Integrand size = 15, antiderivative size = 81 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=-\frac {5 b}{3 a^2 (-a+b x)^{3/2}}+\frac {1}{a x (-a+b x)^{3/2}}+\frac {5 b}{a^3 \sqrt {-a+b x}}+\frac {5 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \] Output:
-5/3*b/a^2/(b*x-a)^(3/2)+1/a/x/(b*x-a)^(3/2)+5*b/a^3/(b*x-a)^(1/2)+5*b*arc tan((b*x-a)^(1/2)/a^(1/2))/a^(7/2)
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {3 a^2-20 a b x+15 b^2 x^2}{3 a^3 x (-a+b x)^{3/2}}+\frac {5 b \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:
Integrate[1/(x^2*(-a + b*x)^(5/2)),x]
Output:
(3*a^2 - 20*a*b*x + 15*b^2*x^2)/(3*a^3*x*(-a + b*x)^(3/2)) + (5*b*ArcTan[S qrt[-a + b*x]/Sqrt[a]])/a^(7/2)
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {52, 61, 61, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^2 (b x-a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 b \int \frac {1}{x (b x-a)^{5/2}}dx}{2 a}+\frac {1}{a x (b x-a)^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 b \left (-\frac {\int \frac {1}{x (b x-a)^{3/2}}dx}{a}-\frac {2}{3 a (b x-a)^{3/2}}\right )}{2 a}+\frac {1}{a x (b x-a)^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 b \left (-\frac {-\frac {\int \frac {1}{x \sqrt {b x-a}}dx}{a}-\frac {2}{a \sqrt {b x-a}}}{a}-\frac {2}{3 a (b x-a)^{3/2}}\right )}{2 a}+\frac {1}{a x (b x-a)^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 b \left (-\frac {-\frac {2 \int \frac {1}{\frac {a}{b}+\frac {b x-a}{b}}d\sqrt {b x-a}}{a b}-\frac {2}{a \sqrt {b x-a}}}{a}-\frac {2}{3 a (b x-a)^{3/2}}\right )}{2 a}+\frac {1}{a x (b x-a)^{3/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 b \left (-\frac {-\frac {2 \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {b x-a}}}{a}-\frac {2}{3 a (b x-a)^{3/2}}\right )}{2 a}+\frac {1}{a x (b x-a)^{3/2}}\) |
Input:
Int[1/(x^2*(-a + b*x)^(5/2)),x]
Output:
1/(a*x*(-a + b*x)^(3/2)) + (5*b*(-2/(3*a*(-a + b*x)^(3/2)) - (-2/(a*Sqrt[- a + b*x]) - (2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/a^(3/2))/a))/(2*a)
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84
method | result | size |
pseudoelliptic | \(\frac {\sqrt {b x -a}}{a^{3} x}+\frac {5 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}-\frac {2 b}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}+\frac {4 b}{a^{3} \sqrt {b x -a}}\) | \(68\) |
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 | \(2 b \left (-\frac {1}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}+\frac {2}{a^{3} \sqrt {b x -a}}+\frac {\frac {\sqrt {b x -a}}{2 b x}+\frac {5 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{3}}\right )\) | \(74\) |
risch | \(-\frac {-b x +a}{a^{3} x \sqrt {b x -a}}+\frac {5 b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}+\frac {4 b}{a^{3} \sqrt {b x -a}}-\frac {2 b}{3 a^{2} \left (b x -a \right )^{\frac {3}{2}}}\) | \(75\) |
Input:
int(1/x^2/(b*x-a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/a^3*(b*x-a)^(1/2)/x+5*b*arctan((b*x-a)^(1/2)/a^(1/2))/a^(7/2)-2/3*b/a^2/ (b*x-a)^(3/2)+4*b/a^3/(b*x-a)^(1/2)
Time = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.80 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x - a}}{6 \, {\left (a^{4} b^{2} x^{3} - 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac {15 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {b x - a}}\right ) - {\left (15 \, a b^{2} x^{2} - 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x - a}}{3 \, {\left (a^{4} b^{2} x^{3} - 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \] Input:
integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="fricas")
Output:
[-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(a)/sqrt(b*x - 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)]
Timed out. \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/x**2/(b*x-a)**(5/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {15 \, {\left (b x - a\right )}^{2} b + 10 \, {\left (b x - a\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (b x - a\right )}^{\frac {5}{2}} a^{3} + {\left (b x - a\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}} \] Input:
integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="maxima")
Output:
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(sqrt(b*x - a)/sqrt(a))/a^(7/2)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {5 \, b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}} + \frac {2 \, {\left (6 \, {\left (b x - a\right )} b - a b\right )}}{3 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{3}} + \frac {\sqrt {b x - a}}{a^{3} x} \] Input:
integrate(1/x^2/(b*x-a)^(5/2),x, algorithm="giac")
Output:
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)
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {1}{a\,x\,{\left (b\,x-a\right )}^{3/2}}-\frac {20\,b}{3\,a^2\,{\left (b\,x-a\right )}^{3/2}}+\frac {5\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5\,b^2\,x}{a^3\,{\left (b\,x-a\right )}^{3/2}} \] Input:
int(1/(x^2*(b*x - a)^(5/2)),x)
Output:
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))
Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^2 (-a+b x)^{5/2}} \, dx=\frac {15 \sqrt {a}\, \sqrt {b x -a}\, \mathit {atan} \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) a b x -15 \sqrt {a}\, \sqrt {b x -a}\, \mathit {atan} \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2}-3 a^{3}+20 a^{2} b x -15 a \,b^{2} x^{2}}{3 \sqrt {b x -a}\, a^{4} x \left (-b x +a \right )} \] Input:
int(1/x^2/(b*x-a)^(5/2),x)
Output:
(15*sqrt(a)*sqrt( - a + b*x)*atan(sqrt( - a + b*x)/sqrt(a))*a*b*x - 15*sqr t(a)*sqrt( - a + b*x)*atan(sqrt( - a + b*x)/sqrt(a))*b**2*x**2 - 3*a**3 + 20*a**2*b*x - 15*a*b**2*x**2)/(3*sqrt( - a + b*x)*a**4*x*(a - b*x))