Integrand size = 20, antiderivative size = 76 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+b x^2}}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \]
1/3*(B*x+A)/a/(b*x^2+a)^(3/2)-A*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+1 /3*(2*B*x+3*A)/a^2/(b*x^2+a)^(1/2)
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {b x^2 (3 A+2 B x)+a (4 A+3 B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}+\frac {2 A \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \]
(b*x^2*(3*A + 2*B*x) + a*(4*A + 3*B*x))/(3*a^2*(a + b*x^2)^(3/2)) + (2*A*A rcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]])/a^(5/2)
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {532, 25, 532, 27, 243, 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.
\(\displaystyle \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}-\frac {\int -\frac {3 A+2 B x}{x \left (b x^2+a\right )^{3/2}}dx}{3 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {3 A+2 B x}{x \left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 532 |
\(\displaystyle \frac {\frac {3 A+2 B x}{a \sqrt {a+b x^2}}-\frac {\int -\frac {3 A}{x \sqrt {b x^2+a}}dx}{a}}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 A \int \frac {1}{x \sqrt {b x^2+a}}dx}{a}+\frac {3 A+2 B x}{a \sqrt {a+b x^2}}}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {3 A \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}+\frac {3 A+2 B x}{a \sqrt {a+b x^2}}}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {3 A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a b}+\frac {3 A+2 B x}{a \sqrt {a+b x^2}}}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {3 A+2 B x}{a \sqrt {a+b x^2}}-\frac {3 A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}}{3 a}+\frac {A+B x}{3 a \left (a+b x^2\right )^{3/2}}\) |
(A + B*x)/(3*a*(a + b*x^2)^(3/2)) + ((3*A + 2*B*x)/(a*Sqrt[a + b*x^2]) - ( 3*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2))/(3*a)
3.1.40.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Time = 3.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.29
method | result | size |
default | \(B \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+A \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )\) | \(98\) |
B*(1/3*x/a/(b*x^2+a)^(3/2)+2/3*x/a^2/(b*x^2+a)^(1/2))+A*(1/3/a/(b*x^2+a)^( 3/2)+1/a*(1/a/(b*x^2+a)^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2)) /x)))
Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.14 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (A b^{2} x^{4} + 2 \, A a b x^{2} + A a^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B a b x^{3} + 3 \, A a b x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}}, \frac {3 \, {\left (A b^{2} x^{4} + 2 \, A a b x^{2} + A a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B a b x^{3} + 3 \, A a b x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}}\right ] \]
[1/6*(3*(A*b^2*x^4 + 2*A*a*b*x^2 + A*a^2)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x ^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*a*b*x^3 + 3*A*a*b*x^2 + 3*B*a^2*x + 4 *A*a^2)*sqrt(b*x^2 + a))/(a^3*b^2*x^4 + 2*a^4*b*x^2 + a^5), 1/3*(3*(A*b^2* x^4 + 2*A*a*b*x^2 + A*a^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2* B*a*b*x^3 + 3*A*a*b*x^2 + 3*B*a^2*x + 4*A*a^2)*sqrt(b*x^2 + a))/(a^3*b^2*x ^4 + 2*a^4*b*x^2 + a^5)]
Leaf count of result is larger than twice the leaf count of optimal. 840 vs. \(2 (65) = 130\).
Time = 8.43 (sec) , antiderivative size = 840, normalized size of antiderivative = 11.05 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=A \left (\frac {8 a^{7} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {3 a^{7} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {6 a^{7} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {14 a^{6} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {9 a^{6} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {18 a^{6} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {6 a^{5} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {9 a^{5} b^{2} x^{4} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {18 a^{5} b^{2} x^{4} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} + \frac {3 a^{4} b^{3} x^{6} \log {\left (\frac {b x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}} - \frac {6 a^{4} b^{3} x^{6} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} b x^{2} + 18 a^{\frac {15}{2}} b^{2} x^{4} + 6 a^{\frac {13}{2}} b^{3} x^{6}}\right ) + B \left (\frac {3 a x}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{3}}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]
A*(8*a**7*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(1 5/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**7*log(b*x**2/a)/(6*a**(19/2 ) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**7*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 14*a**6*b*x**2*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6* a**(13/2)*b**3*x**6) + 9*a**6*b*x**2*log(b*x**2/a)/(6*a**(19/2) + 18*a**(1 7/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 18*a**6*b* x**2*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a **(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 6*a**5*b**2*x**4*sqrt(1 + b* x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a* *(13/2)*b**3*x**6) + 9*a**5*b**2*x**4*log(b*x**2/a)/(6*a**(19/2) + 18*a**( 17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 18*a**5*b **2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**4*b**3*x**6*log(b*x **2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a** (13/2)*b**3*x**6) - 6*a**4*b**3*x**6*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19 /2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6 )) + B*(3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x**2/a)) + 2*b*x**3/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**...
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {2 \, B x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {A}{\sqrt {b x^{2} + a} a^{2}} + \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} \]
2/3*B*x/(sqrt(b*x^2 + a)*a^2) + 1/3*B*x/((b*x^2 + a)^(3/2)*a) - A*arcsinh( a/(sqrt(a*b)*abs(x)))/a^(5/2) + A/(sqrt(b*x^2 + a)*a^2) + 1/3*A/((b*x^2 + a)^(3/2)*a)
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (\frac {2 \, B b x}{a^{2}} + \frac {3 \, A b}{a^{2}}\right )} x + \frac {3 \, B}{a}\right )} x + \frac {4 \, A}{a}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {2 \, A \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} \]
1/3*(((2*B*b*x/a^2 + 3*A*b/a^2)*x + 3*B/a)*x + 4*A/a)/(b*x^2 + a)^(3/2) + 2*A*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2)
Time = 6.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x}{x \left (a+b x^2\right )^{5/2}} \, dx=\frac {\frac {A}{3\,a}+\frac {A\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}+\frac {2\,B\,x\,\left (b\,x^2+a\right )+B\,a\,x}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}} \]