\(\int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx\) [366]

Optimal result

Integrand size = 15, antiderivative size = 81 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=-\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}+\frac {5 b}{a^3 \sqrt {-a+b x}}+\frac {5 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

-5/3*b/a^2/(b*x-a)^(3/2)+1/a/x/(b*x-a)^(3/2)+5*b*arctan((b*x-a)^(1/2)/a^(1/2))/a^(7/2)+5*b/a^3/(b*x-a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {44, 53, 65, 211} \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {5 b \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b}{a^3 \sqrt {b x-a}}-\frac {5 b}{3 a^2 (b x-a)^{3/2}}+\frac {1}{a x (b x-a)^{3/2}} \]

[In]

Int[1/(x^2*(-a + b*x)^(5/2)),x]

[Out]

(-5*b)/(3*a^2*(-a + b*x)^(3/2)) + 1/(a*x*(-a + b*x)^(3/2)) + (5*b)/(a^3*Sqrt[-a + b*x]) + (5*b*ArcTan[Sqrt[-a
+ b*x]/Sqrt[a]])/a^(7/2)

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 (-a+b x)^{3/2}}+\frac {(5 b) \int \frac {1}{x (-a+b x)^{5/2}} \, dx}{2 a} \\ & = -\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}-\frac {(5 b) \int \frac {1}{x (-a+b x)^{3/2}} \, dx}{2 a^2} \\ & = -\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}+\frac {5 b}{a^3 \sqrt {-a+b x}}+\frac {(5 b) \int \frac {1}{x \sqrt {-a+b x}} \, dx}{2 a^3} \\ & = -\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}+\frac {5 b}{a^3 \sqrt {-a+b x}}+\frac {5 \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )}{a^3} \\ & = -\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}+\frac {5 b}{a^3 \sqrt {-a+b x}}+\frac {5 b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {3 a^2-20 a b x+15 b^2 x^2}{3 a^3 x (-a+b x)^{3/2}}+\frac {5 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[In]

Integrate[1/(x^2*(-a + b*x)^(5/2)),x]

[Out]

(3*a^2 - 20*a*b*x + 15*b^2*x^2)/(3*a^3*x*(-a + b*x)^(3/2)) + (5*b*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/a^(7/2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(2 b \left (-\frac {1}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}+\frac {2}{a^{3} \sqrt {b x -a}}+\frac {\frac {\sqrt {b x -a}}{2 b x}+\frac {5 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{3}}\right )\)	\(74\)
default	\(2 b \left (-\frac {1}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}+\frac {2}{a^{3} \sqrt {b x -a}}+\frac {\frac {\sqrt {b x -a}}{2 b x}+\frac {5 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{3}}\right )\)	\(74\)
pseudoelliptic	\(-\frac {5 \left (\sqrt {b x -a}\, b x \left (-b x +a \right ) \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )-\sqrt {a}\, b^{2} x^{2}+\frac {4 a^{\frac {3}{2}} b x}{3}-\frac {a^{\frac {5}{2}}}{5}\right )}{\left (b x -a \right )^{\frac {3}{2}} a^{\frac {7}{2}} x}\)	\(74\)
risch	\(-\frac {-b x +a}{a^{3} x \sqrt {b x -a}}+\frac {5 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}+\frac {4 b}{a^{3} \sqrt {b x -a}}-\frac {2 b}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}\)	\(75\)

[In]

int(1/x^2/(b*x-a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.79 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x - a}}{6 \, {\left (a^{4} b^{2} x^{3} - 2 \, a^{5} b x^{2} + a^{6} x\right )}}, \frac {15 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (15 \, a b^{2} x^{2} - 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x - a}}{3 \, {\left (a^{4} b^{2} x^{3} - 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \]

[In]

integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(15*(b^3*x^3 - 2*a*b^2*x^2 + a^2*b*x)*sqrt(-a)*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) - 2*(15*a*b
^2*x^2 - 20*a^2*b*x + 3*a^3)*sqrt(b*x - a))/(a^4*b^2*x^3 - 2*a^5*b*x^2 + a^6*x), 1/3*(15*(b^3*x^3 - 2*a*b^2*x^
2 + a^2*b*x)*sqrt(a)*arctan(sqrt(b*x - a)/sqrt(a)) + (15*a*b^2*x^2 - 20*a^2*b*x + 3*a^3)*sqrt(b*x - a))/(a^4*b
^2*x^3 - 2*a^5*b*x^2 + a^6*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/x**2/(b*x-a)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {15 \, {\left (b x - a\right )}^{2} b + 10 \, {\left (b x - a\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (b x - a\right )}^{\frac {5}{2}} a^{3} + {\left (b x - a\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}} \]

[In]

integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="maxima")

[Out]

1/3*(15*(b*x - a)^2*b + 10*(b*x - a)*a*b - 2*a^2*b)/((b*x - a)^(5/2)*a^3 + (b*x - a)^(3/2)*a^4) + 5*b*arctan(s
qrt(b*x - a)/sqrt(a))/a^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {5 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}} + \frac {2 \, {\left (6 \, {\left (b x - a\right )} b - a b\right )}}{3 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{3}} + \frac {\sqrt {b x - a}}{a^{3} x} \]

[In]

integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="giac")

[Out]

5*b*arctan(sqrt(b*x - a)/sqrt(a))/a^(7/2) + 2/3*(6*(b*x - a)*b - a*b)/((b*x - a)^(3/2)*a^3) + sqrt(b*x - a)/(a
^3*x)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {1}{a\,x\,{\left (b\,x-a\right )}^{3/2}}-\frac {20\,b}{3\,a^2\,{\left (b\,x-a\right )}^{3/2}}+\frac {5\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5\,b^2\,x}{a^3\,{\left (b\,x-a\right )}^{3/2}} \]

[In]

int(1/(x^2*(b*x - a)^(5/2)),x)

[Out]

1/(a*x*(b*x - a)^(3/2)) - (20*b)/(3*a^2*(b*x - a)^(3/2)) + (5*b*atan((b*x - a)^(1/2)/a^(1/2)))/a^(7/2) + (5*b^
2*x)/(a^3*(b*x - a)^(3/2))