Integrand size = 22, antiderivative size = 110 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {A b-a B}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac {2 A b-a B}{a^3 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^3 x^2}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \] Output:
-1/3*(A*b-B*a)/a^2/(b*x^2+a)^(3/2)-(2*A*b-B*a)/a^3/(b*x^2+a)^(1/2)-1/2*A*( b*x^2+a)^(1/2)/a^3/x^2+1/2*(5*A*b-2*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/ a^(7/2)
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 a^2 A-20 a A b x^2+8 a^2 B x^2-15 A b^2 x^4+6 a b B x^4}{6 a^3 x^2 \left (a+b x^2\right )^{3/2}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \] Input:
Integrate[(A + B*x^2)/(x^3*(a + b*x^2)^(5/2)),x]
Output:
(-3*a^2*A - 20*a*A*b*x^2 + 8*a^2*B*x^2 - 15*A*b^2*x^4 + 6*a*b*B*x^4)/(6*a^ 3*x^2*(a + b*x^2)^(3/2)) + ((5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a ]])/(2*a^(7/2))
Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {354, 87, 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.
\(\displaystyle \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^4 \left (b x^2+a\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-2 a B) \int \frac {1}{x^2 \left (b x^2+a\right )^{5/2}}dx^2}{2 a}-\frac {A}{a x^2 \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-2 a B) \left (\frac {\int \frac {1}{x^2 \left (b x^2+a\right )^{3/2}}dx^2}{a}+\frac {2}{3 a \left (a+b x^2\right )^{3/2}}\right )}{2 a}-\frac {A}{a x^2 \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-2 a B) \left (\frac {\frac {\int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{a}+\frac {2}{a \sqrt {a+b x^2}}}{a}+\frac {2}{3 a \left (a+b x^2\right )^{3/2}}\right )}{2 a}-\frac {A}{a x^2 \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-2 a B) \left (\frac {\frac {2 \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a b}+\frac {2}{a \sqrt {a+b x^2}}}{a}+\frac {2}{3 a \left (a+b x^2\right )^{3/2}}\right )}{2 a}-\frac {A}{a x^2 \left (a+b x^2\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-2 a B) \left (\frac {\frac {2}{a \sqrt {a+b x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}}{a}+\frac {2}{3 a \left (a+b x^2\right )^{3/2}}\right )}{2 a}-\frac {A}{a x^2 \left (a+b x^2\right )^{3/2}}\right )\) |
Input:
Int[(A + B*x^2)/(x^3*(a + b*x^2)^(5/2)),x]
Output:
(-(A/(a*x^2*(a + b*x^2)^(3/2))) - ((5*A*b - 2*a*B)*(2/(3*a*(a + b*x^2)^(3/ 2)) + (2/(a*Sqrt[a + b*x^2]) - (2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2 ))/a))/(2*a))/2
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.52 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(-\frac {-5 \left (A b -\frac {2 B a}{5}\right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}} x^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\frac {20 \left (-\frac {3 x^{2} B}{10}+A \right ) b \,x^{2} a^{\frac {3}{2}}}{3}+5 A \sqrt {a}\, b^{2} x^{4}+\left (-\frac {8 x^{2} B}{3}+A \right ) a^{\frac {5}{2}}}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{2}}\) | \(96\) |
default | \(A \left (-\frac {1}{2 a \,x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 b \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )}{2 a}\right )+B \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )\) | \(152\) |
risch | \(-\frac {A \sqrt {b \,x^{2}+a}}{2 a^{3} x^{2}}+\frac {5 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A b}{2 a^{\frac {7}{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{a^{\frac {5}{2}}}-\frac {13 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {7 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {13 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {7 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} b \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} b \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}\) | \(595\) |
Input:
int((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/2*(-5*(A*b-2/5*B*a)*(b*x^2+a)^(3/2)*x^2*arctanh((b*x^2+a)^(1/2)/a^(1/2) )+20/3*(-3/10*x^2*B+A)*b*x^2*a^(3/2)+5*A*a^(1/2)*b^2*x^4+(-8/3*x^2*B+A)*a^ (5/2))/(b*x^2+a)^(3/2)/a^(7/2)/x^2
Time = 0.10 (sec) , antiderivative size = 352, normalized size of antiderivative = 3.20 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \] Input:
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[-1/12*(3*((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^4 + (2* B*a^3 - 5*A*a^2*b)*x^2)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*(2*B*a^2*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*A* a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), 1/6*(3 *((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^4 + (2*B*a^3 - 5 *A*a^2*b)*x^2)*sqrt(-a)*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (3*(2*B*a^2*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a) )/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2)]
Leaf count of result is larger than twice the leaf count of optimal. 1608 vs. \(2 (97) = 194\).
Time = 22.28 (sec) , antiderivative size = 1608, normalized size of antiderivative = 14.62 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((B*x**2+A)/x**3/(b*x**2+a)**(5/2),x)
Output:
A*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x **6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2 )*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3* x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 3 6*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70 *a**15*b**2*x**4*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x* *4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b**2*x**4 *log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b** 2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/ a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**( 39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b **3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(3 7/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14* b**3*x**6*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b* x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**13*b**4*x* *8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b **2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*...
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {5 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a^{2}} + \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {5 \, A b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, A b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {A}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} \] Input:
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-B*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 5/2*A*b*arcsinh(a/(sqrt(a*b)*ab s(x)))/a^(7/2) + B/(sqrt(b*x^2 + a)*a^2) + 1/3*B/((b*x^2 + a)^(3/2)*a) - 5 /2*A*b/(sqrt(b*x^2 + a)*a^3) - 5/6*A*b/((b*x^2 + a)^(3/2)*a^2) - 1/2*A/((b *x^2 + a)^(3/2)*a*x^2)
Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{3}} + \frac {3 \, {\left (b x^{2} + a\right )} B a + B a^{2} - 6 \, {\left (b x^{2} + a\right )} A b - A a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x^{2} + a} A}{2 \, a^{3} x^{2}} \] Input:
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
1/2*(2*B*a - 5*A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 1/3* (3*(b*x^2 + a)*B*a + B*a^2 - 6*(b*x^2 + a)*A*b - A*a*b)/((b*x^2 + a)^(3/2) *a^3) - 1/2*sqrt(b*x^2 + a)*A/(a^3*x^2)
Time = 1.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {\frac {B}{3\,a}+\frac {B\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {10\,A\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {5\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}-\frac {5\,A\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \] Input:
int((A + B*x^2)/(x^3*(a + b*x^2)^(5/2)),x)
Output:
(B/(3*a) + (B*(a + b*x^2))/a^2)/(a + b*x^2)^(3/2) - (B*atanh((a + b*x^2)^( 1/2)/a^(1/2)))/a^(5/2) - (10*A*b)/(3*a^2*(a + b*x^2)^(3/2)) - A/(2*a*x^2*( a + b*x^2)^(3/2)) + (5*A*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(7/2)) - (5*A*b^2*x^2)/(2*a^3*(a + b*x^2)^(3/2))
Time = 0.22 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{2}+a}\, a^{2}-3 \sqrt {b \,x^{2}+a}\, a b \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )} \] Input:
int((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x)
Output:
( - sqrt(a + b*x**2)*a**2 - 3*sqrt(a + b*x**2)*a*b*x**2 - 3*sqrt(a)*log((s qrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*x**2 - 3*sqrt(a)*log(( sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4 + 3*sqrt(a)*log ((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*x**2 + 3*sqrt(a)*lo g((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*x**4)/(2*a**3*x** 2*(a + b*x**2))