Integrand size = 19, antiderivative size = 117 \[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}+\frac {5 c}{4 b^2 \sqrt {x} \sqrt {b x+c x^2}}+\frac {15 c^2 \sqrt {x}}{4 b^3 \sqrt {b x+c x^2}}-\frac {15 c^2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{7/2}} \]
-15/4*c^2*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-1/2/b/x^(3/2) /(c*x^2+b*x)^(1/2)+5/4*c/b^2/x^(1/2)/(c*x^2+b*x)^(1/2)+15/4*c^2*x^(1/2)/b^ 3/(c*x^2+b*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \left (-2 b^2+5 b c x+15 c^2 x^2\right )-15 c^2 x^2 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{4 b^{7/2} x^{3/2} \sqrt {x (b+c x)}} \]
(Sqrt[b]*(-2*b^2 + 5*b*c*x + 15*c^2*x^2) - 15*c^2*x^2*Sqrt[b + c*x]*ArcTan h[Sqrt[b + c*x]/Sqrt[b]])/(4*b^(7/2)*x^(3/2)*Sqrt[x*(b + c*x)])
Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1135, 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.
\(\displaystyle \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle -\frac {5 c \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{3/2}}dx}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle -\frac {5 c \left (-\frac {3 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1132 |
\(\displaystyle -\frac {5 c \left (-\frac {3 c \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle -\frac {5 c \left (-\frac {3 c \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {5 c \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\) |
-1/2*1/(b*x^(3/2)*Sqrt[b*x + c*x^2]) - (5*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)))/(4*b)
3.2.10.3.1 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]
Time = 2.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65
method | result | size |
default | \(-\frac {\sqrt {x \left (c x +b \right )}\, \left (15 \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-5 b^{\frac {3}{2}} c x -15 c^{2} x^{2} \sqrt {b}+2 b^{\frac {5}{2}}\right )}{4 x^{\frac {5}{2}} \left (c x +b \right ) b^{\frac {7}{2}}}\) | \(76\) |
risch | \(-\frac {\left (c x +b \right ) \left (-7 c x +2 b \right )}{4 b^{3} x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c^{2} \left (\frac {16}{\sqrt {c x +b}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{8 b^{3} \sqrt {x \left (c x +b \right )}}\) | \(86\) |
-1/4/x^(5/2)*(x*(c*x+b))^(1/2)*(15*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^( 1/2))*c^2*x^2-5*b^(3/2)*c*x-15*c^2*x^2*b^(1/2)+2*b^(5/2))/(c*x+b)/b^(7/2)
Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.86 \[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (c^{3} x^{4} + b c^{2} x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, \frac {15 \, {\left (c^{3} x^{4} + b c^{2} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \]
[1/8*(15*(c^3*x^4 + b*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*(15*b*c^2*x^2 + 5*b^2*c*x - 2*b^3)*sqrt(c *x^2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3), 1/4*(15*(c^3*x^4 + b*c^2*x^3)* sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^2*x^2 + 5*b^ 2*c*x - 2*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3)]
\[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {15 \, c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{3}} + \frac {2 \, c^{2}}{\sqrt {c x + b} b^{3}} + \frac {7 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{2} - 9 \, \sqrt {c x + b} b c^{2}}{4 \, b^{3} c^{2} x^{2}} \]
15/4*c^2*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + 2*c^2/(sqrt(c*x + b)*b^3) + 1/4*(7*(c*x + b)^(3/2)*c^2 - 9*sqrt(c*x + b)*b*c^2)/(b^3*c^2*x^ 2)
Timed out. \[ \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]