Integrand size = 22, antiderivative size = 123 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=-\frac {A \sqrt {a+b x^2}}{6 a x^6}+\frac {(5 A b-6 a B) \sqrt {a+b x^2}}{24 a^2 x^4}-\frac {b (5 A b-6 a B) \sqrt {a+b x^2}}{16 a^3 x^2}+\frac {b^2 (5 A b-6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{7/2}} \]
1/16*b^2*(5*A*b-6*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(7/2)-1/6*A*(b*x ^2+a)^(1/2)/a/x^6+1/24*(5*A*b-6*B*a)*(b*x^2+a)^(1/2)/a^2/x^4-1/16*b*(5*A*b -6*B*a)*(b*x^2+a)^(1/2)/a^3/x^2
Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a^2 A+10 a A b x^2-12 a^2 B x^2-15 A b^2 x^4+18 a b B x^4\right )}{48 a^3 x^6}-\frac {b^2 (-5 A b+6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{7/2}} \]
(Sqrt[a + b*x^2]*(-8*a^2*A + 10*a*A*b*x^2 - 12*a^2*B*x^2 - 15*A*b^2*x^4 + 18*a*b*B*x^4))/(48*a^3*x^6) - (b^2*(-5*A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x^2 ]/Sqrt[a]])/(16*a^(7/2))
Time = 0.22 (sec) , antiderivative size = 118, 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, 52, 52, 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.
\(\displaystyle \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^8 \sqrt {b x^2+a}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-6 a B) \int \frac {1}{x^6 \sqrt {b x^2+a}}dx^2}{6 a}-\frac {A \sqrt {a+b x^2}}{3 a x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-6 a B) \left (-\frac {3 b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {A \sqrt {a+b x^2}}{3 a x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-6 a B) \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {A \sqrt {a+b x^2}}{3 a x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-6 a B) \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {A \sqrt {a+b x^2}}{3 a x^6}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-6 a B) \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {A \sqrt {a+b x^2}}{3 a x^6}\right )\) |
(-1/3*(A*Sqrt[a + b*x^2])/(a*x^6) - ((5*A*b - 6*a*B)*(-1/2*Sqrt[a + b*x^2] /(a*x^4) - (3*b*(-(Sqrt[a + b*x^2]/(a*x^2)) + (b*ArcTanh[Sqrt[a + b*x^2]/S qrt[a]])/a^(3/2)))/(4*a)))/(6*a))/2
3.6.67.3.1 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] :> 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 = 2.94 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(-\frac {-\frac {15 x^{6} b^{2} \left (A b -\frac {6 B a}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8}+\sqrt {b \,x^{2}+a}\, \left (-\frac {5 x^{2} \left (\frac {9 x^{2} B}{5}+A \right ) b \,a^{\frac {3}{2}}}{4}+\left (\frac {3 x^{2} B}{2}+A \right ) a^{\frac {5}{2}}+\frac {15 A \sqrt {a}\, b^{2} x^{4}}{8}\right )}{6 a^{\frac {7}{2}} x^{6}}\) | \(92\) |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 A \,b^{2} x^{4}-18 B a b \,x^{4}-10 a A b \,x^{2}+12 a^{2} B \,x^{2}+8 a^{2} A \right )}{48 a^{3} x^{6}}+\frac {\left (5 A b -6 B a \right ) b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{16 a^{\frac {7}{2}}}\) | \(100\) |
default | \(A \left (-\frac {\sqrt {b \,x^{2}+a}}{6 a \,x^{6}}-\frac {5 b \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )}{6 a}\right )+B \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\) | \(172\) |
-1/6/a^(7/2)*(-15/8*x^6*b^2*(A*b-6/5*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2)) +(b*x^2+a)^(1/2)*(-5/4*x^2*(9/5*x^2*B+A)*b*a^(3/2)+(3/2*x^2*B+A)*a^(5/2)+1 5/8*A*a^(1/2)*b^2*x^4))/x^6
Time = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.81 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {a} x^{6} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, a^{4} x^{6}}, \frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a^{4} x^{6}}\right ] \]
[-1/96*(3*(6*B*a*b^2 - 5*A*b^3)*sqrt(a)*x^6*log(-(b*x^2 + 2*sqrt(b*x^2 + a )*sqrt(a) + 2*a)/x^2) - 2*(3*(6*B*a^2*b - 5*A*a*b^2)*x^4 - 8*A*a^3 - 2*(6* B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^6), 1/48*(3*(6*B*a*b^2 - 5 *A*b^3)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(6*B*a^2*b - 5* A*a*b^2)*x^4 - 8*A*a^3 - 2*(6*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^ 4*x^6)]
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (114) = 228\).
Time = 21.16 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.91 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=- \frac {A}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A \sqrt {b}}{24 a x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A b^{\frac {3}{2}}}{48 a^{2} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A b^{\frac {5}{2}}}{16 a^{3} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {7}{2}}} - \frac {B}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B \sqrt {b}}{8 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 B b^{\frac {3}{2}}}{8 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 B b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {5}{2}}} \]
-A/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) + A*sqrt(b)/(24*a*x**5*sqrt(a/(b* x**2) + 1)) - 5*A*b**(3/2)/(48*a**2*x**3*sqrt(a/(b*x**2) + 1)) - 5*A*b**(5 /2)/(16*a**3*x*sqrt(a/(b*x**2) + 1)) + 5*A*b**3*asinh(sqrt(a)/(sqrt(b)*x)) /(16*a**(7/2)) - B/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + B*sqrt(b)/(8*a* x**3*sqrt(a/(b*x**2) + 1)) + 3*B*b**(3/2)/(8*a**2*x*sqrt(a/(b*x**2) + 1)) - 3*B*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2))
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=-\frac {3 \, B b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {5 \, A b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {7}{2}}} + \frac {3 \, \sqrt {b x^{2} + a} B b}{8 \, a^{2} x^{2}} - \frac {5 \, \sqrt {b x^{2} + a} A b^{2}}{16 \, a^{3} x^{2}} - \frac {\sqrt {b x^{2} + a} B}{4 \, a x^{4}} + \frac {5 \, \sqrt {b x^{2} + a} A b}{24 \, a^{2} x^{4}} - \frac {\sqrt {b x^{2} + a} A}{6 \, a x^{6}} \]
-3/8*B*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 5/16*A*b^3*arcsinh(a/(s qrt(a*b)*abs(x)))/a^(7/2) + 3/8*sqrt(b*x^2 + a)*B*b/(a^2*x^2) - 5/16*sqrt( b*x^2 + a)*A*b^2/(a^3*x^2) - 1/4*sqrt(b*x^2 + a)*B/(a*x^4) + 5/24*sqrt(b*x ^2 + a)*A*b/(a^2*x^4) - 1/6*sqrt(b*x^2 + a)*A/(a*x^6)
Time = 0.30 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=\frac {\frac {3 \, {\left (6 \, B a b^{3} - 5 \, A b^{4}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {18 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a b^{3} - 48 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 30 \, \sqrt {b x^{2} + a} B a^{3} b^{3} - 15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4} + 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{4} - 33 \, \sqrt {b x^{2} + a} A a^{2} b^{4}}{a^{3} b^{3} x^{6}}}{48 \, b} \]
1/48*(3*(6*B*a*b^3 - 5*A*b^4)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a ^3) + (18*(b*x^2 + a)^(5/2)*B*a*b^3 - 48*(b*x^2 + a)^(3/2)*B*a^2*b^3 + 30* sqrt(b*x^2 + a)*B*a^3*b^3 - 15*(b*x^2 + a)^(5/2)*A*b^4 + 40*(b*x^2 + a)^(3 /2)*A*a*b^4 - 33*sqrt(b*x^2 + a)*A*a^2*b^4)/(a^3*b^3*x^6))/b
Time = 6.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2}{x^7 \sqrt {a+b x^2}} \, dx=\frac {5\,A\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a^2\,x^6}-\frac {11\,A\,\sqrt {b\,x^2+a}}{16\,a\,x^6}-\frac {3\,B\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {5\,A\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^3\,x^6}-\frac {5\,B\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,B\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,a^{7/2}} \]
(5*A*(a + b*x^2)^(3/2))/(6*a^2*x^6) - (3*B*b^2*atanh((a + b*x^2)^(1/2)/a^( 1/2)))/(8*a^(5/2)) - (11*A*(a + b*x^2)^(1/2))/(16*a*x^6) - (A*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(16*a^(7/2)) - (5*A*(a + b*x^2)^(5/2))/(1 6*a^3*x^6) - (5*B*(a + b*x^2)^(1/2))/(8*a*x^4) + (3*B*(a + b*x^2)^(3/2))/( 8*a^2*x^4)