Integrand size = 13, antiderivative size = 69 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=-\frac {5}{3 a^2 x^{3/2}}+\frac {5 b}{a^3 \sqrt {x}}+\frac {1}{a x^{3/2} (a+b x)}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]
-5/3/a^2/x^(3/2)+1/a/x^(3/2)/(b*x+a)+5*b^(3/2)*arctan(b^(1/2)*x^(1/2)/a^(1 /2))/a^(7/2)+5*b/a^3/x^(1/2)
Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\frac {-2 a^2+10 a b x+15 b^2 x^2}{3 a^3 x^{3/2} (a+b x)}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]
(-2*a^2 + 10*a*b*x + 15*b^2*x^2)/(3*a^3*x^(3/2)*(a + b*x)) + (5*b^(3/2)*Ar cTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2)
Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {52, 61, 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.
\(\displaystyle \int \frac {1}{x^{5/2} (a+b x)^2} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5 \int \frac {1}{x^{5/2} (a+b x)}dx}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5 \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\) |
1/(a*x^(3/2)*(a + b*x)) + (5*(-2/(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*S qrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/(2*a)
3.5.61.3.1 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 = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {2 \left (-6 b x +a \right )}{3 a^{3} x^{\frac {3}{2}}}+\frac {b^{2} \left (\frac {\sqrt {x}}{b x +a}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{3}}\) | \(53\) |
derivativedivides | \(\frac {2 b^{2} \left (\frac {\sqrt {x}}{2 b x +2 a}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}}+\frac {4 b}{a^{3} \sqrt {x}}\) | \(58\) |
default | \(\frac {2 b^{2} \left (\frac {\sqrt {x}}{2 b x +2 a}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {2}{3 a^{2} x^{\frac {3}{2}}}+\frac {4 b}{a^{3} \sqrt {x}}\) | \(58\) |
-2/3*(-6*b*x+a)/a^3/x^(3/2)+1/a^3*b^2*(x^(1/2)/(b*x+a)+5/(a*b)^(1/2)*arcta n(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.67 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\left [\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]
[1/6*(15*(b^2*x^3 + a*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2), -1/3*(15*(b^2*x^3 + a*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqr t(x))) - (15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]
Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (65) = 130\).
Time = 20.42 (sec) , antiderivative size = 452, normalized size of antiderivative = 6.55 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 a^{2} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {2}{7 b^{2} x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {4 a^{2} \sqrt {- \frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {15 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {15 a b x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {20 a b x \sqrt {- \frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} - \frac {15 b^{2} x^{\frac {5}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} + \frac {30 b^{2} x^{2} \sqrt {- \frac {a}{b}}}{6 a^{4} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}} + 6 a^{3} b x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(3*a**2*x**(3/2)), Eq(b , 0)), (-2/(7*b**2*x**(7/2)), Eq(a, 0)), (-4*a**2*sqrt(-a/b)/(6*a**4*x**(3 /2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 15*a*b*x**(3/2)*log(sqrt( x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b )) - 15*a*b*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 20*a*b*x*sqrt(-a/b)/(6*a**4*x**(3/2)*sq rt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 15*b**2*x**(5/2)*log(sqrt(x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) - 15*b**2*x**(5/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6 *a**3*b*x**(5/2)*sqrt(-a/b)) + 30*b**2*x**2*sqrt(-a/b)/(6*a**4*x**(3/2)*sq rt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)), True))
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\frac {15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \]
1/3*(15*b^2*x^2 + 10*a*b*x - 2*a^2)/(a^3*b*x^(5/2) + a^4*x^(3/2)) + 5*b^2* arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} + \frac {2 \, {\left (6 \, b x - a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \]
5*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + b^2*sqrt(x)/((b*x + a) *a^3) + 2/3*(6*b*x - a)/(a^3*x^(3/2))
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\frac {\frac {5\,b^2\,x^2}{a^3}-\frac {2}{3\,a}+\frac {10\,b\,x}{3\,a^2}}{a\,x^{3/2}+b\,x^{5/2}}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]
((5*b^2*x^2)/a^3 - 2/(3*a) + (10*b*x)/(3*a^2))/(a*x^(3/2) + b*x^(5/2)) + ( 5*b^(3/2)*atan((b^(1/2)*x^(1/2))/a^(1/2)))/a^(7/2)
Time = 0.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^{5/2} (a+b x)^2} \, dx=\frac {15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a b x +15 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}-2 a^{3}+10 a^{2} b x +15 a \,b^{2} x^{2}}{3 \sqrt {x}\, a^{4} x \left (b x +a \right )} \]