Integrand size = 15, antiderivative size = 88 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=2 b^2 \sqrt {-a+b x}-\frac {a^2 \sqrt {-a+b x}}{2 x^2}+\frac {9 a b \sqrt {-a+b x}}{4 x}-\frac {15}{4} \sqrt {a} b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \] Output:
2*b^2*(b*x-a)^(1/2)-1/2*a^2*(b*x-a)^(1/2)/x^2+9/4*a*b*(b*x-a)^(1/2)/x-15/4 *a^(1/2)*b^2*arctan((b*x-a)^(1/2)/a^(1/2))
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {-a+b x} \left (-2 a^2+9 a b x+8 b^2 x^2\right )}{x^2}-15 \sqrt {a} b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )\right ) \] Input:
Integrate[(-a + b*x)^(5/2)/x^3,x]
Output:
((Sqrt[-a + b*x]*(-2*a^2 + 9*a*b*x + 8*b^2*x^2))/x^2 - 15*Sqrt[a]*b^2*ArcT an[Sqrt[-a + b*x]/Sqrt[a]])/4
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {51, 51, 60, 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.
\(\displaystyle \int \frac {(b x-a)^{5/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5}{4} b \int \frac {(b x-a)^{3/2}}{x^2}dx-\frac {(b x-a)^{5/2}}{2 x^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \int \frac {\sqrt {b x-a}}{x}dx-\frac {(b x-a)^{3/2}}{x}\right )-\frac {(b x-a)^{5/2}}{2 x^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {b x-a}-a \int \frac {1}{x \sqrt {b x-a}}dx\right )-\frac {(b x-a)^{3/2}}{x}\right )-\frac {(b x-a)^{5/2}}{2 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {b x-a}-\frac {2 a \int \frac {1}{\frac {a}{b}+\frac {b x-a}{b}}d\sqrt {b x-a}}{b}\right )-\frac {(b x-a)^{3/2}}{x}\right )-\frac {(b x-a)^{5/2}}{2 x^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {b x-a}-2 \sqrt {a} \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )\right )-\frac {(b x-a)^{3/2}}{x}\right )-\frac {(b x-a)^{5/2}}{2 x^2}\) |
Input:
Int[(-a + b*x)^(5/2)/x^3,x]
Output:
-1/2*(-a + b*x)^(5/2)/x^2 + (5*b*(-((-a + b*x)^(3/2)/x) + (3*b*(2*Sqrt[-a + b*x] - 2*Sqrt[a]*ArcTan[Sqrt[-a + b*x]/Sqrt[a]]))/2))/4
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(\frac {-15 \sqrt {a}\, b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) x^{2}-\left (-8 b^{2} x^{2}-9 a b x +2 a^{2}\right ) \sqrt {b x -a}}{4 x^{2}}\) | \(62\) |
risch | \(\frac {a \left (-b x +a \right ) \left (-9 b x +2 a \right )}{4 x^{2} \sqrt {b x -a}}-\frac {15 \sqrt {a}\, b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4}+2 b^{2} \sqrt {b x -a}\) | \(67\) |
derivativedivides | \(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(70\) |
default | \(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\) | \(70\) |
Input:
int((b*x-a)^(5/2)/x^3,x,method=_RETURNVERBOSE)
Output:
1/4*(-15*a^(1/2)*b^2*arctan((b*x-a)^(1/2)/a^(1/2))*x^2-(-8*b^2*x^2-9*a*b*x +2*a^2)*(b*x-a)^(1/2))/x^2
Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.57 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\left [\frac {15 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, x^{2}}, \frac {15 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b x - a}}\right ) + {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, x^{2}}\right ] \] Input:
integrate((b*x-a)^(5/2)/x^3,x, algorithm="fricas")
Output:
[1/8*(15*sqrt(-a)*b^2*x^2*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*(8*b^2*x^2 + 9*a*b*x - 2*a^2)*sqrt(b*x - a))/x^2, 1/4*(15*sqrt(a)*b^2*x^ 2*arctan(sqrt(a)/sqrt(b*x - a)) + (8*b^2*x^2 + 9*a*b*x - 2*a^2)*sqrt(b*x - a))/x^2]
Result contains complex when optimal does not.
Time = 2.10 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.03 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\begin {cases} - \frac {15 i \sqrt {a} b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {i a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {11 i a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {15 \sqrt {a} b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} + \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {- \frac {a}{b x} + 1}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x-a)**(5/2)/x**3,x)
Output:
Piecewise((-15*I*sqrt(a)*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - I*a**3/ (2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) + 11*I*a**2*sqrt(b)/(4*x**(3/2)*sqr t(a/(b*x) - 1)) - I*a*b**(3/2)/(4*sqrt(x)*sqrt(a/(b*x) - 1)) - 2*I*b**(5/2 )*sqrt(x)/sqrt(a/(b*x) - 1), Abs(a/(b*x)) > 1), (15*sqrt(a)*b**2*asin(sqrt (a)/(sqrt(b)*sqrt(x)))/4 + a**3/(2*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) - 11*a**2*sqrt(b)/(4*x**(3/2)*sqrt(-a/(b*x) + 1)) + a*b**(3/2)/(4*sqrt(x)*sq rt(-a/(b*x) + 1)) + 2*b**(5/2)*sqrt(x)/sqrt(-a/(b*x) + 1), True))
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=-\frac {15}{4} \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b^{2} + \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{2} + 7 \, \sqrt {b x - a} a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \] Input:
integrate((b*x-a)^(5/2)/x^3,x, algorithm="maxima")
Output:
-15/4*sqrt(a)*b^2*arctan(sqrt(b*x - a)/sqrt(a)) + 2*sqrt(b*x - a)*b^2 + 1/ 4*(9*(b*x - a)^(3/2)*a*b^2 + 7*sqrt(b*x - a)*a^2*b^2)/((b*x - a)^2 + 2*(b* x - a)*a + a^2)
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=-\frac {15 \, \sqrt {a} b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - 8 \, \sqrt {b x - a} b^{3} - \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{3} + 7 \, \sqrt {b x - a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \] Input:
integrate((b*x-a)^(5/2)/x^3,x, algorithm="giac")
Output:
-1/4*(15*sqrt(a)*b^3*arctan(sqrt(b*x - a)/sqrt(a)) - 8*sqrt(b*x - a)*b^3 - (9*(b*x - a)^(3/2)*a*b^3 + 7*sqrt(b*x - a)*a^2*b^3)/(b^2*x^2))/b
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=2\,b^2\,\sqrt {b\,x-a}-\frac {15\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4}+\frac {9\,a\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}+\frac {7\,a^2\,\sqrt {b\,x-a}}{4\,x^2} \] Input:
int((b*x - a)^(5/2)/x^3,x)
Output:
2*b^2*(b*x - a)^(1/2) - (15*a^(1/2)*b^2*atan((b*x - a)^(1/2)/a^(1/2)))/4 + (9*a*(b*x - a)^(3/2))/(4*x^2) + (7*a^2*(b*x - a)^(1/2))/(4*x^2)
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {(-a+b x)^{5/2}}{x^3} \, dx=\frac {-15 \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2}-2 \sqrt {b x -a}\, a^{2}+9 \sqrt {b x -a}\, a b x +8 \sqrt {b x -a}\, b^{2} x^{2}}{4 x^{2}} \] Input:
int((b*x-a)^(5/2)/x^3,x)
Output:
( - 15*sqrt(a)*atan(sqrt( - a + b*x)/sqrt(a))*b**2*x**2 - 2*sqrt( - a + b* x)*a**2 + 9*sqrt( - a + b*x)*a*b*x + 8*sqrt( - a + b*x)*b**2*x**2)/(4*x**2 )