\(\int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx\) [363]

Optimal result

Integrand size = 15, antiderivative size = 62 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=-\frac {3 b}{a^2 \sqrt {-a+b x}}+\frac {1}{a x \sqrt {-a+b x}}-\frac {3 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, 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 = {44, 53, 65, 211} \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=-\frac {3 b \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {3 b}{a^2 \sqrt {b x-a}}+\frac {1}{a x \sqrt {b x-a}} \]

[In]

Int[1/(x^2*(-a + b*x)^(3/2)),x]

[Out]

(-3*b)/(a^2*Sqrt[-a + b*x]) + 1/(a*x*Sqrt[-a + b*x]) - (3*b*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/a^(5/2)

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 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 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 {1}{a x \sqrt {-a+b x}}+\frac {(3 b) \int \frac {1}{x (-a+b x)^{3/2}} \, dx}{2 a} \\ & = -\frac {3 b}{a^2 \sqrt {-a+b x}}+\frac {1}{a x \sqrt {-a+b x}}-\frac {(3 b) \int \frac {1}{x \sqrt {-a+b x}} \, dx}{2 a^2} \\ & = -\frac {3 b}{a^2 \sqrt {-a+b x}}+\frac {1}{a x \sqrt {-a+b x}}-\frac {3 \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )}{a^2} \\ & = -\frac {3 b}{a^2 \sqrt {-a+b x}}+\frac {1}{a x \sqrt {-a+b x}}-\frac {3 b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=\frac {a-3 b x}{a^2 x \sqrt {-a+b x}}-\frac {3 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \]

[In]

Integrate[1/(x^2*(-a + b*x)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(\frac {-3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b x \sqrt {b x -a}-3 \sqrt {a}\, b x +a^{\frac {3}{2}}}{a^{\frac {5}{2}} x \sqrt {b x -a}}\)	\(55\)
risch	\(\frac {-b x +a}{a^{2} x \sqrt {b x -a}}-\frac {2 b}{a^{2} \sqrt {b x -a}}-\frac {3 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\)	\(59\)
derivativedivides	\(2 b \left (-\frac {1}{a^{2} \sqrt {b x -a}}-\frac {\frac {\sqrt {b x -a}}{2 b x}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )\)	\(61\)
default	\(2 b \left (-\frac {1}{a^{2} \sqrt {b x -a}}-\frac {\frac {\sqrt {b x -a}}{2 b x}+\frac {3 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{2}}\right )\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.65 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (b^{2} x^{2} - a b x\right )} \sqrt {-a} \log \left (\frac {b x + 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x - a^{2}\right )} \sqrt {b x - a}}{2 \, {\left (a^{3} b x^{2} - a^{4} x\right )}}, -\frac {3 \, {\left (b^{2} x^{2} - a b x\right )} \sqrt {a} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (3 \, a b x - a^{2}\right )} \sqrt {b x - a}}{a^{3} b x^{2} - a^{4} x}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.42 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.52 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=\begin {cases} - \frac {i}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {3 i b \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {3 b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I/(a*sqrt(b)*x**(3/2)*sqrt(a/(b*x) - 1)) + 3*I*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) - 1)) - 3*I*b*ac
osh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(5/2), Abs(a/(b*x)) > 1), (1/(a*sqrt(b)*x**(3/2)*sqrt(-a/(b*x) + 1)) - 3*sqr
t(b)/(a**2*sqrt(x)*sqrt(-a/(b*x) + 1)) + 3*b*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=-\frac {3 \, {\left (b x - a\right )} b + 2 \, a b}{{\left (b x - a\right )}^{\frac {3}{2}} a^{2} + \sqrt {b x - a} a^{3}} - \frac {3 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {3 \, {\left (b x - a\right )} b + 2 \, a b}{{\left ({\left (b x - a\right )}^{\frac {3}{2}} + \sqrt {b x - a} a\right )} a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 (-a+b x)^{3/2}} \, dx=\frac {1}{a\,x\,\sqrt {b\,x-a}}-\frac {3\,b}{a^2\,\sqrt {b\,x-a}}-\frac {3\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{a^{5/2}} \]

[In]

int(1/(x^2*(b*x - a)^(3/2)),x)

[Out]

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