Integrand size = 15, antiderivative size = 70 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\frac {5 a \sqrt {x}}{b^3}+\frac {5 x^{3/2}}{3 b^2}+\frac {x^{5/2}}{b (a-b x)}-\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]
5/3*x^(3/2)/b^2+x^(5/2)/b/(-b*x+a)-5*a^(3/2)*arctanh(b^(1/2)*x^(1/2)/a^(1/ 2))/b^(7/2)+5*a*x^(1/2)/b^3
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\frac {\sqrt {x} \left (-15 a^2+10 a b x+2 b^2 x^2\right )}{3 b^3 (-a+b x)}-\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]
(Sqrt[x]*(-15*a^2 + 10*a*b*x + 2*b^2*x^2))/(3*b^3*(-a + b*x)) - (5*a^(3/2) *ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)
Time = 0.18 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {51, 25, 60, 60, 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.
\(\displaystyle \int \frac {x^{5/2}}{(b x-a)^2} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \int -\frac {x^{3/2}}{a-b x}dx}{2 b}+\frac {x^{5/2}}{b (a-b x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {x^{5/2}}{b (a-b x)}-\frac {5 \int \frac {x^{3/2}}{a-b x}dx}{2 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2}}{b (a-b x)}-\frac {5 \left (\frac {a \int \frac {\sqrt {x}}{a-b x}dx}{b}-\frac {2 x^{3/2}}{3 b}\right )}{2 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2}}{b (a-b x)}-\frac {5 \left (\frac {a \left (\frac {a \int \frac {1}{\sqrt {x} (a-b x)}dx}{b}-\frac {2 \sqrt {x}}{b}\right )}{b}-\frac {2 x^{3/2}}{3 b}\right )}{2 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{5/2}}{b (a-b x)}-\frac {5 \left (\frac {a \left (\frac {2 a \int \frac {1}{a-b x}d\sqrt {x}}{b}-\frac {2 \sqrt {x}}{b}\right )}{b}-\frac {2 x^{3/2}}{3 b}\right )}{2 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {x^{5/2}}{b (a-b x)}-\frac {5 \left (\frac {a \left (\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b}\right )}{b}-\frac {2 x^{3/2}}{3 b}\right )}{2 b}\) |
x^(5/2)/(b*(a - b*x)) - (5*((-2*x^(3/2))/(3*b) + (a*((-2*Sqrt[x])/b + (2*S qrt[a]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/(2*b)
3.5.76.3.1 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/(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)*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 = 0.81
method | result | size |
risch | \(\frac {2 \left (b x +6 a \right ) \sqrt {x}}{3 b^{3}}+\frac {a^{2} \left (-\frac {\sqrt {x}}{b x -a}-\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b^{3}}\) | \(57\) |
derivativedivides | \(\frac {\frac {2 b \,x^{\frac {3}{2}}}{3}+4 a \sqrt {x}}{b^{3}}-\frac {2 a^{2} \left (-\frac {\sqrt {x}}{2 \left (-b x +a \right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(60\) |
default | \(\frac {\frac {2 b \,x^{\frac {3}{2}}}{3}+4 a \sqrt {x}}{b^{3}}-\frac {2 a^{2} \left (-\frac {\sqrt {x}}{2 \left (-b x +a \right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(60\) |
2/3*(b*x+6*a)*x^(1/2)/b^3+a^2/b^3*(-x^(1/2)/(b*x-a)-5/(a*b)^(1/2)*arctanh( b*x^(1/2)/(a*b)^(1/2)))
Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.39 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\left [\frac {15 \, {\left (a b x - a^{2}\right )} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (b^{4} x - a b^{3}\right )}}, \frac {15 \, {\left (a b x - a^{2}\right )} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (2 \, b^{2} x^{2} + 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (b^{4} x - a b^{3}\right )}}\right ] \]
[1/6*(15*(a*b*x - a^2)*sqrt(a/b)*log((b*x - 2*b*sqrt(x)*sqrt(a/b) + a)/(b* x - a)) + 2*(2*b^2*x^2 + 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x - a*b^3), 1/3* (15*(a*b*x - a^2)*sqrt(-a/b)*arctan(b*sqrt(x)*sqrt(-a/b)/a) + (2*b^2*x^2 + 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x - a*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (63) = 126\).
Time = 12.46 (sec) , antiderivative size = 354, normalized size of antiderivative = 5.06 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{2}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b^{2}} & \text {for}\: a = 0 \\- \frac {15 a^{3} \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} + \frac {15 a^{3} \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} - \frac {30 a^{2} b \sqrt {x} \sqrt {\frac {a}{b}}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} + \frac {15 a^{2} b x \log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} - \frac {15 a^{2} b x \log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} + \frac {20 a b^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b}}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} + \frac {4 b^{3} x^{\frac {5}{2}} \sqrt {\frac {a}{b}}}{- 6 a b^{4} \sqrt {\frac {a}{b}} + 6 b^{5} x \sqrt {\frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**2), Eq(b, 0)), (2*x**(3/2)/(3*b**2), Eq(a, 0)), (-15*a**3*log(sqrt(x) - sqrt(a/b))/ (-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) + 15*a**3*log(sqrt(x) + sqrt(a/ b))/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) - 30*a**2*b*sqrt(x)*sqrt(a/ b)/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) + 15*a**2*b*x*log(sqrt(x) - sqrt(a/b))/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) - 15*a**2*b*x*log(sq rt(x) + sqrt(a/b))/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) + 20*a*b**2* x**(3/2)*sqrt(a/b)/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)) + 4*b**3*x** (5/2)*sqrt(a/b)/(-6*a*b**4*sqrt(a/b) + 6*b**5*x*sqrt(a/b)), True))
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.16 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=-\frac {a^{2} \sqrt {x}}{b^{4} x - a b^{3}} + \frac {5 \, a^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + 6 \, a \sqrt {x}\right )}}{3 \, b^{3}} \]
-a^2*sqrt(x)/(b^4*x - a*b^3) + 5/2*a^2*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt (x) + sqrt(a*b)))/(sqrt(a*b)*b^3) + 2/3*(b*x^(3/2) + 6*a*sqrt(x))/b^3
Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b^{3}} - \frac {a^{2} \sqrt {x}}{{\left (b x - a\right )} b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {3}{2}} + 6 \, a b^{3} \sqrt {x}\right )}}{3 \, b^{6}} \]
5*a^2*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*b^3) - a^2*sqrt(x)/((b*x - a)*b^3) + 2/3*(b^4*x^(3/2) + 6*a*b^3*sqrt(x))/b^6
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int \frac {x^{5/2}}{(-a+b x)^2} \, dx=\frac {2\,x^{3/2}}{3\,b^2}+\frac {4\,a\,\sqrt {x}}{b^3}+\frac {a^2\,\sqrt {x}}{a\,b^3-b^4\,x}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{b^{7/2}} \]