\(\int \frac {(-a+b x)^{3/2}}{x^3} \, dx\) [330]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=-\frac {3 b \sqrt {-a+b x}}{4 x}-\frac {(-a+b x)^{3/2}}{2 x^2}+\frac {3 b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {43, 65, 211} \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\frac {3 b^2 \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {(b x-a)^{3/2}}{2 x^2}-\frac {3 b \sqrt {b x-a}}{4 x} \]

[In]

Int[(-a + b*x)^(3/2)/x^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {(2 a-5 b x) \sqrt {-a+b x}}{x^2}+\frac {3 b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \]

[In]

Integrate[(-a + b*x)^(3/2)/x^3,x]

[Out]

(((2*a - 5*b*x)*Sqrt[-a + b*x])/x^2 + (3*b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/Sqrt[a])/4

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (-b x +a \right ) \left (-5 b x +2 a \right )}{4 x^{2} \sqrt {b x -a}}+\frac {3 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\)	\(52\)
derivativedivides	\(2 b^{2} \left (\frac {-\frac {5 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\)	\(57\)
default	\(2 b^{2} \left (\frac {-\frac {5 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\)	\(57\)
pseudoelliptic	\(\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2}-5 b x \sqrt {b x -a}\, \sqrt {a}+2 \sqrt {b x -a}\, a^{\frac {3}{2}}}{4 x^{2} \sqrt {a}}\)	\(62\)

[In]

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

[Out]

-1/4*(-b*x+a)*(-5*b*x+2*a)/x^2/(b*x-a)^(1/2)+3/4*b^2*arctan((b*x-a)^(1/2)/a^(1/2))/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\left [-\frac {3 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (5 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a x^{2}}, \frac {3 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (5 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a x^{2}}\right ] \]

[In]

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

[Out]

[-1/8*(3*sqrt(-a)*b^2*x^2*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*(5*a*b*x - 2*a^2)*sqrt(b*x - a))/(
a*x^2), 1/4*(3*sqrt(a)*b^2*x^2*arctan(sqrt(b*x - a)/sqrt(a)) - (5*a*b*x - 2*a^2)*sqrt(b*x - a))/(a*x^2)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.39 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.78 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\begin {cases} \frac {i a \sqrt {b} \sqrt {\frac {a}{b x} - 1}}{2 x^{\frac {3}{2}}} - \frac {5 i b^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}}{4 \sqrt {x}} + \frac {3 i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 \sqrt {a}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {a^{2}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {7 a \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {5 b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 \sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((I*a*sqrt(b)*sqrt(a/(b*x) - 1)/(2*x**(3/2)) - 5*I*b**(3/2)*sqrt(a/(b*x) - 1)/(4*sqrt(x)) + 3*I*b**2*
acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*sqrt(a)), Abs(a/(b*x)) > 1), (-a**2/(2*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)
) + 7*a*sqrt(b)/(4*x**(3/2)*sqrt(-a/(b*x) + 1)) - 5*b**(3/2)/(4*sqrt(x)*sqrt(-a/(b*x) + 1)) - 3*b**2*asin(sqrt
(a)/(sqrt(b)*sqrt(x)))/(4*sqrt(a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\frac {3 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, \sqrt {a}} - \frac {5 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \]

[In]

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

[Out]

3/4*b^2*arctan(sqrt(b*x - a)/sqrt(a))/sqrt(a) - 1/4*(5*(b*x - a)^(3/2)*b^2 + 3*sqrt(b*x - a)*a*b^2)/((b*x - a)
^2 + 2*(b*x - a)*a + a^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {5 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{3} + 3 \, \sqrt {b x - a} a b^{3}}{b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

1/4*(3*b^3*arctan(sqrt(b*x - a)/sqrt(a))/sqrt(a) - (5*(b*x - a)^(3/2)*b^3 + 3*sqrt(b*x - a)*a*b^3)/(b^2*x^2))/
b

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {(-a+b x)^{3/2}}{x^3} \, dx=\frac {3\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,\sqrt {a}}-\frac {5\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}-\frac {3\,a\,\sqrt {b\,x-a}}{4\,x^2} \]

[In]

int((b*x - a)^(3/2)/x^3,x)

[Out]

(3*b^2*atan((b*x - a)^(1/2)/a^(1/2)))/(4*a^(1/2)) - (5*(b*x - a)^(3/2))/(4*x^2) - (3*a*(b*x - a)^(1/2))/(4*x^2
)