Integrand size = 20, antiderivative size = 147 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=-\frac {A \sqrt {a+c x^2}}{5 a x^5}-\frac {B \sqrt {a+c x^2}}{4 a x^4}+\frac {4 A c \sqrt {a+c x^2}}{15 a^2 x^3}+\frac {3 B c \sqrt {a+c x^2}}{8 a^2 x^2}-\frac {8 A c^2 \sqrt {a+c x^2}}{15 a^3 x}-\frac {3 B c^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
-3/8*B*c^2*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(5/2)-1/5*A*(c*x^2+a)^(1/2)/ a/x^5-1/4*B*(c*x^2+a)^(1/2)/a/x^4+4/15*A*c*(c*x^2+a)^(1/2)/a^2/x^3+3/8*B*c *(c*x^2+a)^(1/2)/a^2/x^2-8/15*A*c^2*(c*x^2+a)^(1/2)/a^3/x
Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.67 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=\frac {\frac {\sqrt {a+c x^2} \left (-64 A c^2 x^4-6 a^2 (4 A+5 B x)+a c x^2 (32 A+45 B x)\right )}{x^5}+90 \sqrt {a} B c^2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{120 a^3} \]
((Sqrt[a + c*x^2]*(-64*A*c^2*x^4 - 6*a^2*(4*A + 5*B*x) + a*c*x^2*(32*A + 4 5*B*x)))/x^5 + 90*Sqrt[a]*B*c^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt [a]])/(120*a^3)
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {539, 25, 539, 27, 539, 25, 539, 27, 534, 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.
| \(\displaystyle \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {\int -\frac {5 a B-4 A c x}{x^5 \sqrt {c x^2+a}}dx}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 a B-4 A c x}{x^5 \sqrt {c x^2+a}}dx}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {-\frac {\int \frac {a c (16 A+15 B x)}{x^4 \sqrt {c x^2+a}}dx}{4 a}-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {1}{4} c \int \frac {16 A+15 B x}{x^4 \sqrt {c x^2+a}}dx-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (-\frac {\int -\frac {45 a B-32 A c x}{x^3 \sqrt {c x^2+a}}dx}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {\int \frac {45 a B-32 A c x}{x^3 \sqrt {c x^2+a}}dx}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {\int \frac {a c (64 A+45 B x)}{x^2 \sqrt {c x^2+a}}dx}{2 a}-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {1}{2} c \int \frac {64 A+45 B x}{x^2 \sqrt {c x^2+a}}dx-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {1}{2} c \left (45 B \int \frac {1}{x \sqrt {c x^2+a}}dx-\frac {64 A \sqrt {a+c x^2}}{a x}\right )-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {1}{2} c \left (\frac {45}{2} B \int \frac {1}{x^2 \sqrt {c x^2+a}}dx^2-\frac {64 A \sqrt {a+c x^2}}{a x}\right )-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {1}{2} c \left (\frac {45 B \int \frac {1}{\frac {x^4}{c}-\frac {a}{c}}d\sqrt {c x^2+a}}{c}-\frac {64 A \sqrt {a+c x^2}}{a x}\right )-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {1}{4} c \left (\frac {-\frac {1}{2} c \left (-\frac {64 A \sqrt {a+c x^2}}{a x}-\frac {45 B \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )-\frac {45 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {16 A \sqrt {a+c x^2}}{3 a x^3}\right )-\frac {5 B \sqrt {a+c x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+c x^2}}{5 a x^5}\) |
-1/5*(A*Sqrt[a + c*x^2])/(a*x^5) + ((-5*B*Sqrt[a + c*x^2])/(4*x^4) - (c*(( -16*A*Sqrt[a + c*x^2])/(3*a*x^3) + ((-45*B*Sqrt[a + c*x^2])/(2*x^2) - (c*( (-64*A*Sqrt[a + c*x^2])/(a*x) - (45*B*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/Sq rt[a]))/2)/(3*a)))/4)/(5*a)
3.4.64.3.1 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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61
| method | result | size |
| risch | \(-\frac {\sqrt {c \,x^{2}+a}\, \left (64 A \,c^{2} x^{4}-45 a B c \,x^{3}-32 a A c \,x^{2}+30 a^{2} B x +24 A \,a^{2}\right )}{120 a^{3} x^{5}}-\frac {3 B \,c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{8 a^{\frac {5}{2}}}\) | \(90\) |
| default | \(B \left (-\frac {\sqrt {c \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 c \left (-\frac {\sqrt {c \,x^{2}+a}}{2 a \,x^{2}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+A \left (-\frac {\sqrt {c \,x^{2}+a}}{5 a \,x^{5}}-\frac {4 c \left (-\frac {\sqrt {c \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 c \sqrt {c \,x^{2}+a}}{3 a^{2} x}\right )}{5 a}\right )\) | \(137\) |
-1/120*(c*x^2+a)^(1/2)*(64*A*c^2*x^4-45*B*a*c*x^3-32*A*a*c*x^2+30*B*a^2*x+ 24*A*a^2)/a^3/x^5-3/8*B/a^(5/2)*c^2*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)
Time = 0.30 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=\left [\frac {45 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (64 \, A c^{2} x^{4} - 45 \, B a c x^{3} - 32 \, A a c x^{2} + 30 \, B a^{2} x + 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, a^{3} x^{5}}, \frac {45 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (64 \, A c^{2} x^{4} - 45 \, B a c x^{3} - 32 \, A a c x^{2} + 30 \, B a^{2} x + 24 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, a^{3} x^{5}}\right ] \]
[1/240*(45*B*sqrt(a)*c^2*x^5*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a )/x^2) - 2*(64*A*c^2*x^4 - 45*B*a*c*x^3 - 32*A*a*c*x^2 + 30*B*a^2*x + 24*A *a^2)*sqrt(c*x^2 + a))/(a^3*x^5), 1/120*(45*B*sqrt(-a)*c^2*x^5*arctan(sqrt (-a)/sqrt(c*x^2 + a)) - (64*A*c^2*x^4 - 45*B*a*c*x^3 - 32*A*a*c*x^2 + 30*B *a^2*x + 24*A*a^2)*sqrt(c*x^2 + a))/(a^3*x^5)]
Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (138) = 276\).
Time = 3.39 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.78 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=- \frac {3 A a^{4} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {2 A a^{3} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {3 A a^{2} c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {12 A a c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {8 A c^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{c x^{2}} + 1}}{15 a^{5} c^{4} x^{4} + 30 a^{4} c^{5} x^{6} + 15 a^{3} c^{6} x^{8}} - \frac {B}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B \sqrt {c}}{8 a x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {3 B c^{\frac {3}{2}}}{8 a^{2} x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 a^{\frac {5}{2}}} \]
-3*A*a**4*c**(9/2)*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5* x**6 + 15*a**3*c**6*x**8) - 2*A*a**3*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/( 15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 3*A*a**2*c**( 13/2)*x**4*sqrt(a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 1 5*a**3*c**6*x**8) - 12*A*a*c**(15/2)*x**6*sqrt(a/(c*x**2) + 1)/(15*a**5*c* *4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x**8) - 8*A*c**(17/2)*x**8*sqrt (a/(c*x**2) + 1)/(15*a**5*c**4*x**4 + 30*a**4*c**5*x**6 + 15*a**3*c**6*x** 8) - B/(4*sqrt(c)*x**5*sqrt(a/(c*x**2) + 1)) + B*sqrt(c)/(8*a*x**3*sqrt(a/ (c*x**2) + 1)) + 3*B*c**(3/2)/(8*a**2*x*sqrt(a/(c*x**2) + 1)) - 3*B*c**2*a sinh(sqrt(a)/(sqrt(c)*x))/(8*a**(5/2))
Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=-\frac {3 \, B c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {8 \, \sqrt {c x^{2} + a} A c^{2}}{15 \, a^{3} x} + \frac {3 \, \sqrt {c x^{2} + a} B c}{8 \, a^{2} x^{2}} + \frac {4 \, \sqrt {c x^{2} + a} A c}{15 \, a^{2} x^{3}} - \frac {\sqrt {c x^{2} + a} B}{4 \, a x^{4}} - \frac {\sqrt {c x^{2} + a} A}{5 \, a x^{5}} \]
-3/8*B*c^2*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(5/2) - 8/15*sqrt(c*x^2 + a)*A* c^2/(a^3*x) + 3/8*sqrt(c*x^2 + a)*B*c/(a^2*x^2) + 4/15*sqrt(c*x^2 + a)*A*c /(a^2*x^3) - 1/4*sqrt(c*x^2 + a)*B/(a*x^4) - 1/5*sqrt(c*x^2 + a)*A/(a*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (119) = 238\).
Time = 0.29 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.64 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=\frac {3 \, B c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B c^{2} - 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a c^{2} - 640 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {5}{2}} + 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{3} c^{2} + 320 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{3} c^{\frac {5}{2}} - 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c^{2} - 64 \, A a^{4} c^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5} a^{2}} \]
3/4*B*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/60*(45*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*c^2 - 210*(sqrt(c)*x - sqrt(c* x^2 + a))^7*B*a*c^2 - 640*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(5/2) + 210*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^3*c^2 + 320*(sqrt(c)*x - sqrt(c*x^ 2 + a))^2*A*a^3*c^(5/2) - 45*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^4*c^2 - 64* A*a^4*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^5*a^2)
Time = 10.58 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.67 \[ \int \frac {A+B x}{x^6 \sqrt {a+c x^2}} \, dx=\frac {3\,B\,{\left (c\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {5\,B\,\sqrt {c\,x^2+a}}{8\,a\,x^4}-\frac {3\,B\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {A\,\sqrt {c\,x^2+a}\,\left (3\,a^2-4\,a\,c\,x^2+8\,c^2\,x^4\right )}{15\,a^3\,x^5} \]