Integrand size = 24, antiderivative size = 146 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 (b B-A c) \sqrt {x}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 b B-5 A c}{3 b^2 c \sqrt {x} \sqrt {b x+c x^2}}+\frac {(2 b B-5 A c) \sqrt {x}}{b^3 \sqrt {b x+c x^2}}-\frac {(2 b B-5 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{7/2}} \]
-(-5*A*c+2*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-2/3*(-A *c+B*b)*x^(1/2)/b/c/(c*x^2+b*x)^(3/2)+1/3*(-5*A*c+2*B*b)/b^2/c/x^(1/2)/(c* x^2+b*x)^(1/2)+(-5*A*c+2*B*b)*x^(1/2)/b^3/(c*x^2+b*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {\sqrt {x} \left (\sqrt {b} \left (2 b B x (4 b+3 c x)-A \left (3 b^2+20 b c x+15 c^2 x^2\right )\right )-3 (2 b B-5 A c) x (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{3 b^{7/2} (x (b+c x))^{3/2}} \]
(Sqrt[x]*(Sqrt[b]*(2*b*B*x*(4*b + 3*c*x) - A*(3*b^2 + 20*b*c*x + 15*c^2*x^ 2)) - 3*(2*b*B - 5*A*c)*x*(b + c*x)^(3/2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])) /(3*b^(7/2)*(x*(b + c*x))^(3/2))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1218, 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.
| \(\displaystyle \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1218 |
| \(\displaystyle -\frac {(2 b B-5 A c) \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{3/2}}dx}{3 b c}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1135 |
| \(\displaystyle -\frac {(2 b B-5 A c) \left (-\frac {3 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b c}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1132 |
| \(\displaystyle -\frac {(2 b B-5 A c) \left (-\frac {3 c \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b c}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1136 |
| \(\displaystyle -\frac {(2 b B-5 A c) \left (-\frac {3 c \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b c}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {(2 b B-5 A c) \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b c}-\frac {2 \sqrt {x} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
(-2*(b*B - A*c)*Sqrt[x])/(3*b*c*(b*x + c*x^2)^(3/2)) - ((2*b*B - 5*A*c)*(- (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]*Sqrt[x])])/b^(3/2)))/(2*b)))/(3* b*c)
3.3.47.3.1 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[(g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(( a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Simp[e*((m*(g*(c* d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e))) I nt[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d , e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(-\frac {A \left (c x +b \right )}{b^{3} \sqrt {x}\, \sqrt {x \left (c x +b \right )}}-\frac {\left (-\frac {2 \left (5 A c -2 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \left (-4 A c +2 B b \right )}{\sqrt {c x +b}}+\frac {4 b \left (A c -B b \right )}{3 \left (c x +b \right )^{\frac {3}{2}}}\right ) \sqrt {c x +b}\, \sqrt {x}}{2 b^{3} \sqrt {x \left (c x +b \right )}}\) | \(112\) |
| default | \(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-6 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c \,x^{2}+15 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c x \sqrt {c x +b}-15 A \sqrt {b}\, c^{2} x^{2}-6 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b^{2} x \sqrt {c x +b}+6 B \,b^{\frac {3}{2}} c \,x^{2}-20 A \,b^{\frac {3}{2}} c x +8 B \,b^{\frac {5}{2}} x -3 A \,b^{\frac {5}{2}}\right )}{3 x^{\frac {3}{2}} \left (c x +b \right )^{2} b^{\frac {7}{2}}}\) | \(175\) |
-1/b^3*A*(c*x+b)/x^(1/2)/(x*(c*x+b))^(1/2)-1/2/b^3*(-2*(5*A*c-2*B*b)/b^(1/ 2)*arctanh((c*x+b)^(1/2)/b^(1/2))-2*(-4*A*c+2*B*b)/(c*x+b)^(1/2)+4/3*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 = 0.29 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.51 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (3 \, A b^{3} - 3 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \, {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{6 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + 2 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (3 \, A b^{3} - 3 \, {\left (2 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} - 4 \, {\left (2 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}\right ] \]
[-1/6*(3*((2*B*b*c^2 - 5*A*c^3)*x^4 + 2*(2*B*b^2*c - 5*A*b*c^2)*x^3 + (2*B *b^3 - 5*A*b^2*c)*x^2)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*s qrt(b)*sqrt(x))/x^2) + 2*(3*A*b^3 - 3*(2*B*b^2*c - 5*A*b*c^2)*x^2 - 4*(2*B *b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2), 1/3*(3*((2*B*b*c^2 - 5*A*c^3)*x^4 + 2*(2*B*b^2*c - 5*A*b*c^2) *x^3 + (2*B*b^3 - 5*A*b^2*c)*x^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c* x^2 + b*x)) - (3*A*b^3 - 3*(2*B*b^2*c - 5*A*b*c^2)*x^2 - 4*(2*B*b^3 - 5*A* b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2) ]
\[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {x} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} \sqrt {x}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, B b - 5 \, A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} - \frac {\sqrt {c x + b} A}{b^{3} x} + \frac {2 \, {\left (3 \, {\left (c x + b\right )} B b + B b^{2} - 6 \, {\left (c x + b\right )} A c - A b c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}} \]
(2*B*b - 5*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - sqrt(c*x + b)*A/(b^3*x) + 2/3*(3*(c*x + b)*B*b + B*b^2 - 6*(c*x + b)*A*c - A*b*c)/(( c*x + b)^(3/2)*b^3)
Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {x}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \]