Integrand size = 22, antiderivative size = 90 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=-\frac {2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}-\frac {2 (7 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}+\frac {4 c (7 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3} \]
-2/7*A*(c*x^2+b*x)^(3/2)/b/x^5-2/35*(-4*A*c+7*B*b)*(c*x^2+b*x)^(3/2)/b^2/x ^4+4/105*c*(-4*A*c+7*B*b)*(c*x^2+b*x)^(3/2)/b^3/x^3
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=\frac {2 (x (b+c x))^{3/2} \left (7 b B x (-3 b+2 c x)+A \left (-15 b^2+12 b c x-8 c^2 x^2\right )\right )}{105 b^3 x^5} \]
(2*(x*(b + c*x))^(3/2)*(7*b*B*x*(-3*b + 2*c*x) + A*(-15*b^2 + 12*b*c*x - 8 *c^2*x^2)))/(105*b^3*x^5)
Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1220, 1129, 1123}
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.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(7 b B-4 A c) \int \frac {\sqrt {c x^2+b x}}{x^4}dx}{7 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {(7 b B-4 A c) \left (-\frac {2 c \int \frac {\sqrt {c x^2+b x}}{x^3}dx}{5 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}\right )}{7 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {\left (\frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}\right ) (7 b B-4 A c)}{7 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{7 b x^5}\) |
(-2*A*(b*x + c*x^2)^(3/2))/(7*b*x^5) + ((7*b*B - 4*A*c)*((-2*(b*x + c*x^2) ^(3/2))/(5*b*x^4) + (4*c*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)))/(7*b)
3.1.75.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b *e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d , e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 2], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.60
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\frac {7 B x}{5}+A \right ) b^{2}-\frac {4 c \left (\frac {7 B x}{6}+A \right ) x b}{5}+\frac {8 A \,c^{2} x^{2}}{15}\right ) \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}{7 x^{4} b^{3}}\) | \(54\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-14 B b c \,x^{2}-12 A b c x +21 b^{2} B x +15 A \,b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 x^{4} b^{3}}\) | \(62\) |
trager | \(-\frac {2 \left (8 A \,c^{3} x^{3}-14 B b \,c^{2} x^{3}-4 A b \,c^{2} x^{2}+7 B \,b^{2} c \,x^{2}+3 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{105 x^{4} b^{3}}\) | \(81\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (8 A \,c^{3} x^{3}-14 B b \,c^{2} x^{3}-4 A b \,c^{2} x^{2}+7 B \,b^{2} c \,x^{2}+3 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right )}{105 x^{3} \sqrt {x \left (c x +b \right )}\, b^{3}}\) | \(84\) |
default | \(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )\) | \(112\) |
-2/7*((7/5*B*x+A)*b^2-4/5*c*(7/6*B*x+A)*x*b+8/15*A*c^2*x^2)*(c*x+b)*(x*(c* x+b))^(1/2)/x^4/b^3
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=-\frac {2 \, {\left (15 \, A b^{3} - 2 \, {\left (7 \, B b c^{2} - 4 \, A c^{3}\right )} x^{3} + {\left (7 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{2} + 3 \, {\left (7 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, b^{3} x^{4}} \]
-2/105*(15*A*b^3 - 2*(7*B*b*c^2 - 4*A*c^3)*x^3 + (7*B*b^2*c - 4*A*b*c^2)*x ^2 + 3*(7*B*b^3 + A*b^2*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^4)
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{5}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=\frac {4 \, \sqrt {c x^{2} + b x} B c^{2}}{15 \, b^{2} x} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{3}}{105 \, b^{3} x} - \frac {2 \, \sqrt {c x^{2} + b x} B c}{15 \, b x^{2}} + \frac {8 \, \sqrt {c x^{2} + b x} A c^{2}}{105 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{5 \, x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A c}{35 \, b x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{7 \, x^{4}} \]
4/15*sqrt(c*x^2 + b*x)*B*c^2/(b^2*x) - 16/105*sqrt(c*x^2 + b*x)*A*c^3/(b^3 *x) - 2/15*sqrt(c*x^2 + b*x)*B*c/(b*x^2) + 8/105*sqrt(c*x^2 + b*x)*A*c^2/( b^2*x^2) - 2/5*sqrt(c*x^2 + b*x)*B/x^3 - 2/35*sqrt(c*x^2 + b*x)*A*c/(b*x^3 ) - 2/7*sqrt(c*x^2 + b*x)*A/x^4
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (78) = 156\).
Time = 0.27 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.79 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=\frac {2 \, {\left (105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B c^{\frac {3}{2}} + 175 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b c + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A c^{2} + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{2} \sqrt {c} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b c^{\frac {3}{2}} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} + 273 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} \sqrt {c} + 15 \, A b^{4}\right )}}{105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]
2/105*(105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*c^(3/2) + 175*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b*c + 140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^2 + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*sqrt(c) + 315*(sqrt(c)*x - sq rt(c*x^2 + b*x))^3*A*b*c^(3/2) + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^ 3 + 273*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c + 105*(sqrt(c)*x - sqrt( c*x^2 + b*x))*A*b^3*sqrt(c) + 15*A*b^4)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^7
Time = 10.61 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^5} \, dx=\frac {8\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^2}-\frac {2\,B\,\sqrt {c\,x^2+b\,x}}{5\,x^3}-\frac {2\,A\,c\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^3}-\frac {2\,B\,c\,\sqrt {c\,x^2+b\,x}}{15\,b\,x^2}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {16\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x}+\frac {4\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{15\,b^2\,x} \]