\(\int \frac {(-a+b x)^{3/2}}{x^2} \, dx\) [329]

Optimal result

Integrand size = 15, antiderivative size = 57 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=3 b \sqrt {-a+b x}-\frac {(-a+b x)^{3/2}}{x}-3 \sqrt {a} b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, 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, 52, 65, 211} \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=-3 \sqrt {a} b \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{3/2}}{x}+3 b \sqrt {b x-a} \]

[In]

Int[(-a + b*x)^(3/2)/x^2,x]

[Out]

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

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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 {(-a+b x)^{3/2}}{x}+\frac {1}{2} (3 b) \int \frac {\sqrt {-a+b x}}{x} \, dx \\ & = 3 b \sqrt {-a+b x}-\frac {(-a+b x)^{3/2}}{x}-\frac {1}{2} (3 a b) \int \frac {1}{x \sqrt {-a+b x}} \, dx \\ & = 3 b \sqrt {-a+b x}-\frac {(-a+b x)^{3/2}}{x}-(3 a) \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right ) \\ & = 3 b \sqrt {-a+b x}-\frac {(-a+b x)^{3/2}}{x}-3 \sqrt {a} b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=\frac {\sqrt {-a+b x} (a+2 b x)}{x}-3 \sqrt {a} b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \]

[In]

Integrate[(-a + b*x)^(3/2)/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (2 b x +a \right ) \sqrt {b x -a}-3 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) \sqrt {a}\, x}{x}\)	\(43\)
derivativedivides	\(2 b \left (\sqrt {b x -a}-a \left (-\frac {\sqrt {b x -a}}{2 b x}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(54\)
default	\(2 b \left (\sqrt {b x -a}-a \left (-\frac {\sqrt {b x -a}}{2 b x}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(54\)
risch	\(-\frac {a \left (-b x +a \right )}{x \sqrt {b x -a}}-3 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) \sqrt {a}+2 b \sqrt {b x -a}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.84 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=\left [\frac {3 \, \sqrt {-a} b x \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (2 \, b x + a\right )} \sqrt {b x - a}}{2 \, x}, -\frac {3 \, \sqrt {a} b x \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (2 \, b x + a\right )} \sqrt {b x - a}}{x}\right ] \]

[In]

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

[Out]

[1/2*(3*sqrt(-a)*b*x*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*(2*b*x + a)*sqrt(b*x - a))/x, -(3*sqrt(
a)*b*x*arctan(sqrt(b*x - a)/sqrt(a)) - (2*b*x + a)*sqrt(b*x - a))/x]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.46 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=\begin {cases} - 3 i \sqrt {a} b \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} + \frac {i a^{2}}{\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {i a \sqrt {b}}{\sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {3}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\3 \sqrt {a} b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {a^{2}}{\sqrt {b} x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {a \sqrt {b}}{\sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {3}{2}} \sqrt {x}}{\sqrt {- \frac {a}{b x} + 1}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=-3 \, \sqrt {a} b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b + \frac {\sqrt {b x - a} a}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=-\frac {3 \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - 2 \, \sqrt {b x - a} b^{2} - \frac {\sqrt {b x - a} a b}{x}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(-a+b x)^{3/2}}{x^2} \, dx=2\,b\,\sqrt {b\,x-a}+\frac {a\,\sqrt {b\,x-a}}{x}-3\,\sqrt {a}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right ) \]

[In]

int((b*x - a)^(3/2)/x^2,x)

[Out]

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