Integrand size = 13, antiderivative size = 82 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \]
-1/2*x^(5/2)/b/(b*x+a)^2-5/4*x^(3/2)/b^2/(b*x+a)-15/4*arctan(b^(1/2)*x^(1/ 2)/a^(1/2))*a^(1/2)/b^(7/2)+15/4*x^(1/2)/b^3
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\frac {\sqrt {x} \left (15 a^2+25 a b x+8 b^2 x^2\right )}{4 b^3 (a+b x)^2}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \]
(Sqrt[x]*(15*a^2 + 25*a*b*x + 8*b^2*x^2))/(4*b^3*(a + b*x)^2) - (15*Sqrt[a ]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(7/2))
Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, 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.
\(\displaystyle \int \frac {x^{5/2}}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{a+b x}dx}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\) |
-1/2*x^(5/2)/(b*(a + b*x)^2) + (5*(-(x^(3/2)/(b*(a + b*x))) + (3*((2*Sqrt[ x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/(2*b)))/(4 *b)
3.5.63.3.1 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)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\) | \(56\) |
default | \(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\) | \(56\) |
risch | \(\frac {2 \sqrt {x}}{b^{3}}-\frac {a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{4}-\frac {7 a \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{3}}\) | \(57\) |
2*x^(1/2)/b^3-2/b^3*a*((-9/8*b*x^(3/2)-7/8*a*x^(1/2))/(b*x+a)^2+15/8/(a*b) ^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.23 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.44 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]
[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt( -a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^ 2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*arc tan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5 *x^2 + 2*a*b^4*x + a^2*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (73) = 146\).
Time = 28.38 (sec) , antiderivative size = 683, normalized size of antiderivative = 8.33 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\- \frac {15 a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {15 a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {30 a^{2} b \sqrt {x} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} - \frac {30 a^{2} b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {30 a^{2} b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {50 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} - \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {16 b^{3} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**3), Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (-15*a**3*log(sqrt(x) - sqrt(-a/b))/(8*a* *2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 15 *a**3*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt (-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 30*a**2*b*sqrt(x)*sqrt(-a/b)/(8*a**2*b* *4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 30*a**2 *b*x*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt( -a/b) + 8*b**6*x**2*sqrt(-a/b)) + 30*a**2*b*x*log(sqrt(x) + sqrt(-a/b))/(8 *a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 50*a*b**2*x**(3/2)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt( -a/b) + 8*b**6*x**2*sqrt(-a/b)) - 15*a*b**2*x**2*log(sqrt(x) - sqrt(-a/b)) /(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b) ) + 15*a*b**2*x**2*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16* a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 16*b**3*x**(5/2)*sqrt(-a/b )/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b )), True))
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} \]
1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 15/4 *a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=-\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \]
-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4* (9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*x + a)^2*b^3)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2}}{(a+b x)^3} \, dx=\frac {\frac {7\,a^2\,\sqrt {x}}{4}+\frac {9\,a\,b\,x^{3/2}}{4}}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,\sqrt {x}}{b^3}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{7/2}} \]