Integrand size = 22, antiderivative size = 57 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}-\frac {2 (7 b B-2 A c) \left (b x+c x^2\right )^{5/2}}{35 b^2 x^5} \] Output:
-2/7*A*(c*x^2+b*x)^(5/2)/b/x^6-2/35*(-2*A*c+7*B*b)*(c*x^2+b*x)^(5/2)/b^2/x ^5
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {2 (x (b+c x))^{5/2} (5 A b+7 b B x-2 A c x)}{35 b^2 x^6} \] Input:
Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]
Output:
(-2*(x*(b + c*x))^(5/2)*(5*A*b + 7*b*B*x - 2*A*c*x))/(35*b^2*x^6)
Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1220, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(7 b B-2 A c) \int \frac {\left (c x^2+b x\right )^{3/2}}{x^5}dx}{7 b}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle -\frac {2 \left (b x+c x^2\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^5}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{7 b x^6}\) |
Input:
Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^6,x]
Output:
(-2*A*(b*x + c*x^2)^(5/2))/(7*b*x^6) - (2*(7*b*B - 2*A*c)*(b*x + c*x^2)^(5 /2))/(35*b^2*x^5)
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_)*((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 = 0.92 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(-\frac {2 \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}\, \left (\left (\frac {7 B x}{5}+A \right ) b -\frac {2 A c x}{5}\right )}{7 x^{4} b^{2}}\) | \(39\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-2 A c x +7 B b x +5 A b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{35 b^{2} x^{5}}\) | \(40\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-2 A c x +7 B b x +5 A b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{35 b^{2} x^{5}}\) | \(40\) |
default | \(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 b \,x^{6}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{35 b^{2} x^{5}}\right )-\frac {2 B \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 b \,x^{5}}\) | \(64\) |
trager | \(-\frac {2 \left (-2 A \,c^{3} x^{3}+7 x^{3} B b \,c^{2}+A b \,c^{2} x^{2}+14 x^{2} B \,b^{2} c +8 A \,b^{2} c x +7 x B \,b^{3}+5 A \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{35 b^{2} x^{4}}\) | \(80\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (-2 A \,c^{3} x^{3}+7 x^{3} B b \,c^{2}+A b \,c^{2} x^{2}+14 x^{2} B \,b^{2} c +8 A \,b^{2} c x +7 x B \,b^{3}+5 A \,b^{3}\right )}{35 x^{3} \sqrt {x \left (c x +b \right )}\, b^{2}}\) | \(83\) |
Input:
int((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-2/7*(c*x+b)^2*(x*(c*x+b))^(1/2)*((7/5*B*x+A)*b-2/5*A*c*x)/x^4/b^2
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {2 \, {\left (5 \, A b^{3} + {\left (7 \, B b c^{2} - 2 \, A c^{3}\right )} x^{3} + {\left (14 \, B b^{2} c + A b c^{2}\right )} x^{2} + {\left (7 \, B b^{3} + 8 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{35 \, b^{2} x^{4}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="fricas")
Output:
-2/35*(5*A*b^3 + (7*B*b*c^2 - 2*A*c^3)*x^3 + (14*B*b^2*c + A*b*c^2)*x^2 + (7*B*b^3 + 8*A*b^2*c)*x)*sqrt(c*x^2 + b*x)/(b^2*x^4)
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{6}}\, dx \] Input:
integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**6,x)
Output:
Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**6, x)
Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (49) = 98\).
Time = 0.03 (sec) , antiderivative size = 176, normalized size of antiderivative = 3.09 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x} B c^{2}}{5 \, b x} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{3}}{35 \, b^{2} x} + \frac {\sqrt {c x^{2} + b x} B c}{5 \, x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{2}}{35 \, b x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} B b}{5 \, x^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A c}{70 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{x^{4}} + \frac {3 \, \sqrt {c x^{2} + b x} A b}{14 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{2 \, x^{5}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="maxima")
Output:
-2/5*sqrt(c*x^2 + b*x)*B*c^2/(b*x) + 4/35*sqrt(c*x^2 + b*x)*A*c^3/(b^2*x) + 1/5*sqrt(c*x^2 + b*x)*B*c/x^2 - 2/35*sqrt(c*x^2 + b*x)*A*c^2/(b*x^2) + 3 /5*sqrt(c*x^2 + b*x)*B*b/x^3 + 3/70*sqrt(c*x^2 + b*x)*A*c/x^3 - (c*x^2 + b *x)^(3/2)*B/x^4 + 3/14*sqrt(c*x^2 + b*x)*A*b/x^4 - 1/2*(c*x^2 + b*x)^(3/2) *A/x^5
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (49) = 98\).
Time = 0.19 (sec) , antiderivative size = 311, normalized size of antiderivative = 5.46 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=\frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B c^{2} + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b c^{\frac {3}{2}} + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A c^{\frac {5}{2}} + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{2} c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b c^{2} + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{3} \sqrt {c} + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{2} c^{\frac {3}{2}} + 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{4} + 98 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{3} c + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{4} \sqrt {c} + 5 \, A b^{5}\right )}}{35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x, algorithm="giac")
Output:
2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*c^2 + 70*(sqrt(c)*x - sqrt(c* x^2 + b*x))^5*B*b*c^(3/2) + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*c^(5/2) + 70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^2*c + 105*(sqrt(c)*x - sqrt(c* x^2 + b*x))^4*A*b*c^2 + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*sqrt(c) + 140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*c^(3/2) + 7*(sqrt(c)*x - sq rt(c*x^2 + b*x))^2*B*b^4 + 98*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*c + 35*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^4*sqrt(c) + 5*A*b^5)/(sqrt(c)*x - s qrt(c*x^2 + b*x))^7
Time = 5.83 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.49 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=\frac {4\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x}-\frac {16\,A\,c\,\sqrt {c\,x^2+b\,x}}{35\,x^3}-\frac {2\,B\,b\,\sqrt {c\,x^2+b\,x}}{5\,x^3}-\frac {4\,B\,c\,\sqrt {c\,x^2+b\,x}}{5\,x^2}-\frac {2\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^2}-\frac {2\,A\,b\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {2\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{5\,b\,x} \] Input:
int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^6,x)
Output:
(4*A*c^3*(b*x + c*x^2)^(1/2))/(35*b^2*x) - (16*A*c*(b*x + c*x^2)^(1/2))/(3 5*x^3) - (2*B*b*(b*x + c*x^2)^(1/2))/(5*x^3) - (4*B*c*(b*x + c*x^2)^(1/2)) /(5*x^2) - (2*A*c^2*(b*x + c*x^2)^(1/2))/(35*b*x^2) - (2*A*b*(b*x + c*x^2) ^(1/2))/(7*x^4) - (2*B*c^2*(b*x + c*x^2)^(1/2))/(5*b*x)
Time = 0.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.60 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^6} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{3}}{7}-\frac {16 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{2} c x}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +b}\, a b \,c^{2} x^{2}}{35}+\frac {4 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{3} x^{3}}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +b}\, b^{4} x}{5}-\frac {4 \sqrt {x}\, \sqrt {c x +b}\, b^{3} c \,x^{2}}{5}-\frac {2 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c^{2} x^{3}}{5}-\frac {4 \sqrt {c}\, a \,c^{3} x^{4}}{35}-\frac {6 \sqrt {c}\, b^{2} c^{2} x^{4}}{35}}{b^{2} x^{4}} \] Input:
int((B*x+A)*(c*x^2+b*x)^(3/2)/x^6,x)
Output:
(2*( - 5*sqrt(x)*sqrt(b + c*x)*a*b**3 - 8*sqrt(x)*sqrt(b + c*x)*a*b**2*c*x - sqrt(x)*sqrt(b + c*x)*a*b*c**2*x**2 + 2*sqrt(x)*sqrt(b + c*x)*a*c**3*x* *3 - 7*sqrt(x)*sqrt(b + c*x)*b**4*x - 14*sqrt(x)*sqrt(b + c*x)*b**3*c*x**2 - 7*sqrt(x)*sqrt(b + c*x)*b**2*c**2*x**3 - 2*sqrt(c)*a*c**3*x**4 - 3*sqrt (c)*b**2*c**2*x**4))/(35*b**2*x**4)