \sqrt {c + \frac {d}{x^{2}}} c x}{{\left (c + \frac {d}{x^{2}}\right )} x^{2} - d} - \frac {c \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{\sqrt {d}}\right )} b \]
1/2*(2*sqrt(c + d/x^2)*x + sqrt(d)*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt (c + d/x^2)*x + sqrt(d))))*a - 1/4*(2*sqrt(c + d/x^2)*c*x/((c + d/x^2)*x^2 - d) - c*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d))) /sqrt(d))*b
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=\frac {2 \, \sqrt {c x^{2} + d} a c \mathrm {sgn}\left (x\right ) + \frac {{\left (b c^{2} \mathrm {sgn}\left (x\right ) + 2 \, a c d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {\sqrt {c x^{2} + d} b c \mathrm {sgn}\left (x\right )}{x^{2}}}{2 \, c} \]
1/2*(2*sqrt(c*x^2 + d)*a*c*sgn(x) + (b*c^2*sgn(x) + 2*a*c*d*sgn(x))*arctan (sqrt(c*x^2 + d)/sqrt(-d))/sqrt(-d) - sqrt(c*x^2 + d)*b*c*sgn(x)/x^2)/c
Time = 9.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx=a\,x\,\sqrt {c+\frac {d}{x^2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{2\,x}-\frac {b\,c\,\ln \left (\sqrt {c+\frac {d}{x^2}}+\frac {\sqrt {d}}{x}\right )}{2\,\sqrt {d}}+\frac {a\,\sqrt {d}\,\mathrm {asin}\left (\frac {\sqrt {d}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}\,\sqrt {\frac {d}{c\,x^2}+1}} \]