Integrand size = 15, antiderivative size = 95 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=-\frac {15 b^2}{4 a^3 \sqrt {-a+b x}}+\frac {1}{2 a x^2 \sqrt {-a+b x}}+\frac {5 b}{4 a^2 x \sqrt {-a+b x}}-\frac {15 b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
-15/4*b^2*arctan((b*x-a)^(1/2)/a^(1/2))/a^(7/2)-15/4*b^2/a^3/(b*x-a)^(1/2) +1/2/a/x^2/(b*x-a)^(1/2)+5/4*b/a^2/x/(b*x-a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=\frac {2 a^2+5 a b x-15 b^2 x^2}{4 a^3 x^2 \sqrt {-a+b x}}-\frac {15 b^2 \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
(2*a^2 + 5*a*b*x - 15*b^2*x^2)/(4*a^3*x^2*Sqrt[-a + b*x]) - (15*b^2*ArcTan [Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(7/2))
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {52, 52, 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.
\(\displaystyle \int \frac {1}{x^3 (b x-a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 b \int \frac {1}{x^2 (b x-a)^{3/2}}dx}{4 a}+\frac {1}{2 a x^2 \sqrt {b x-a}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 b \left (\frac {3 b \int \frac {1}{x (b x-a)^{3/2}}dx}{2 a}+\frac {1}{a x \sqrt {b x-a}}\right )}{4 a}+\frac {1}{2 a x^2 \sqrt {b x-a}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 b \left (\frac {3 b \left (-\frac {\int \frac {1}{x \sqrt {b x-a}}dx}{a}-\frac {2}{a \sqrt {b x-a}}\right )}{2 a}+\frac {1}{a x \sqrt {b x-a}}\right )}{4 a}+\frac {1}{2 a x^2 \sqrt {b x-a}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 b \left (\frac {3 b \left (-\frac {2 \int \frac {1}{\frac {a}{b}+\frac {b x-a}{b}}d\sqrt {b x-a}}{a b}-\frac {2}{a \sqrt {b x-a}}\right )}{2 a}+\frac {1}{a x \sqrt {b x-a}}\right )}{4 a}+\frac {1}{2 a x^2 \sqrt {b x-a}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 b \left (\frac {3 b \left (-\frac {2 \arctan \left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {b x-a}}\right )}{2 a}+\frac {1}{a x \sqrt {b x-a}}\right )}{4 a}+\frac {1}{2 a x^2 \sqrt {b x-a}}\) |
1/(2*a*x^2*Sqrt[-a + b*x]) + (5*b*(1/(a*x*Sqrt[-a + b*x]) + (3*b*(-2/(a*Sq rt[-a + b*x]) - (2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a)
3.4.64.3.1 Defintions of rubi rules used
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 = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(-\frac {15 \left (\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2} \sqrt {b x -a}+\sqrt {a}\, b^{2} x^{2}-\frac {a^{\frac {3}{2}} b x}{3}-\frac {2 a^{\frac {5}{2}}}{15}\right )}{4 \sqrt {b x -a}\, a^{\frac {7}{2}} x^{2}}\) | \(71\) |
risch | \(\frac {\left (-b x +a \right ) \left (7 b x +2 a \right )}{4 a^{3} x^{2} \sqrt {b x -a}}-\frac {2 b^{2}}{a^{3} \sqrt {b x -a}}-\frac {15 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\) | \(72\) |
derivativedivides | \(2 b^{2} \left (-\frac {1}{a^{3} \sqrt {b x -a}}-\frac {\frac {\frac {7 \left (b x -a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(77\) |
default | \(2 b^{2} \left (-\frac {1}{a^{3} \sqrt {b x -a}}-\frac {\frac {\frac {7 \left (b x -a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(77\) |
-15/4/(b*x-a)^(1/2)/a^(7/2)*(arctan((b*x-a)^(1/2)/a^(1/2))*b^2*x^2*(b*x-a) ^(1/2)+a^(1/2)*b^2*x^2-1/3*a^(3/2)*b*x-2/15*a^(5/2))/x^2
Time = 0.23 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.08 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt {-a} \log \left (\frac {b x + 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x - a}}{8 \, {\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}, -\frac {15 \, {\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x - a}}{4 \, {\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}\right ] \]
[-1/8*(15*(b^3*x^3 - a*b^2*x^2)*sqrt(-a)*log((b*x + 2*sqrt(b*x - a)*sqrt(- a) - 2*a)/x) + 2*(15*a*b^2*x^2 - 5*a^2*b*x - 2*a^3)*sqrt(b*x - a))/(a^4*b* x^3 - a^5*x^2), -1/4*(15*(b^3*x^3 - a*b^2*x^2)*sqrt(a)*arctan(sqrt(b*x - a )/sqrt(a)) + (15*a*b^2*x^2 - 5*a^2*b*x - 2*a^3)*sqrt(b*x - a))/(a^4*b*x^3 - a^5*x^2)]
Result contains complex when optimal does not.
Time = 5.50 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.38 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=\begin {cases} - \frac {i}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {5 i \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {15 i b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {15 i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {15 b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-I/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) - 5*I*sqrt(b)/(4*a* *2*x**(3/2)*sqrt(a/(b*x) - 1)) + 15*I*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x ) - 1)) - 15*I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2)), Abs(a/( b*x)) > 1), (1/(2*a*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) + 5*sqrt(b)/(4*a* *2*x**(3/2)*sqrt(-a/(b*x) + 1)) - 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(-a/(b*x ) + 1)) + 15*b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2)), True))
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=-\frac {15 \, {\left (b x - a\right )}^{2} b^{2} + 25 \, {\left (b x - a\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{\frac {5}{2}} a^{3} + 2 \, {\left (b x - a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x - a} a^{5}\right )}} - \frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {7}{2}}} \]
-1/4*(15*(b*x - a)^2*b^2 + 25*(b*x - a)*a*b^2 + 8*a^2*b^2)/((b*x - a)^(5/2 )*a^3 + 2*(b*x - a)^(3/2)*a^4 + sqrt(b*x - a)*a^5) - 15/4*b^2*arctan(sqrt( b*x - a)/sqrt(a))/a^(7/2)
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=-\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {7}{2}}} - \frac {2 \, b^{2}}{\sqrt {b x - a} a^{3}} - \frac {7 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} + 9 \, \sqrt {b x - a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]
-15/4*b^2*arctan(sqrt(b*x - a)/sqrt(a))/a^(7/2) - 2*b^2/(sqrt(b*x - a)*a^3 ) - 1/4*(7*(b*x - a)^(3/2)*b^2 + 9*sqrt(b*x - a)*a*b^2)/(a^3*b^2*x^2)
Time = 0.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=-\frac {\frac {2\,b^2}{a}+\frac {15\,b^2\,{\left (a-b\,x\right )}^2}{4\,a^3}-\frac {25\,b^2\,\left (a-b\,x\right )}{4\,a^2}}{2\,a\,{\left (b\,x-a\right )}^{3/2}+{\left (b\,x-a\right )}^{5/2}+a^2\,\sqrt {b\,x-a}}-\frac {15\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]
- ((2*b^2)/a + (15*b^2*(a - b*x)^2)/(4*a^3) - (25*b^2*(a - b*x))/(4*a^2))/ (2*a*(b*x - a)^(3/2) + (b*x - a)^(5/2) + a^2*(b*x - a)^(1/2)) - (15*b^2*at an((b*x - a)^(1/2)/a^(1/2)))/(4*a^(7/2))
Time = 0.00 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 (-a+b x)^{3/2}} \, dx=\frac {-15 \sqrt {a}\, \sqrt {b x -a}\, \mathit {atan} \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) b^{2} x^{2}+2 a^{3}+5 a^{2} b x -15 a \,b^{2} x^{2}}{4 \sqrt {b x -a}\, a^{4} x^{2}} \]