Integrand size = 20, antiderivative size = 93 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {3 B c \sqrt {a+c x^2}}{8 x^2}-\frac {B \left (a+c x^2\right )^{3/2}}{4 x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}-\frac {3 B c^2 \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]
-1/4*B*(c*x^2+a)^(3/2)/x^4-1/5*A*(c*x^2+a)^(5/2)/a/x^5-3/8*B*c^2*arctanh(( c*x^2+a)^(1/2)/a^(1/2))/a^(1/2)-3/8*B*c*(c*x^2+a)^(1/2)/x^2
Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {\frac {\sqrt {a+c x^2} \left (8 A c^2 x^4+2 a^2 (4 A+5 B x)+a c x^2 (16 A+25 B x)\right )}{x^5}+15 \sqrt {a} B c^2 \log (x)-15 \sqrt {a} B c^2 \log \left (-\sqrt {a}+\sqrt {a+c x^2}\right )}{40 a} \]
-1/40*((Sqrt[a + c*x^2]*(8*A*c^2*x^4 + 2*a^2*(4*A + 5*B*x) + a*c*x^2*(16*A + 25*B*x)))/x^5 + 15*Sqrt[a]*B*c^2*Log[x] - 15*Sqrt[a]*B*c^2*Log[-Sqrt[a] + Sqrt[a + c*x^2]])/a
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {534, 243, 51, 51, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right )^{3/2} (A+B x)}{x^6} \, dx\) |
\(\Big \downarrow \) 534 |
\(\displaystyle B \int \frac {\left (c x^2+a\right )^{3/2}}{x^5}dx-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} B \int \frac {\left (c x^2+a\right )^{3/2}}{x^6}dx^2-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} B \left (\frac {3}{4} c \int \frac {\sqrt {c x^2+a}}{x^4}dx^2-\frac {\left (a+c x^2\right )^{3/2}}{2 x^4}\right )-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} B \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{x^2 \sqrt {c x^2+a}}dx^2-\frac {\sqrt {a+c x^2}}{x^2}\right )-\frac {\left (a+c x^2\right )^{3/2}}{2 x^4}\right )-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} B \left (\frac {3}{4} c \left (\int \frac {1}{\frac {x^4}{c}-\frac {a}{c}}d\sqrt {c x^2+a}-\frac {\sqrt {a+c x^2}}{x^2}\right )-\frac {\left (a+c x^2\right )^{3/2}}{2 x^4}\right )-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} B \left (\frac {3}{4} c \left (-\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{x^2}\right )-\frac {\left (a+c x^2\right )^{3/2}}{2 x^4}\right )-\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}\) |
-1/5*(A*(a + c*x^2)^(5/2))/(a*x^5) + (B*(-1/2*(a + c*x^2)^(3/2)/x^4 + (3*c *(-(Sqrt[a + c*x^2]/x^2) - (c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/Sqrt[a]))/ 4))/2
3.4.36.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[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_)^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]
Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {\sqrt {c \,x^{2}+a}\, \left (8 A \,c^{2} x^{4}+25 a B c \,x^{3}+16 a A c \,x^{2}+10 a^{2} B x +8 A \,a^{2}\right )}{40 x^{5} a}-\frac {3 B \,c^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{8 \sqrt {a}}\) | \(90\) |
default | \(B \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 c \left (\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )-\frac {A \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 a \,x^{5}}\) | \(122\) |
-1/40*(c*x^2+a)^(1/2)*(8*A*c^2*x^4+25*B*a*c*x^3+16*A*a*c*x^2+10*B*a^2*x+8* A*a^2)/x^5/a-3/8*B*c^2/a^(1/2)*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 = 2.04 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=\left [\frac {15 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (8 \, A c^{2} x^{4} + 25 \, B a c x^{3} + 16 \, A a c x^{2} + 10 \, B a^{2} x + 8 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{80 \, a x^{5}}, \frac {15 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (8 \, A c^{2} x^{4} + 25 \, B a c x^{3} + 16 \, A a c x^{2} + 10 \, B a^{2} x + 8 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{40 \, a x^{5}}\right ] \]
[1/80*(15*B*sqrt(a)*c^2*x^5*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a) /x^2) - 2*(8*A*c^2*x^4 + 25*B*a*c*x^3 + 16*A*a*c*x^2 + 10*B*a^2*x + 8*A*a^ 2)*sqrt(c*x^2 + a))/(a*x^5), 1/40*(15*B*sqrt(-a)*c^2*x^5*arctan(sqrt(-a)/s qrt(c*x^2 + a)) - (8*A*c^2*x^4 + 25*B*a*c*x^3 + 16*A*a*c*x^2 + 10*B*a^2*x + 8*A*a^2)*sqrt(c*x^2 + a))/(a*x^5)]
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (87) = 174\).
Time = 4.35 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.14 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=- \frac {A a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {2 A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{2}} - \frac {A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 a} - \frac {B a^{2}}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B a \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {B c^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 \sqrt {a}} \]
-A*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 2*A*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**2) - A*c**(5/2)*sqrt(a/(c*x**2) + 1)/(5*a) - B*a**2/(4*sqrt(c)* x**5*sqrt(a/(c*x**2) + 1)) - 3*B*a*sqrt(c)/(8*x**3*sqrt(a/(c*x**2) + 1)) - B*c**(3/2)*sqrt(a/(c*x**2) + 1)/(2*x) - B*c**(3/2)/(8*x*sqrt(a/(c*x**2) + 1)) - 3*B*c**2*asinh(sqrt(a)/(sqrt(c)*x))/(8*sqrt(a))
Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {3 \, B c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {c x^{2} + a} B c^{2}}{8 \, a} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{8 \, a^{2} x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{4 \, a x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, a x^{5}} \]
-3/8*B*c^2*arcsinh(a/(sqrt(a*c)*abs(x)))/sqrt(a) + 1/8*(c*x^2 + a)^(3/2)*B *c^2/a^2 + 3/8*sqrt(c*x^2 + a)*B*c^2/a - 1/8*(c*x^2 + a)^(5/2)*B*c/(a^2*x^ 2) - 1/4*(c*x^2 + a)^(5/2)*B/(a*x^4) - 1/5*(c*x^2 + a)^(5/2)*A/(a*x^5)
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (73) = 146\).
Time = 0.30 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.49 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=\frac {3 \, B c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} + \frac {25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B c^{2} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} A c^{\frac {5}{2}} - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a c^{2} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {5}{2}} + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{3} c^{2} - 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c^{2} + 8 \, A a^{4} c^{\frac {5}{2}}}{20 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5}} \]
3/4*B*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) + 1/20* (25*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*c^2 + 40*(sqrt(c)*x - sqrt(c*x^2 + a ))^8*A*c^(5/2) - 10*(sqrt(c)*x - sqrt(c*x^2 + a))^7*B*a*c^2 + 80*(sqrt(c)* x - sqrt(c*x^2 + a))^4*A*a^2*c^(5/2) + 10*(sqrt(c)*x - sqrt(c*x^2 + a))^3* B*a^3*c^2 - 25*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^4*c^2 + 8*A*a^4*c^(5/2))/ ((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^5
Time = 11.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx=\frac {3\,B\,a\,\sqrt {c\,x^2+a}}{8\,x^4}-\frac {3\,B\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {5\,B\,{\left (c\,x^2+a\right )}^{3/2}}{8\,x^4}-\frac {A\,{\left (c\,x^2+a\right )}^{5/2}}{5\,a\,x^5} \]