\(\int \frac {\sqrt {a+b x}}{x^3} \, dx\) [290]

Optimal result

Integrand size = 13, antiderivative size = 65 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=-\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

1/4*b^2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)-1/2*(b*x+a)^(1/2)/x^2-1/4*b*(b*x+a)^(1/2)/a/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 44, 65, 214} \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x} \]

[In]

Int[Sqrt[a + b*x]/x^3,x]

[Out]

-1/2*Sqrt[a + b*x]/x^2 - (b*Sqrt[a + b*x])/(4*a*x) + (b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(3/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 {\sqrt {a+b x}}{2 x^2}+\frac {1}{4} b \int \frac {1}{x^2 \sqrt {a+b x}} \, dx \\ & = -\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x}-\frac {b^2 \int \frac {1}{x \sqrt {a+b x}} \, dx}{8 a} \\ & = -\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x}-\frac {b \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{4 a} \\ & = -\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=-\frac {\sqrt {a+b x} (2 a+b x)}{4 a x^2}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[In]

Integrate[Sqrt[a + b*x]/x^3,x]

[Out]

-1/4*(Sqrt[a + b*x]*(2*a + b*x))/(a*x^2) + (b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(3/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (b x +2 a \right )}{4 x^{2} a}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\)	\(44\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-\left (2 a^{\frac {3}{2}}+\sqrt {a}\, b x \right ) \sqrt {b x +a}}{4 a^{\frac {3}{2}} x^{2}}\)	\(50\)
derivativedivides	\(2 b^{2} \left (-\frac {\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(54\)
default	\(2 b^{2} \left (-\frac {\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(54\)

[In]

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

[Out]

-1/4*(b*x+a)^(1/2)*(b*x+2*a)/x^2/a+1/4*b^2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=\left [\frac {\sqrt {a} b^{2} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{2} x^{2}}, -\frac {\sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{2} x^{2}}\right ] \]

[In]

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

[Out]

[1/8*(sqrt(a)*b^2*x^2*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2
), -1/4*(sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2)]

Sympy [A] (verification not implemented)

Time = 2.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=- \frac {a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} \]

[In]

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

[Out]

-a/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1)) - b**(3/2)/(4*a*sqrt(x)*s
qrt(a/(b*x) + 1)) + b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=-\frac {b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b x + a} a b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} a - 2 \, {\left (b x + a\right )} a^{2} + a^{3}\right )}} \]

[In]

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

[Out]

-1/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2) - 1/4*((b*x + a)^(3/2)*b^2 + sqrt(b*
x + a)*a*b^2)/((b*x + a)^2*a - 2*(b*x + a)*a^2 + a^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=-\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x + a} a b^{3}}{a b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

-1/4*(b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + ((b*x + a)^(3/2)*b^3 + sqrt(b*x + a)*a*b^3)/(a*b^2*x^2
))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {a+b x}}{x^3} \, dx=\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {{\left (a+b\,x\right )}^{3/2}}{4\,a\,x^2}-\frac {\sqrt {a+b\,x}}{4\,x^2} \]

[In]

int((a + b*x)^(1/2)/x^3,x)

[Out]

(b^2*atanh((a + b*x)^(1/2)/a^(1/2)))/(4*a^(3/2)) - (a + b*x)^(3/2)/(4*a*x^2) - (a + b*x)^(1/2)/(4*x^2)