Integrand size = 13, antiderivative size = 57 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=-\frac {3 b}{a^2 \sqrt {a+b x}}-\frac {1}{a x \sqrt {a+b x}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \] Output:
-3*b/a^2/(b*x+a)^(1/2)-1/a/x/(b*x+a)^(1/2)+3*b*arctanh((b*x+a)^(1/2)/a^(1/ 2))/a^(5/2)
Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=\frac {-a-3 b x}{a^2 x \sqrt {a+b x}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \] Input:
Integrate[1/(x^2*(a + b*x)^(3/2)),x]
Output:
(-a - 3*b*x)/(a^2*x*Sqrt[a + b*x]) + (3*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/ a^(5/2)
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {52, 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.
\(\displaystyle \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {3 b \int \frac {1}{x (a+b x)^{3/2}}dx}{2 a}-\frac {1}{a x \sqrt {a+b x}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {3 b \left (\frac {\int \frac {1}{x \sqrt {a+b x}}dx}{a}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {3 b \left (\frac {2 \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a b}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {3 b \left (\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\) |
Input:
Int[1/(x^2*(a + b*x)^(3/2)),x]
Output:
-(1/(a*x*Sqrt[a + b*x])) - (3*b*(2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/(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] && ILtQ[m, -1] && FractionQ[n] && LtQ[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] && 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]
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-\frac {\sqrt {b x +a}}{a^{2} x}-\frac {b \left (\frac {4}{\sqrt {b x +a}}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{2 a^{2}}\) | \(50\) |
pseudoelliptic | \(-\frac {-3 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x +\sqrt {a}\, \left (3 b x +a \right )}{\sqrt {b x +a}\, x \,a^{\frac {5}{2}}}\) | \(50\) |
derivativedivides | \(2 b \left (\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}-\frac {1}{a^{2} \sqrt {b x +a}}\right )\) | \(54\) |
default | \(2 b \left (\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}-\frac {1}{a^{2} \sqrt {b x +a}}\right )\) | \(54\) |
Input:
int(1/x^2/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/a^2*(b*x+a)^(1/2)/x-1/2*b/a^2*(4/(b*x+a)^(1/2)-6*arctanh((b*x+a)^(1/2)/ a^(1/2))/a^(1/2))
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.60 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, -\frac {3 \, {\left (b^{2} x^{2} + a b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (3 \, a b x + a^{2}\right )} \sqrt {b x + a}}{a^{3} b x^{2} + a^{4} x}\right ] \] Input:
integrate(1/x^2/(b*x+a)^(3/2),x, algorithm="fricas")
Output:
[1/2*(3*(b^2*x^2 + a*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a )/x) - 2*(3*a*b*x + a^2)*sqrt(b*x + a))/(a^3*b*x^2 + a^4*x), -(3*(b^2*x^2 + a*b*x)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x + a)) + (3*a*b*x + a^2)*sqrt(b* x + a))/(a^3*b*x^2 + a^4*x)]
Time = 1.49 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=- \frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}} \] Input:
integrate(1/x**2/(b*x+a)**(3/2),x)
Output:
-1/(a*sqrt(b)*x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a /(b*x) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(5/2)
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=-\frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left (b x + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x + a} a^{3}} - \frac {3 \, b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} \] Input:
integrate(1/x^2/(b*x+a)^(3/2),x, algorithm="maxima")
Output:
-(3*(b*x + a)*b - 2*a*b)/((b*x + a)^(3/2)*a^2 - sqrt(b*x + a)*a^3) - 3/2*b *log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(5/2)
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2}} \] Input:
integrate(1/x^2/(b*x+a)^(3/2),x, algorithm="giac")
Output:
-3*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) - (3*(b*x + a)*b - 2*a* b)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)*a^2)
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {2\,b}{a}-\frac {3\,b\,\left (a+b\,x\right )}{a^2}}{a\,\sqrt {a+b\,x}-{\left (a+b\,x\right )}^{3/2}} \] Input:
int(1/(x^2*(a + b*x)^(3/2)),x)
Output:
(3*b*atanh((a + b*x)^(1/2)/a^(1/2)))/a^(5/2) - ((2*b)/a - (3*b*(a + b*x))/ a^2)/(a*(a + b*x)^(1/2) - (a + b*x)^(3/2))
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx=\frac {-3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b x +3 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b x -2 a^{2}-6 a b x}{2 \sqrt {b x +a}\, a^{3} x} \] Input:
int(1/x^2/(b*x+a)^(3/2),x)
Output:
( - 3*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b*x + 3*sqrt(a)*s qrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b*x - 2*a**2 - 6*a*b*x)/(2*sqrt( a + b*x)*a**3*x)