Integrand size = 22, antiderivative size = 135 \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b B-A c) x^3}{b c \sqrt {b x+c x^2}}-\frac {3 (5 b B-4 A c) \sqrt {b x+c x^2}}{4 c^3}+\frac {(5 b B-4 A c) x \sqrt {b x+c x^2}}{2 b c^2}+\frac {3 b (5 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}} \]
3/4*b*(-4*A*c+5*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-2*(-A*c+ B*b)*x^3/b/c/(c*x^2+b*x)^(1/2)-3/4*(-4*A*c+5*B*b)*(c*x^2+b*x)^(1/2)/c^3+1/ 2*(-4*A*c+5*B*b)*x*(c*x^2+b*x)^(1/2)/b/c^2
Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x^{3/2} \left (\sqrt {c} \sqrt {x} (b+c x) \left (-15 b^2 B+b c (12 A-5 B x)+2 c^2 x (2 A+B x)\right )+6 b (5 b B-4 A c) (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{4 c^{7/2} (x (b+c x))^{3/2}} \]
(x^(3/2)*(Sqrt[c]*Sqrt[x]*(b + c*x)*(-15*b^2*B + b*c*(12*A - 5*B*x) + 2*c^ 2*x*(2*A + B*x)) + 6*b*(5*b*B - 4*A*c)*(b + c*x)^(3/2)*ArcTanh[(Sqrt[c]*Sq rt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(4*c^(7/2)*(x*(b + c*x))^(3/2))
Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1211, 2192, 27, 1160, 1091, 219}
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.
| \(\displaystyle \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1211 |
| \(\displaystyle \frac {\int \frac {B c^2 x^2-c (b B-A c) x+b (b B-A c)}{\sqrt {c x^2+b x}}dx}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
| \(\Big \downarrow \) 2192 |
| \(\displaystyle \frac {\frac {\int \frac {c (4 b (b B-A c)-c (7 b B-4 A c) x)}{2 \sqrt {c x^2+b x}}dx}{2 c}+\frac {1}{2} B c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {4 b (b B-A c)-c (7 b B-4 A c) x}{\sqrt {c x^2+b x}}dx+\frac {1}{2} B c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {3}{2} b (5 b B-4 A c) \int \frac {1}{\sqrt {c x^2+b x}}dx-\sqrt {b x+c x^2} (7 b B-4 A c)\right )+\frac {1}{2} B c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {\frac {1}{4} \left (3 b (5 b B-4 A c) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}-\sqrt {b x+c x^2} (7 b B-4 A c)\right )+\frac {1}{2} B c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {3 b (5 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\sqrt {b x+c x^2} (7 b B-4 A c)\right )+\frac {1}{2} B c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b x (b B-A c)}{c^3 \sqrt {b x+c x^2}}\) |
(-2*b*(b*B - A*c)*x)/(c^3*Sqrt[b*x + c*x^2]) + ((B*c*x*Sqrt[b*x + c*x^2])/ 2 + (-((7*b*B - 4*A*c)*Sqrt[b*x + c*x^2]) + (3*b*(5*b*B - 4*A*c)*ArcTanh[( Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/Sqrt[c])/4)/c^3
3.2.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2)) Int[ExpandToSum[((2*c*d - b *e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) *(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64
| method | result | size |
| pseudoelliptic | \(\frac {-3 \left (A c -\frac {5 B b}{4}\right ) \sqrt {x \left (c x +b \right )}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+x \left (3 \left (-\frac {5 B x}{12}+A \right ) b \,c^{\frac {3}{2}}+\left (\frac {B x}{2}+A \right ) x \,c^{\frac {5}{2}}-\frac {15 B \sqrt {c}\, b^{2}}{4}\right )}{\sqrt {x \left (c x +b \right )}\, c^{\frac {7}{2}}}\) | \(87\) |
| risch | \(\frac {\left (2 B c x +4 A c -7 B b \right ) x \left (c x +b \right )}{4 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {b \left (12 A \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {15 B b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {16 \left (A c -B b \right ) \sqrt {c \left (x +\frac {b}{c}\right )^{2}-b \left (x +\frac {b}{c}\right )}}{c \left (x +\frac {b}{c}\right )}\right )}{8 c^{3}}\) | \(150\) |
| default | \(B \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+A \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(268\) |
(-3*(A*c-5/4*B*b)*(x*(c*x+b))^(1/2)*b*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2)) +x*(3*(-5/12*B*x+A)*b*c^(3/2)+(1/2*B*x+A)*x*c^(5/2)-15/4*B*c^(1/2)*b^2))/( x*(c*x+b))^(1/2)/c^(7/2)
Time = 0.27 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.94 \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B c^{3} x^{2} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \]
[-1/8*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(c)*log(2*c *x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*B*c^3*x^2 - 15*B*b^2*c + 12*A *b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4), -1/4 *(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x)*sqrt(-c)*arctan(sqrt (c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*B*c^3*x^2 - 15*B*b^2*c + 12*A*b*c^2 - ( 5*B*b*c^2 - 4*A*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4)]
\[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.21 \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {B x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, B b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {A x^{2}}{\sqrt {c x^{2} + b x} c} - \frac {15 \, B b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {3 \, A b x}{\sqrt {c x^{2} + b x} c^{2}} + \frac {15 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} - \frac {3 \, A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} \]
1/2*B*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*B*b*x^2/(sqrt(c*x^2 + b*x)*c^2) + A* x^2/(sqrt(c*x^2 + b*x)*c) - 15/4*B*b^2*x/(sqrt(c*x^2 + b*x)*c^3) + 3*A*b*x /(sqrt(c*x^2 + b*x)*c^2) + 15/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)* sqrt(c))/c^(7/2) - 3/2*A*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^ (5/2)
Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B x}{c^{2}} - \frac {7 \, B b c^{5} - 4 \, A c^{6}}{c^{8}}\right )} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, {\left (B b^{3} \sqrt {c} - A b^{2} c^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{4}} \]
1/4*sqrt(c*x^2 + b*x)*(2*B*x/c^2 - (7*B*b*c^5 - 4*A*c^6)/c^8) - 3/8*(5*B*b ^2 - 4*A*b*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7 /2) - 2*(B*b^3*sqrt(c) - A*b^2*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x))* sqrt(c) + b)*c^4)
Timed out. \[ \int \frac {x^3 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]