Integrand size = 13, antiderivative size = 74 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=-\frac {5 b}{3 a^2 (a+b x)^{3/2}}-\frac {1}{a x (a+b x)^{3/2}}-\frac {5 b}{a^3 \sqrt {a+b x}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]
-5/3*b/a^2/(b*x+a)^(3/2)-1/a/x/(b*x+a)^(3/2)+5*b*arctanh((b*x+a)^(1/2)/a^( 1/2))/a^(7/2)-5*b/a^3/(b*x+a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=\frac {-3 a^2-20 a b x-15 b^2 x^2}{3 a^3 x (a+b x)^{3/2}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]
(-3*a^2 - 20*a*b*x - 15*b^2*x^2)/(3*a^3*x*(a + b*x)^(3/2)) + (5*b*ArcTanh[ Sqrt[a + b*x]/Sqrt[a]])/a^(7/2)
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15, 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, 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.
\(\displaystyle \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 b \int \frac {1}{x (a+b x)^{5/2}}dx}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 b \left (\frac {\int \frac {1}{x (a+b x)^{3/2}}dx}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 b \left (\frac {\frac {\int \frac {1}{x \sqrt {a+b x}}dx}{a}+\frac {2}{a \sqrt {a+b x}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 b \left (\frac {\frac {2 \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a b}+\frac {2}{a \sqrt {a+b x}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {5 b \left (\frac {\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\) |
-(1/(a*x*(a + b*x)^(3/2))) - (5*b*(2/(3*a*(a + b*x)^(3/2)) + (2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2))/a))/(2*a)
3.4.57.3.1 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 = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {\sqrt {b x +a}}{a^{3} x}-\frac {b \left (-\frac {10 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {8}{\sqrt {b x +a}}+\frac {4 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right )}{2 a^{3}}\) | \(60\) |
pseudoelliptic | \(\frac {5 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x -5 \sqrt {a}\, b^{2} x^{2}-\frac {20 a^{\frac {3}{2}} b x}{3}-a^{\frac {5}{2}}}{x \,a^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}}}\) | \(62\) |
derivativedivides | \(2 b \left (-\frac {1}{3 a^{2} \left (b x +a \right )^{\frac {3}{2}}}-\frac {2}{a^{3} \sqrt {b x +a}}+\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{3}}\right )\) | \(66\) |
default | \(2 b \left (-\frac {1}{3 a^{2} \left (b x +a \right )^{\frac {3}{2}}}-\frac {2}{a^{3} \sqrt {b x +a}}+\frac {-\frac {\sqrt {b x +a}}{2 b x}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{3}}\right )\) | \(66\) |
-1/a^3*(b*x+a)^(1/2)/x-1/2/a^3*b*(-10*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/ 2)+8/(b*x+a)^(1/2)+4/3*a/(b*x+a)^(3/2))
Time = 0.23 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.99 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{6 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac {15 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \]
[1/6*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a ))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), -1/3*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 + 20*a^ 2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]
Leaf count of result is larger than twice the leaf count of optimal. 818 vs. \(2 (68) = 136\).
Time = 4.23 (sec) , antiderivative size = 818, normalized size of antiderivative = 11.05 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=- \frac {6 a^{17} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {46 a^{16} b x \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {15 a^{16} b x \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {30 a^{16} b x \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {70 a^{15} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {45 a^{15} b^{2} x^{2} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {90 a^{15} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {30 a^{14} b^{3} x^{3} \sqrt {1 + \frac {b x}{a}}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {45 a^{14} b^{3} x^{3} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {90 a^{14} b^{3} x^{3} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} - \frac {15 a^{13} b^{4} x^{4} \log {\left (\frac {b x}{a} \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} + \frac {30 a^{13} b^{4} x^{4} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{6 a^{\frac {39}{2}} x + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x^{3} + 6 a^{\frac {33}{2}} b^{3} x^{4}} \]
-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/ 2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a* *(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b** 3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1 8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b**2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2 )*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4 ) - 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1 8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**15*b**2*x**2*log(sq rt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b** 2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b*x/a)/(6*a* *(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b** 3*x**4) - 45*a**14*b**3*x**3*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x* *2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**14*b**3*x**3* log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/ 2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)/(6*a **(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b* *3*x**4) + 30*a**13*b**4*x**4*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18 *a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=-\frac {15 \, {\left (b x + a\right )}^{2} b - 10 \, {\left (b x + a\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} a^{3} - {\left (b x + a\right )}^{\frac {3}{2}} a^{4}\right )}} - \frac {5 \, b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, a^{\frac {7}{2}}} \]
-1/3*(15*(b*x + a)^2*b - 10*(b*x + a)*a*b - 2*a^2*b)/((b*x + a)^(5/2)*a^3 - (b*x + a)^(3/2)*a^4) - 5/2*b*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a ) + sqrt(a)))/a^(7/2)
Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=-\frac {5 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} b + a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x + a}}{a^{3} x} \]
-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(6*(b*x + a)*b + a*b)/((b*x + a)^(3/2)*a^3) - sqrt(b*x + a)/(a^3*x)
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx=\frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {\frac {2\,b}{3\,a}+\frac {10\,b\,\left (a+b\,x\right )}{3\,a^2}-\frac {5\,b\,{\left (a+b\,x\right )}^2}{a^3}}{a\,{\left (a+b\,x\right )}^{3/2}-{\left (a+b\,x\right )}^{5/2}} \]