Integrand size = 22, antiderivative size = 90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac {2 (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}+\frac {4 c (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5} \]
-2/9*A*(c*x^2+b*x)^(5/2)/b/x^7-2/63*(-4*A*c+9*B*b)*(c*x^2+b*x)^(5/2)/b^2/x ^6+4/315*c*(-4*A*c+9*B*b)*(c*x^2+b*x)^(5/2)/b^3/x^5
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {2 (x (b+c x))^{5/2} \left (9 b B x (-5 b+2 c x)+A \left (-35 b^2+20 b c x-8 c^2 x^2\right )\right )}{315 b^3 x^7} \]
(2*(x*(b + c*x))^(5/2)*(9*b*B*x*(-5*b + 2*c*x) + A*(-35*b^2 + 20*b*c*x - 8 *c^2*x^2)))/(315*b^3*x^7)
Time = 0.23 (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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(9 b B-4 A c) \int \frac {\left (c x^2+b x\right )^{3/2}}{x^6}dx}{9 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {(9 b B-4 A c) \left (-\frac {2 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^5}dx}{7 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{7 b x^6}\right )}{9 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {\left (\frac {4 c \left (b x+c x^2\right )^{5/2}}{35 b^2 x^5}-\frac {2 \left (b x+c x^2\right )^{5/2}}{7 b x^6}\right ) (9 b B-4 A c)}{9 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}\) |
(-2*A*(b*x + c*x^2)^(5/2))/(9*b*x^7) + ((9*b*B - 4*A*c)*((-2*(b*x + c*x^2) ^(5/2))/(7*b*x^6) + (4*c*(b*x + c*x^2)^(5/2))/(35*b^2*x^5)))/(9*b)
3.1.89.3.1 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 = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\frac {9 B x}{7}+A \right ) b^{2}-\frac {4 c \left (\frac {9 B x}{10}+A \right ) x b}{7}+\frac {8 A \,c^{2} x^{2}}{35}\right ) \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}}{9 x^{5} b^{3}}\) | \(56\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-18 B b c \,x^{2}-20 A b c x +45 b^{2} B x +35 A \,b^{2}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{315 x^{6} b^{3}}\) | \(62\) |
trager | \(-\frac {2 \left (8 A \,c^{4} x^{4}-18 B b \,c^{3} x^{4}-4 A b \,c^{3} x^{3}+9 B \,b^{2} c^{2} x^{3}+3 A \,b^{2} c^{2} x^{2}+72 B \,b^{3} c \,x^{2}+50 A \,b^{3} c x +45 b^{4} B x +35 A \,b^{4}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{3} x^{5}}\) | \(105\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (8 A \,c^{4} x^{4}-18 B b \,c^{3} x^{4}-4 A b \,c^{3} x^{3}+9 B \,b^{2} c^{2} x^{3}+3 A \,b^{2} c^{2} x^{2}+72 B \,b^{3} c \,x^{2}+50 A \,b^{3} c x +45 b^{4} B x +35 A \,b^{4}\right )}{315 x^{4} \sqrt {x \left (c x +b \right )}\, b^{3}}\) | \(108\) |
default | \(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{9 b \,x^{7}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 b \,x^{6}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{35 b^{2} x^{5}}\right )}{9 b}\right )+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 b \,x^{6}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{35 b^{2} x^{5}}\right )\) | \(112\) |
-2/9*((9/7*B*x+A)*b^2-4/7*c*(9/10*B*x+A)*x*b+8/35*A*c^2*x^2)*(c*x+b)^2*(x* (c*x+b))^(1/2)/x^5/b^3
Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=-\frac {2 \, {\left (35 \, A b^{4} - 2 \, {\left (9 \, B b c^{3} - 4 \, A c^{4}\right )} x^{4} + {\left (9 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{3} + 3 \, {\left (24 \, B b^{3} c + A b^{2} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{4} + 10 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, b^{3} x^{5}} \]
-2/315*(35*A*b^4 - 2*(9*B*b*c^3 - 4*A*c^4)*x^4 + (9*B*b^2*c^2 - 4*A*b*c^3) *x^3 + 3*(24*B*b^3*c + A*b^2*c^2)*x^2 + 5*(9*B*b^4 + 10*A*b^3*c)*x)*sqrt(c *x^2 + b*x)/(b^3*x^5)
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{7}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (78) = 156\).
Time = 0.18 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.47 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {4 \, \sqrt {c x^{2} + b x} B c^{3}}{35 \, b^{2} x} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{4}}{315 \, b^{3} x} - \frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{35 \, b x^{2}} + \frac {8 \, \sqrt {c x^{2} + b x} A c^{3}}{315 \, b^{2} x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} B c}{70 \, x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{105 \, b x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{14 \, x^{4}} + \frac {\sqrt {c x^{2} + b x} A c}{63 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{2 \, x^{5}} + \frac {\sqrt {c x^{2} + b x} A b}{9 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, x^{6}} \]
4/35*sqrt(c*x^2 + b*x)*B*c^3/(b^2*x) - 16/315*sqrt(c*x^2 + b*x)*A*c^4/(b^3 *x) - 2/35*sqrt(c*x^2 + b*x)*B*c^2/(b*x^2) + 8/315*sqrt(c*x^2 + b*x)*A*c^3 /(b^2*x^2) + 3/70*sqrt(c*x^2 + b*x)*B*c/x^3 - 2/105*sqrt(c*x^2 + b*x)*A*c^ 2/(b*x^3) + 3/14*sqrt(c*x^2 + b*x)*B*b/x^4 + 1/63*sqrt(c*x^2 + b*x)*A*c/x^ 4 - 1/2*(c*x^2 + b*x)^(3/2)*B/x^5 + 1/9*sqrt(c*x^2 + b*x)*A*b/x^5 - 1/3*(c *x^2 + b*x)^(3/2)*A/x^6
Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (78) = 156\).
Time = 0.27 (sec) , antiderivative size = 371, normalized size of antiderivative = 4.12 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {2 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B c^{\frac {5}{2}} + 945 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b c^{2} + 420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A c^{3} + 1260 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{2} c^{\frac {3}{2}} + 1575 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b c^{\frac {5}{2}} + 882 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{3} c + 2583 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{2} c^{2} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{4} \sqrt {c} + 2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{3} c^{\frac {3}{2}} + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{5} + 1170 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{4} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{5} \sqrt {c} + 35 \, A b^{6}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]
2/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*c^(5/2) + 945*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b*c^2 + 420*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*c^3 + 1260*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c^(3/2) + 1575*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b*c^(5/2) + 882*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4 *B*b^3*c + 2583*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^2 + 315*(sqrt(c) *x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 2310*(sqrt(c)*x - sqrt(c*x^2 + b *x))^3*A*b^3*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^5 + 1170*( sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^4*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b *x))*A*b^5*sqrt(c) + 35*A*b^6)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9
Time = 11.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx=\frac {8\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^2\,x^2}-\frac {20\,A\,c\,\sqrt {c\,x^2+b\,x}}{63\,x^4}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {16\,B\,c\,\sqrt {c\,x^2+b\,x}}{35\,x^3}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b\,x^3}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^2}+\frac {4\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x} \]
(8*A*c^3*(b*x + c*x^2)^(1/2))/(315*b^2*x^2) - (20*A*c*(b*x + c*x^2)^(1/2)) /(63*x^4) - (2*B*b*(b*x + c*x^2)^(1/2))/(7*x^4) - (16*B*c*(b*x + c*x^2)^(1 /2))/(35*x^3) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(105*b*x^3) - (2*A*b*(b*x + c*x^2)^(1/2))/(9*x^5) - (16*A*c^4*(b*x + c*x^2)^(1/2))/(315*b^3*x) - (2*B* c^2*(b*x + c*x^2)^(1/2))/(35*b*x^2) + (4*B*c^3*(b*x + c*x^2)^(1/2))/(35*b^ 2*x)