\(\int \frac {1}{x (a+b x)^{3/2}} \, dx\) [348]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

-2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)+2/a/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {53, 65, 214} \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

Int[1/(x*(a + b*x)^(3/2)),x]

[Out]

2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{a \sqrt {a+b x}}+\frac {\int \frac {1}{x \sqrt {a+b x}} \, dx}{a} \\ & = \frac {2}{a \sqrt {a+b x}}+\frac {2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{a b} \\ & = \frac {2}{a \sqrt {a+b x}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

Integrate[1/(x*(a + b*x)^(3/2)),x]

[Out]

2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {b x +a}}\)	\(31\)
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {b x +a}}\)	\(31\)
pseudoelliptic	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {b x +a}}\)	\(31\)

[In]

int(1/x/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)+2/a/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.89 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\left [\frac {{\left (b x + a\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a} a}{a^{2} b x + a^{3}}, \frac {2 \, {\left ({\left (b x + a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \sqrt {b x + a} a\right )}}{a^{2} b x + a^{3}}\right ] \]

[In]

integrate(1/x/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[((b*x + a)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*sqrt(b*x + a)*a)/(a^2*b*x + a^3), 2*((b*x
+ a)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + sqrt(b*x + a)*a)/(a^2*b*x + a^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (32) = 64\).

Time = 1.32 (sec) , antiderivative size = 146, normalized size of antiderivative = 3.84 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2 a^{3} \sqrt {1 + \frac {b x}{a}}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} + \frac {a^{3} \log {\left (\frac {b x}{a} \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} + \frac {a^{2} b x \log {\left (\frac {b x}{a} \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} - \frac {2 a^{2} b x \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} \]

[In]

integrate(1/x/(b*x+a)**(3/2),x)

[Out]

2*a**3*sqrt(1 + b*x/a)/(a**(9/2) + a**(7/2)*b*x) + a**3*log(b*x/a)/(a**(9/2) + a**(7/2)*b*x) - 2*a**3*log(sqrt
(1 + b*x/a) + 1)/(a**(9/2) + a**(7/2)*b*x) + a**2*b*x*log(b*x/a)/(a**(9/2) + a**(7/2)*b*x) - 2*a**2*b*x*log(sq
rt(1 + b*x/a) + 1)/(a**(9/2) + a**(7/2)*b*x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {\log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {b x + a} a} \]

[In]

integrate(1/x/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2) + 2/(sqrt(b*x + a)*a)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {2}{\sqrt {b x + a} a} \]

[In]

integrate(1/x/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + 2/(sqrt(b*x + a)*a)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x (a+b x)^{3/2}} \, dx=\frac {2}{a\,\sqrt {a+b\,x}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[In]

int(1/(x*(a + b*x)^(3/2)),x)

[Out]

2/(a*(a + b*x)^(1/2)) - (2*atanh((a + b*x)^(1/2)/a^(1/2)))/a^(3/2)