\(\int \frac {\sqrt {-a+b x}}{x^3} \, dx\) [327]

Optimal result

Integrand size = 15, antiderivative size = 71 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=-\frac {\sqrt {-a+b x}}{2 x^2}+\frac {b \sqrt {-a+b x}}{4 a x}+\frac {b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

1/4*b^2*arctan((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 = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {43, 44, 65, 211} \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {b x-a}}{2 x^2}+\frac {b \sqrt {b x-a}}{4 a x} \]

[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*ArcTan[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=-\frac {(2 a-b x) \sqrt {-a+b x}}{4 a x^2}+\frac {b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[In]

Integrate[Sqrt[-a + b*x]/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (-2 a^{\frac {3}{2}}+\sqrt {a}\, b x \right ) \sqrt {b x -a}+\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2}}{4 a^{\frac {3}{2}} x^{2}}\)	\(53\)
risch	\(\frac {\left (-b x +a \right ) \left (-b x +2 a \right )}{4 x^{2} \sqrt {b x -a}\, a}+\frac {b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\)	\(55\)
derivativedivides	\(2 b^{2} \left (\frac {\frac {\left (b x -a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(59\)
default	\(2 b^{2} \left (\frac {\frac {\left (b x -a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\left [-\frac {\sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a^{2} x^{2}}, \frac {\sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a^{2} x^{2}}\right ] \]

[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*b*x - 2*a^2)*sqrt(b*x - a))/(a^2*x^2)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.92 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\begin {cases} - \frac {i a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I*a/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) + 3*I*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) - 1)) - I*b**(3/
2)/(4*a*sqrt(x)*sqrt(a/(b*x) - 1)) + I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2)), Abs(a/(b*x)) > 1),
(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)
*sqrt(-a/(b*x) + 1)) - b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {3}{2}}} + \frac {{\left (b x - a\right )}^{\frac {3}{2}} b^{2} - \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} a + 2 \, {\left (b x - a\right )} a^{2} + a^{3}\right )}} \]

[In]

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

[Out]

1/4*b^2*arctan(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 = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {{\left (b x - a\right )}^{\frac {3}{2}} b^{3} - \sqrt {b x - a} a b^{3}}{a b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

1/4*(b^3*arctan(sqrt(b*x - a)/sqrt(a))/a^(3/2) + ((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 = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {-a+b x}}{x^3} \, dx=\frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {\sqrt {b\,x-a}}{4\,x^2}+\frac {{\left (b\,x-a\right )}^{3/2}}{4\,a\,x^2} \]

[In]

int((b*x - a)^(1/2)/x^3,x)

[Out]

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