Integrand size = 19, antiderivative size = 71 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\frac {c \sqrt {\frac {c}{a+b x^2}}}{a}-\frac {c \sqrt {\frac {c}{a+b x^2}} \sqrt {1+\frac {b x^2}{a}} \text {arctanh}\left (\sqrt {1+\frac {b x^2}{a}}\right )}{a} \] Output:
c*(c/(b*x^2+a))^(1/2)/a-c*(c/(b*x^2+a))^(1/2)*(1+b*x^2/a)^(1/2)*arctanh((1 +b*x^2/a)^(1/2))/a
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\frac {c \sqrt {\frac {c}{a+b x^2}} \left (\sqrt {a}-\sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )}{a^{3/2}} \] Input:
Integrate[(c/(a + b*x^2))^(3/2)/x,x]
Output:
(c*Sqrt[c/(a + b*x^2)]*(Sqrt[a] - Sqrt[a + b*x^2]*ArcTanh[Sqrt[a + b*x^2]/ Sqrt[a]]))/a^(3/2)
Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2045, 243, 61, 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.
\(\displaystyle \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \int \frac {1}{x \left (\frac {b x^2}{a}+1\right )^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \int \frac {1}{x^2 \left (\frac {b x^2}{a}+1\right )^{3/2}}dx^2}{2 a}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \left (\int \frac {1}{x^2 \sqrt {\frac {b x^2}{a}+1}}dx^2+\frac {2}{\sqrt {\frac {b x^2}{a}+1}}\right )}{2 a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {c}{a+b x^2}} \left (\frac {2 a \int \frac {1}{\frac {a x^4}{b}-\frac {a}{b}}d\sqrt {\frac {b x^2}{a}+1}}{b}+\frac {2}{\sqrt {\frac {b x^2}{a}+1}}\right )}{2 a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \left (\frac {2}{\sqrt {\frac {b x^2}{a}+1}}-2 \text {arctanh}\left (\sqrt {\frac {b x^2}{a}+1}\right )\right ) \sqrt {\frac {c}{a+b x^2}}}{2 a}\) |
Input:
Int[(c/(a + b*x^2))^(3/2)/x,x]
Output:
(c*Sqrt[c/(a + b*x^2)]*Sqrt[1 + (b*x^2)/a]*(2/Sqrt[1 + (b*x^2)/a] - 2*ArcT anh[Sqrt[1 + (b*x^2)/a]]))/(2*a)
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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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]
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {\left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}} \left (b \,x^{2}+a \right ) \left (\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) a \sqrt {b \,x^{2}+a}-a^{\frac {3}{2}}\right )}{a^{\frac {5}{2}}}\) | \(64\) |
Input:
int((c/(b*x^2+a))^(3/2)/x,x,method=_RETURNVERBOSE)
Output:
-(c/(b*x^2+a))^(3/2)*(b*x^2+a)*(ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*a*(b*x ^2+a)^(1/2)-a^(3/2))/a^(5/2)
Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.94 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\left [\frac {c \sqrt {\frac {c}{a}} \log \left (-\frac {b c x^{2} + 2 \, a c - 2 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {\frac {c}{b x^{2} + a}} \sqrt {\frac {c}{a}}}{x^{2}}\right ) + 2 \, c \sqrt {\frac {c}{b x^{2} + a}}}{2 \, a}, \frac {c \sqrt {-\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} \sqrt {-\frac {c}{a}}}{c}\right ) + c \sqrt {\frac {c}{b x^{2} + a}}}{a}\right ] \] Input:
integrate((c/(b*x^2+a))^(3/2)/x,x, algorithm="fricas")
Output:
[1/2*(c*sqrt(c/a)*log(-(b*c*x^2 + 2*a*c - 2*(a*b*x^2 + a^2)*sqrt(c/(b*x^2 + a))*sqrt(c/a))/x^2) + 2*c*sqrt(c/(b*x^2 + a)))/a, (c*sqrt(-c/a)*arctan(a *sqrt(c/(b*x^2 + a))*sqrt(-c/a)/c) + c*sqrt(c/(b*x^2 + a)))/a]
\[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\int \frac {\left (\frac {c}{a + b x^{2}}\right )^{\frac {3}{2}}}{x}\, dx \] Input:
integrate((c/(b*x**2+a))**(3/2)/x,x)
Output:
Integral((c/(a + b*x**2))**(3/2)/x, x)
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\frac {1}{2} \, c {\left (\frac {c \log \left (\frac {a \sqrt {\frac {c}{b x^{2} + a}} - \sqrt {a c}}{a \sqrt {\frac {c}{b x^{2} + a}} + \sqrt {a c}}\right )}{\sqrt {a c} a} + \frac {2 \, \sqrt {\frac {c}{b x^{2} + a}}}{a}\right )} \] Input:
integrate((c/(b*x^2+a))^(3/2)/x,x, algorithm="maxima")
Output:
1/2*c*(c*log((a*sqrt(c/(b*x^2 + a)) - sqrt(a*c))/(a*sqrt(c/(b*x^2 + a)) + sqrt(a*c)))/(sqrt(a*c)*a) + 2*sqrt(c/(b*x^2 + a))/a)
Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=c {\left (\frac {c \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a} + \frac {c}{\sqrt {b c x^{2} + a c} a}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate((c/(b*x^2+a))^(3/2)/x,x, algorithm="giac")
Output:
c*(c*arctan(sqrt(b*c*x^2 + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a) + c/(sqrt(b*c*x ^2 + a*c)*a))*sgn(b*x^2 + a)
Timed out. \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\int \frac {{\left (\frac {c}{b\,x^2+a}\right )}^{3/2}}{x} \,d x \] Input:
int((c/(a + b*x^2))^(3/2)/x,x)
Output:
int((c/(a + b*x^2))^(3/2)/x, x)
Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.96 \[ \int \frac {\left (\frac {c}{a+b x^2}\right )^{3/2}}{x} \, dx=\frac {\sqrt {c}\, c \left (\sqrt {b \,x^{2}+a}\, a +\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a +\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,x^{2}-\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a -\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,x^{2}\right )}{a^{2} \left (b \,x^{2}+a \right )} \] Input:
int((c/(b*x^2+a))^(3/2)/x,x)
Output:
(sqrt(c)*c*(sqrt(a + b*x**2)*a + sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a + sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b) *x)/sqrt(a))*b*x**2 - sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x) /sqrt(a))*a - sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a) )*b*x**2))/(a**2*(a + b*x**2))