Integrand size = 24, antiderivative size = 105 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=-\frac {(4 b B-A c) \sqrt {b x+c x^2}}{4 b x^{3/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}-\frac {c (4 b B-A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{3/2}} \]
-1/2*A*(c*x^2+b*x)^(3/2)/b/x^(7/2)-1/4*c*(-A*c+4*B*b)*arctanh((c*x^2+b*x)^ (1/2)/b^(1/2)/x^(1/2))/b^(3/2)-1/4*(-A*c+4*B*b)*(c*x^2+b*x)^(1/2)/b/x^(3/2 )
Time = 0.13 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} (2 A b+4 b B x+A c x)+c (4 b B-A c) x^2 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{4 b^{3/2} x^{5/2} \sqrt {b+c x}} \]
-1/4*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(2*A*b + 4*b*B*x + A*c*x) + c*(4*b*B - A*c)*x^2*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(b^(3/2)*x^(5/2)*Sqr t[b + c*x])
Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1220, 1130, 1136, 220}
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.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(4 b B-A c) \int \frac {\sqrt {c x^2+b x}}{x^{5/2}}dx}{4 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {(4 b B-A c) \left (\frac {1}{2} c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )}{4 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {(4 b B-A c) \left (c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )}{4 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {(4 b B-A c) \left (-\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}}-\frac {\sqrt {b x+c x^2}}{x^{3/2}}\right )}{4 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{2 b x^{7/2}}\) |
-1/2*(A*(b*x + c*x^2)^(3/2))/(b*x^(7/2)) + ((4*b*B - A*c)*(-(Sqrt[b*x + c* x^2]/x^(3/2)) - (c*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/Sqrt[b])) /(4*b)
3.3.1.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^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 = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {\left (c x +b \right ) \left (A c x +4 B b x +2 A b \right )}{4 x^{\frac {3}{2}} b \sqrt {x \left (c x +b \right )}}+\frac {c \left (A c -4 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{4 b^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}\) | \(83\) |
default | \(\frac {\sqrt {x \left (c x +b \right )}\, \left (A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-4 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c \,x^{2}-A c x \sqrt {b}\, \sqrt {c x +b}-4 B \,b^{\frac {3}{2}} x \sqrt {c x +b}-2 A \,b^{\frac {3}{2}} \sqrt {c x +b}\right )}{4 b^{\frac {3}{2}} x^{\frac {5}{2}} \sqrt {c x +b}}\) | \(108\) |
-1/4*(c*x+b)*(A*c*x+4*B*b*x+2*A*b)/x^(3/2)/b/(x*(c*x+b))^(1/2)+1/4*c*(A*c- 4*B*b)/b^(3/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c* x+b))^(1/2)
Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.78 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=\left [-\frac {{\left (4 \, B b c - A c^{2}\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (2 \, A b^{2} + {\left (4 \, B b^{2} + A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b^{2} x^{3}}, \frac {{\left (4 \, B b c - A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (2 \, A b^{2} + {\left (4 \, B b^{2} + A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b^{2} x^{3}}\right ] \]
[-1/8*((4*B*b*c - A*c^2)*sqrt(b)*x^3*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*A*b^2 + (4*B*b^2 + A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^3), 1/4*((4*B*b*c - A*c^2)*sqrt(-b)*x^3*arctan(sqrt (-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (2*A*b^2 + (4*B*b^2 + A*b*c)*x)*sqrt(c*x ^2 + b*x)*sqrt(x))/(b^2*x^3)]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac {7}{2}}}\, dx \]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x} {\left (B x + A\right )}}{x^{\frac {7}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=\frac {\frac {{\left (4 \, B b c^{2} - A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{2} - 4 \, \sqrt {c x + b} B b^{2} c^{2} + {\left (c x + b\right )}^{\frac {3}{2}} A c^{3} + \sqrt {c x + b} A b c^{3}}{b c^{2} x^{2}}}{4 \, c} \]
1/4*((4*B*b*c^2 - A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - (4* (c*x + b)^(3/2)*B*b*c^2 - 4*sqrt(c*x + b)*B*b^2*c^2 + (c*x + b)^(3/2)*A*c^ 3 + sqrt(c*x + b)*A*b*c^3)/(b*c^2*x^2))/c
Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{7/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^{7/2}} \,d x \]