Integrand size = 24, antiderivative size = 174 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}+\frac {(4 b B-7 A c) \sqrt {x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {5 (4 b B-7 A c)}{12 b^3 \sqrt {x} \sqrt {b x+c x^2}}-\frac {5 c (4 b B-7 A c) \sqrt {x}}{4 b^4 \sqrt {b x+c x^2}}+\frac {5 c (4 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{9/2}} \]
5/4*c*(-7*A*c+4*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(9/2)-1/ 2*A/b/(c*x^2+b*x)^(3/2)/x^(1/2)+1/6*(-7*A*c+4*B*b)*x^(1/2)/b^2/(c*x^2+b*x) ^(3/2)-5/12*(-7*A*c+4*B*b)/b^3/x^(1/2)/(c*x^2+b*x)^(1/2)-5/4*c*(-7*A*c+4*B *b)*x^(1/2)/b^4/(c*x^2+b*x)^(1/2)
Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=\frac {\sqrt {b} \left (-4 b B x \left (3 b^2+20 b c x+15 c^2 x^2\right )+A \left (-6 b^3+21 b^2 c x+140 b c^2 x^2+105 c^3 x^3\right )\right )+15 c (4 b B-7 A c) x^2 (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{12 b^{9/2} \sqrt {x} (x (b+c x))^{3/2}} \]
(Sqrt[b]*(-4*b*B*x*(3*b^2 + 20*b*c*x + 15*c^2*x^2) + A*(-6*b^3 + 21*b^2*c* x + 140*b*c^2*x^2 + 105*c^3*x^3)) + 15*c*(4*b*B - 7*A*c)*x^2*(b + c*x)^(3/ 2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(12*b^(9/2)*Sqrt[x]*(x*(b + c*x))^(3/2) )
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1220, 1132, 1135, 1132, 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.
| \(\displaystyle \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(4 b B-7 A c) \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{5/2}}dx}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1132 |
| \(\displaystyle \frac {(4 b B-7 A c) \left (\frac {5 \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{3/2}}dx}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1135 |
| \(\displaystyle \frac {(4 b B-7 A c) \left (\frac {5 \left (-\frac {3 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1132 |
| \(\displaystyle \frac {(4 b B-7 A c) \left (\frac {5 \left (-\frac {3 c \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1136 |
| \(\displaystyle \frac {(4 b B-7 A c) \left (\frac {5 \left (-\frac {3 c \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {(4 b B-7 A c) \left (\frac {5 \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {A}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\) |
-1/2*A/(b*Sqrt[x]*(b*x + c*x^2)^(3/2)) + ((4*b*B - 7*A*c)*((2*Sqrt[x])/(3* b*(b*x + c*x^2)^(3/2)) + (5*(-(1/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) - (3*c*((2 *Sqrt[x])/(b*Sqrt[b*x + c*x^2]) - (2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sq rt[x])])/b^(3/2)))/(2*b)))/(3*b)))/(4*b)
3.3.48.3.1 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[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ [0, m, 1] && IntegerQ[2*p]
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*((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, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[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 = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.73
| method | result | size |
| risch | \(-\frac {\left (c x +b \right ) \left (-11 A c x +4 B b x +2 A b \right )}{4 b^{4} x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c \left (-\frac {2 \left (35 A c -20 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \left (-24 A c +16 B b \right )}{\sqrt {c x +b}}+\frac {16 b \left (A c -B b \right )}{3 \left (c x +b \right )^{\frac {3}{2}}}\right ) \sqrt {c x +b}\, \sqrt {x}}{8 b^{4} \sqrt {x \left (c x +b \right )}}\) | \(127\) |
| default | \(-\frac {\sqrt {x \left (c x +b \right )}\, \left (105 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{3} x^{3}-60 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{2} x^{3}+105 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{2} x^{2} \sqrt {c x +b}-105 A \sqrt {b}\, c^{3} x^{3}-60 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b^{2} c \,x^{2} \sqrt {c x +b}+60 B \,b^{\frac {3}{2}} c^{2} x^{3}-140 A \,b^{\frac {3}{2}} c^{2} x^{2}+80 B \,b^{\frac {5}{2}} c \,x^{2}-21 A \,b^{\frac {5}{2}} c x +12 B \,b^{\frac {7}{2}} x +6 A \,b^{\frac {7}{2}}\right )}{12 x^{\frac {5}{2}} \left (c x +b \right )^{2} b^{\frac {9}{2}}}\) | \(208\) |
-1/4*(c*x+b)*(-11*A*c*x+4*B*b*x+2*A*b)/b^4/x^(3/2)/(x*(c*x+b))^(1/2)+1/8/b ^4*c*(-2*(35*A*c-20*B*b)/b^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))-2*(-24*A*c +16*B*b)/(c*x+b)^(1/2)+16/3*b*(A*c-B*b)/(c*x+b)^(3/2))*(c*x+b)^(1/2)*x^(1/ 2)/(x*(c*x+b))^(1/2)
Time = 0.33 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.44 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} + {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (6 \, A b^{4} + 15 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \, {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \, {\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac {15 \, {\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} + {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (6 \, A b^{4} + 15 \, {\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \, {\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \, {\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{12 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \]
[-1/24*(15*((4*B*b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c - 7*A*b^2*c^2)*x^3)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(6*A*b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^3)*x ^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c* x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5 + 2*b^6*c*x^4 + b^7*x^3), -1/12*(15*((4*B *b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c - 7*A *b^2*c^2)*x^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (6*A* b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^3)*x^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5 + 2*b ^6*c*x^4 + b^7*x^3)]
Timed out. \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}} \sqrt {x}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {5 \, {\left (4 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{4}} - \frac {2 \, {\left (6 \, {\left (c x + b\right )} B b c + B b^{2} c - 9 \, {\left (c x + b\right )} A c^{2} - A b c^{2}\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c - 4 \, \sqrt {c x + b} B b^{2} c - 11 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{2} + 13 \, \sqrt {c x + b} A b c^{2}}{4 \, b^{4} c^{2} x^{2}} \]
-5/4*(4*B*b*c - 7*A*c^2)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) - 2 /3*(6*(c*x + b)*B*b*c + B*b^2*c - 9*(c*x + b)*A*c^2 - A*b*c^2)/((c*x + b)^ (3/2)*b^4) - 1/4*(4*(c*x + b)^(3/2)*B*b*c - 4*sqrt(c*x + b)*B*b^2*c - 11*( c*x + b)^(3/2)*A*c^2 + 13*sqrt(c*x + b)*A*b*c^2)/(b^4*c^2*x^2)
Timed out. \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x}{\sqrt {x}\,{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \]