Integrand size = 22, antiderivative size = 125 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=-\frac {2 A \sqrt {b x+c x^2}}{7 b x^4}-\frac {2 (7 b B-6 A c) \sqrt {b x+c x^2}}{35 b^2 x^3}+\frac {8 c (7 b B-6 A c) \sqrt {b x+c x^2}}{105 b^3 x^2}-\frac {16 c^2 (7 b B-6 A c) \sqrt {b x+c x^2}}{105 b^4 x} \]
-2/7*A*(c*x^2+b*x)^(1/2)/b/x^4-2/35*(-6*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/b^2/x ^3+8/105*c*(-6*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^2-16/105*c^2*(-6*A*c+7*B *b)*(c*x^2+b*x)^(1/2)/b^4/x
Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (7 b B x \left (3 b^2-4 b c x+8 c^2 x^2\right )+3 A \left (5 b^3-6 b^2 c x+8 b c^2 x^2-16 c^3 x^3\right )\right )}{105 b^4 x^4} \]
(-2*Sqrt[x*(b + c*x)]*(7*b*B*x*(3*b^2 - 4*b*c*x + 8*c^2*x^2) + 3*A*(5*b^3 - 6*b^2*c*x + 8*b*c^2*x^2 - 16*c^3*x^3)))/(105*b^4*x^4)
Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1220, 1129, 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.
\(\displaystyle \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(7 b B-6 A c) \int \frac {1}{x^3 \sqrt {c x^2+b x}}dx}{7 b}-\frac {2 A \sqrt {b x+c x^2}}{7 b x^4}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {(7 b B-6 A c) \left (-\frac {4 c \int \frac {1}{x^2 \sqrt {c x^2+b x}}dx}{5 b}-\frac {2 \sqrt {b x+c x^2}}{5 b x^3}\right )}{7 b}-\frac {2 A \sqrt {b x+c x^2}}{7 b x^4}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {(7 b B-6 A c) \left (-\frac {4 c \left (-\frac {2 c \int \frac {1}{x \sqrt {c x^2+b x}}dx}{3 b}-\frac {2 \sqrt {b x+c x^2}}{3 b x^2}\right )}{5 b}-\frac {2 \sqrt {b x+c x^2}}{5 b x^3}\right )}{7 b}-\frac {2 A \sqrt {b x+c x^2}}{7 b x^4}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {\left (-\frac {4 c \left (\frac {4 c \sqrt {b x+c x^2}}{3 b^2 x}-\frac {2 \sqrt {b x+c x^2}}{3 b x^2}\right )}{5 b}-\frac {2 \sqrt {b x+c x^2}}{5 b x^3}\right ) (7 b B-6 A c)}{7 b}-\frac {2 A \sqrt {b x+c x^2}}{7 b x^4}\) |
(-2*A*Sqrt[b*x + c*x^2])/(7*b*x^4) + ((7*b*B - 6*A*c)*((-2*Sqrt[b*x + c*x^ 2])/(5*b*x^3) - (4*c*((-2*Sqrt[b*x + c*x^2])/(3*b*x^2) + (4*c*Sqrt[b*x + c *x^2])/(3*b^2*x)))/(5*b)))/(7*b)
3.2.17.3.1 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 = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.53
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\frac {7 B x}{5}+A \right ) b^{3}-\frac {6 c x \left (\frac {14 B x}{9}+A \right ) b^{2}}{5}+\frac {8 c^{2} \left (\frac {7 B x}{3}+A \right ) x^{2} b}{5}-\frac {16 A \,c^{3} x^{3}}{5}\right ) \sqrt {x \left (c x +b \right )}}{7 x^{4} b^{4}}\) | \(66\) |
trager | \(-\frac {2 \left (-48 A \,c^{3} x^{3}+56 B b \,c^{2} x^{3}+24 A b \,c^{2} x^{2}-28 B \,b^{2} c \,x^{2}-18 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right ) \sqrt {c \,x^{2}+b x}}{105 x^{4} b^{4}}\) | \(81\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (-48 A \,c^{3} x^{3}+56 B b \,c^{2} x^{3}+24 A b \,c^{2} x^{2}-28 B \,b^{2} c \,x^{2}-18 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right )}{105 b^{4} x^{3} \sqrt {x \left (c x +b \right )}}\) | \(84\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-48 A \,c^{3} x^{3}+56 B b \,c^{2} x^{3}+24 A b \,c^{2} x^{2}-28 B \,b^{2} c \,x^{2}-18 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right )}{105 x^{3} b^{4} \sqrt {c \,x^{2}+b x}}\) | \(86\) |
default | \(B \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}-\frac {4 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{3 b \,x^{2}}+\frac {4 c \sqrt {c \,x^{2}+b x}}{3 b^{2} x}\right )}{5 b}\right )+A \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{7 b \,x^{4}}-\frac {6 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}-\frac {4 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{3 b \,x^{2}}+\frac {4 c \sqrt {c \,x^{2}+b x}}{3 b^{2} x}\right )}{5 b}\right )}{7 b}\right )\) | \(164\) |
-2/7*((7/5*B*x+A)*b^3-6/5*c*x*(14/9*B*x+A)*b^2+8/5*c^2*(7/3*B*x+A)*x^2*b-1 6/5*A*c^3*x^3)*(x*(c*x+b))^(1/2)/x^4/b^4
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=-\frac {2 \, {\left (15 \, A b^{3} + 8 \, {\left (7 \, B b c^{2} - 6 \, A c^{3}\right )} x^{3} - 4 \, {\left (7 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{2} + 3 \, {\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, b^{4} x^{4}} \]
-2/105*(15*A*b^3 + 8*(7*B*b*c^2 - 6*A*c^3)*x^3 - 4*(7*B*b^2*c - 6*A*b*c^2) *x^2 + 3*(7*B*b^3 - 6*A*b^2*c)*x)*sqrt(c*x^2 + b*x)/(b^4*x^4)
\[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=\int \frac {A + B x}{x^{4} \sqrt {x \left (b + c x\right )}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=-\frac {16 \, \sqrt {c x^{2} + b x} B c^{2}}{15 \, b^{3} x} + \frac {32 \, \sqrt {c x^{2} + b x} A c^{3}}{35 \, b^{4} x} + \frac {8 \, \sqrt {c x^{2} + b x} B c}{15 \, b^{2} x^{2}} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{2}}{35 \, b^{3} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{5 \, b x^{3}} + \frac {12 \, \sqrt {c x^{2} + b x} A c}{35 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{7 \, b x^{4}} \]
-16/15*sqrt(c*x^2 + b*x)*B*c^2/(b^3*x) + 32/35*sqrt(c*x^2 + b*x)*A*c^3/(b^ 4*x) + 8/15*sqrt(c*x^2 + b*x)*B*c/(b^2*x^2) - 16/35*sqrt(c*x^2 + b*x)*A*c^ 2/(b^3*x^2) - 2/5*sqrt(c*x^2 + b*x)*B/(b*x^3) + 12/35*sqrt(c*x^2 + b*x)*A* c/(b^2*x^3) - 2/7*sqrt(c*x^2 + b*x)*A/(b*x^4)
Time = 0.29 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b \sqrt {c} + 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A c^{\frac {3}{2}} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} + 252 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} \sqrt {c} + 15 \, A b^{3}\right )}}{105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]
2/105*(140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*c + 105*(sqrt(c)*x - sqrt(c *x^2 + b*x))^3*B*b*sqrt(c) + 210*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^(3/ 2) + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2 + 252*(sqrt(c)*x - sqrt(c* x^2 + b*x))^2*A*b*c + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*sqrt(c) + 15*A*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^7
Time = 10.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}\,\left (96\,A\,c^3-112\,B\,b\,c^2\right )}{105\,b^4\,x}-\frac {\left (48\,A\,c^2-56\,B\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x^2}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{7\,b\,x^4}+\frac {\sqrt {c\,x^2+b\,x}\,\left (12\,A\,c-14\,B\,b\right )}{35\,b^2\,x^3} \]