Integrand size = 13, antiderivative size = 97 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=2 a^4 \sqrt {a+b x}+\frac {2}{3} a^3 (a+b x)^{3/2}+\frac {2}{5} a^2 (a+b x)^{5/2}+\frac {2}{7} a (a+b x)^{7/2}+\frac {2}{9} (a+b x)^{9/2}-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
2/3*a^3*(b*x+a)^(3/2)+2/5*a^2*(b*x+a)^(5/2)+2/7*a*(b*x+a)^(7/2)+2/9*(b*x+a )^(9/2)-2*a^(9/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2*a^4*(b*x+a)^(1/2)
Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2}{315} \sqrt {a+b x} \left (563 a^4+506 a^3 b x+408 a^2 b^2 x^2+185 a b^3 x^3+35 b^4 x^4\right )-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
(2*Sqrt[a + b*x]*(563*a^4 + 506*a^3*b*x + 408*a^2*b^2*x^2 + 185*a*b^3*x^3 + 35*b^4*x^4))/315 - 2*a^(9/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {60, 60, 60, 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.
\(\displaystyle \int \frac {(a+b x)^{9/2}}{x} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle a \int \frac {(a+b x)^{7/2}}{x}dx+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle a \left (a \int \frac {(a+b x)^{5/2}}{x}dx+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle a \left (a \left (a \int \frac {(a+b x)^{3/2}}{x}dx+\frac {2}{5} (a+b x)^{5/2}\right )+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle a \left (a \left (a \left (a \int \frac {\sqrt {a+b x}}{x}dx+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle a \left (a \left (a \left (a \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle a \left (a \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle a \left (a \left (a \left (a \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a+b x)^{3/2}\right )+\frac {2}{5} (a+b x)^{5/2}\right )+\frac {2}{7} (a+b x)^{7/2}\right )+\frac {2}{9} (a+b x)^{9/2}\) |
(2*(a + b*x)^(9/2))/9 + a*((2*(a + b*x)^(7/2))/7 + a*((2*(a + b*x)^(5/2))/ 5 + a*((2*(a + b*x)^(3/2))/3 + a*(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt [a + b*x]/Sqrt[a]]))))
3.4.17.3.1 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 + 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 = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (35 b^{4} x^{4}+185 a \,b^{3} x^{3}+408 a^{2} b^{2} x^{2}+506 a^{3} b x +563 a^{4}\right )}{315}\) | \(69\) |
derivativedivides | \(\frac {2 a^{3} \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 a^{2} \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x +a \right )^{\frac {9}{2}}}{9}-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{4} \sqrt {b x +a}\) | \(74\) |
default | \(\frac {2 a^{3} \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 a^{2} \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x +a \right )^{\frac {9}{2}}}{9}-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{4} \sqrt {b x +a}\) | \(74\) |
-2*a^(9/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2/315*(b*x+a)^(1/2)*(35*b^4*x^4+ 185*a*b^3*x^3+408*a^2*b^2*x^2+506*a^3*b*x+563*a^4)
Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\left [a^{\frac {9}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{315} \, {\left (35 \, b^{4} x^{4} + 185 \, a b^{3} x^{3} + 408 \, a^{2} b^{2} x^{2} + 506 \, a^{3} b x + 563 \, a^{4}\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \frac {2}{315} \, {\left (35 \, b^{4} x^{4} + 185 \, a b^{3} x^{3} + 408 \, a^{2} b^{2} x^{2} + 506 \, a^{3} b x + 563 \, a^{4}\right )} \sqrt {b x + a}\right ] \]
[a^(9/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/315*(35*b^4*x^4 + 185*a*b^3*x^3 + 408*a^2*b^2*x^2 + 506*a^3*b*x + 563*a^4)*sqrt(b*x + a), 2*sqrt(-a)*a^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) + 2/315*(35*b^4*x^4 + 185* a*b^3*x^3 + 408*a^2*b^2*x^2 + 506*a^3*b*x + 563*a^4)*sqrt(b*x + a)]
Time = 15.93 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {1126 a^{\frac {9}{2}} \sqrt {1 + \frac {b x}{a}}}{315} + a^{\frac {9}{2}} \log {\left (\frac {b x}{a} \right )} - 2 a^{\frac {9}{2}} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {1012 a^{\frac {7}{2}} b x \sqrt {1 + \frac {b x}{a}}}{315} + \frac {272 a^{\frac {5}{2}} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{105} + \frac {74 a^{\frac {3}{2}} b^{3} x^{3} \sqrt {1 + \frac {b x}{a}}}{63} + \frac {2 \sqrt {a} b^{4} x^{4} \sqrt {1 + \frac {b x}{a}}}{9} \]
1126*a**(9/2)*sqrt(1 + b*x/a)/315 + a**(9/2)*log(b*x/a) - 2*a**(9/2)*log(s qrt(1 + b*x/a) + 1) + 1012*a**(7/2)*b*x*sqrt(1 + b*x/a)/315 + 272*a**(5/2) *b**2*x**2*sqrt(1 + b*x/a)/105 + 74*a**(3/2)*b**3*x**3*sqrt(1 + b*x/a)/63 + 2*sqrt(a)*b**4*x**4*sqrt(1 + b*x/a)/9
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=a^{\frac {9}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{9} \, {\left (b x + a\right )}^{\frac {9}{2}} + \frac {2}{7} \, {\left (b x + a\right )}^{\frac {7}{2}} a + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 2 \, \sqrt {b x + a} a^{4} \]
a^(9/2)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/9*(b* x + a)^(9/2) + 2/7*(b*x + a)^(7/2)*a + 2/5*(b*x + a)^(5/2)*a^2 + 2/3*(b*x + a)^(3/2)*a^3 + 2*sqrt(b*x + a)*a^4
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2 \, a^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2}{9} \, {\left (b x + a\right )}^{\frac {9}{2}} + \frac {2}{7} \, {\left (b x + a\right )}^{\frac {7}{2}} a + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 2 \, \sqrt {b x + a} a^{4} \]
2*a^5*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/9*(b*x + a)^(9/2) + 2/7* (b*x + a)^(7/2)*a + 2/5*(b*x + a)^(5/2)*a^2 + 2/3*(b*x + a)^(3/2)*a^3 + 2* sqrt(b*x + a)*a^4
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2\,a\,{\left (a+b\,x\right )}^{7/2}}{7}+\frac {2\,{\left (a+b\,x\right )}^{9/2}}{9}+2\,a^4\,\sqrt {a+b\,x}+\frac {2\,a^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {2\,a^2\,{\left (a+b\,x\right )}^{5/2}}{5}+a^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]
(2*a*(a + b*x)^(7/2))/7 + (2*(a + b*x)^(9/2))/9 + 2*a^4*(a + b*x)^(1/2) + (2*a^3*(a + b*x)^(3/2))/3 + (2*a^2*(a + b*x)^(5/2))/5 + a^(9/2)*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*2i
Time = 0.00 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {1126 \sqrt {b x +a}\, a^{4}}{315}+\frac {1012 \sqrt {b x +a}\, a^{3} b x}{315}+\frac {272 \sqrt {b x +a}\, a^{2} b^{2} x^{2}}{105}+\frac {74 \sqrt {b x +a}\, a \,b^{3} x^{3}}{63}+\frac {2 \sqrt {b x +a}\, b^{4} x^{4}}{9}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{4}-\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{4} \]