Integrand size = 15, antiderivative size = 84 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\frac {15}{4 a^3 \sqrt {x}}-\frac {1}{2 a \sqrt {x} (a-b x)^2}-\frac {5}{4 a^2 \sqrt {x} (a-b x)}-\frac {15 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
-15/4*arctanh(b^(1/2)*x^(1/2)/a^(1/2))*b^(1/2)/a^(7/2)+15/4/a^3/x^(1/2)-1/ 2/a/(-b*x+a)^2/x^(1/2)-5/4/a^2/(-b*x+a)/x^(1/2)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\frac {8 a^2-25 a b x+15 b^2 x^2}{4 a^3 \sqrt {x} (a-b x)^2}-\frac {15 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
(8*a^2 - 25*a*b*x + 15*b^2*x^2)/(4*a^3*Sqrt[x]*(a - b*x)^2) - (15*Sqrt[b]* ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(7/2))
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11, 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, 221}
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.
\(\displaystyle \int \frac {1}{x^{3/2} (b x-a)^3} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 \int \frac {1}{x^{3/2} (a-b x)^2}dx}{4 a}-\frac {1}{2 a \sqrt {x} (a-b x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 \left (\frac {3 \int \frac {1}{x^{3/2} (a-b x)}dx}{2 a}+\frac {1}{a \sqrt {x} (a-b x)}\right )}{4 a}-\frac {1}{2 a \sqrt {x} (a-b x)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 \left (\frac {3 \left (\frac {b \int \frac {1}{\sqrt {x} (a-b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a-b x)}\right )}{4 a}-\frac {1}{2 a \sqrt {x} (a-b x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 \left (\frac {3 \left (\frac {2 b \int \frac {1}{a-b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a-b x)}\right )}{4 a}-\frac {1}{2 a \sqrt {x} (a-b x)^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a-b x)}\right )}{4 a}-\frac {1}{2 a \sqrt {x} (a-b x)^2}\) |
-1/2*1/(a*Sqrt[x]*(a - b*x)^2) - (5*(1/(a*Sqrt[x]*(a - b*x)) + (3*(-2/(a*S qrt[x]) + (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/(2*a))) /(4*a)
3.5.87.3.1 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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(-\frac {2 b \left (\frac {-\frac {7 b \,x^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}+\frac {2}{a^{3} \sqrt {x}}\) | \(57\) |
default | \(-\frac {2 b \left (\frac {-\frac {7 b \,x^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {x}}{8}}{\left (-b x +a \right )^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}+\frac {2}{a^{3} \sqrt {x}}\) | \(57\) |
risch | \(\frac {2}{a^{3} \sqrt {x}}+\frac {b \left (\frac {\frac {7 b \,x^{\frac {3}{2}}}{4}-\frac {9 a \sqrt {x}}{4}}{\left (b x -a \right )^{2}}-\frac {15 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{3}}\) | \(58\) |
-2/a^3*b*((-7/8*b*x^(3/2)+9/8*a*x^(1/2))/(-b*x+a)^2+15/8/(a*b)^(1/2)*arcta nh(b*x^(1/2)/(a*b)^(1/2)))+2/a^3/x^(1/2)
Time = 0.24 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.54 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]
[1/8*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sqrt(b/a)*log((b*x - 2*a*sqrt(x)*sq rt(b/a) + a)/(b*x - a)) + 2*(15*b^2*x^2 - 25*a*b*x + 8*a^2)*sqrt(x))/(a^3* b^2*x^3 - 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sqrt (-b/a)*arctan(a*sqrt(-b/a)/(b*sqrt(x))) + (15*b^2*x^2 - 25*a*b*x + 8*a^2)* sqrt(x))/(a^3*b^2*x^3 - 2*a^4*b*x^2 + a^5*x)]
Leaf count of result is larger than twice the leaf count of optimal. 716 vs. \(2 (75) = 150\).
Time = 24.37 (sec) , antiderivative size = 716, normalized size of antiderivative = 8.52 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2}{a^{3} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{7 b^{3} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\\frac {15 a^{2} \sqrt {x} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {15 a^{2} \sqrt {x} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {16 a^{2} \sqrt {\frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {30 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {30 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {50 a b x \sqrt {\frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} + \frac {30 b^{2} x^{2} \sqrt {\frac {a}{b}}}{8 a^{5} \sqrt {x} \sqrt {\frac {a}{b}} - 16 a^{4} b x^{\frac {3}{2}} \sqrt {\frac {a}{b}} + 8 a^{3} b^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (2/(a**3*sqrt(x)), Eq(b, 0) ), (-2/(7*b**3*x**(7/2)), Eq(a, 0)), (15*a**2*sqrt(x)*log(sqrt(x) - sqrt(a /b))/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8*a**3*b** 2*x**(5/2)*sqrt(a/b)) - 15*a**2*sqrt(x)*log(sqrt(x) + sqrt(a/b))/(8*a**5*s qrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8*a**3*b**2*x**(5/2)*sqr t(a/b)) + 16*a**2*sqrt(a/b)/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2) *sqrt(a/b) + 8*a**3*b**2*x**(5/2)*sqrt(a/b)) - 30*a*b*x**(3/2)*log(sqrt(x) - sqrt(a/b))/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8 *a**3*b**2*x**(5/2)*sqrt(a/b)) + 30*a*b*x**(3/2)*log(sqrt(x) + sqrt(a/b))/ (8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8*a**3*b**2*x** (5/2)*sqrt(a/b)) - 50*a*b*x*sqrt(a/b)/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4* b*x**(3/2)*sqrt(a/b) + 8*a**3*b**2*x**(5/2)*sqrt(a/b)) + 15*b**2*x**(5/2)* log(sqrt(x) - sqrt(a/b))/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sq rt(a/b) + 8*a**3*b**2*x**(5/2)*sqrt(a/b)) - 15*b**2*x**(5/2)*log(sqrt(x) + sqrt(a/b))/(8*a**5*sqrt(x)*sqrt(a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8*a **3*b**2*x**(5/2)*sqrt(a/b)) + 30*b**2*x**2*sqrt(a/b)/(8*a**5*sqrt(x)*sqrt (a/b) - 16*a**4*b*x**(3/2)*sqrt(a/b) + 8*a**3*b**2*x**(5/2)*sqrt(a/b)), Tr ue))
Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\frac {15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} - 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} + \frac {15 \, b \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]
1/4*(15*b^2*x^2 - 25*a*b*x + 8*a^2)/(a^3*b^2*x^(5/2) - 2*a^4*b*x^(3/2) + a ^5*sqrt(x)) + 15/8*b*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/ (sqrt(a*b)*a^3)
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} a^{3}} + \frac {2}{a^{3} \sqrt {x}} + \frac {7 \, b^{2} x^{\frac {3}{2}} - 9 \, a b \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} a^{3}} \]
15/4*b*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a^3) + 2/(a^3*sqrt(x)) + 1 /4*(7*b^2*x^(3/2) - 9*a*b*sqrt(x))/((b*x - a)^2*a^3)
Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^{3/2} (-a+b x)^3} \, dx=\frac {\frac {2}{a}+\frac {15\,b^2\,x^2}{4\,a^3}-\frac {25\,b\,x}{4\,a^2}}{a^2\,\sqrt {x}+b^2\,x^{5/2}-2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]