Integrand size = 31, antiderivative size = 94 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 (b d-a e)^3}{e^4 \sqrt {d+e x}}+\frac {6 b (b d-a e)^2 \sqrt {d+e x}}{e^4}-\frac {2 b^2 (b d-a e) (d+e x)^{3/2}}{e^4}+\frac {2 b^3 (d+e x)^{5/2}}{5 e^4} \]
-2*b^2*(-a*e+b*d)*(e*x+d)^(3/2)/e^4+2/5*b^3*(e*x+d)^(5/2)/e^4+2*(-a*e+b*d) ^3/e^4/(e*x+d)^(1/2)+6*b*(-a*e+b*d)^2*(e*x+d)^(1/2)/e^4
Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \left (-5 a^3 e^3+15 a^2 b e^2 (2 d+e x)+5 a b^2 e \left (-8 d^2-4 d e x+e^2 x^2\right )+b^3 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{5 e^4 \sqrt {d+e x}} \]
(2*(-5*a^3*e^3 + 15*a^2*b*e^2*(2*d + e*x) + 5*a*b^2*e*(-8*d^2 - 4*d*e*x + e^2*x^2) + b^3*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)))/(5*e^4*Sqrt[ d + e*x])
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 53, 2009}
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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{(d+e x)^{3/2}}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{(d+e x)^{3/2}}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 \sqrt {d+e x} (b d-a e)}{e^3}+\frac {3 b (b d-a e)^2}{e^3 \sqrt {d+e x}}+\frac {(a e-b d)^3}{e^3 (d+e x)^{3/2}}+\frac {b^3 (d+e x)^{3/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^2 (d+e x)^{3/2} (b d-a e)}{e^4}+\frac {6 b \sqrt {d+e x} (b d-a e)^2}{e^4}+\frac {2 (b d-a e)^3}{e^4 \sqrt {d+e x}}+\frac {2 b^3 (d+e x)^{5/2}}{5 e^4}\) |
(2*(b*d - a*e)^3)/(e^4*Sqrt[d + e*x]) + (6*b*(b*d - a*e)^2*Sqrt[d + e*x])/ e^4 - (2*b^2*(b*d - a*e)*(d + e*x)^(3/2))/e^4 + (2*b^3*(d + e*x)^(5/2))/(5 *e^4)
3.21.47.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (e^{3} x^{3}-2 d \,e^{2} x^{2}+8 d^{2} e x +16 d^{3}\right ) b^{3}}{5}-16 e a \left (-\frac {1}{8} e^{2} x^{2}+\frac {1}{2} d e x +d^{2}\right ) b^{2}+12 e^{2} \left (\frac {e x}{2}+d \right ) a^{2} b -2 a^{3} e^{3}}{\sqrt {e x +d}\, e^{4}}\) | \(94\) |
risch | \(\frac {2 b \left (b^{2} e^{2} x^{2}+5 a b \,e^{2} x -3 b^{2} d e x +15 e^{2} a^{2}-25 a b d e +11 b^{2} d^{2}\right ) \sqrt {e x +d}}{5 e^{4}}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{e^{4} \sqrt {e x +d}}\) | \(112\) |
gosper | \(-\frac {2 \left (-b^{3} x^{3} e^{3}-5 x^{2} a \,b^{2} e^{3}+2 x^{2} b^{3} d \,e^{2}-15 x \,a^{2} b \,e^{3}+20 x a \,b^{2} d \,e^{2}-8 x \,b^{3} d^{2} e +5 a^{3} e^{3}-30 a^{2} b d \,e^{2}+40 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{5 \sqrt {e x +d}\, e^{4}}\) | \(116\) |
trager | \(-\frac {2 \left (-b^{3} x^{3} e^{3}-5 x^{2} a \,b^{2} e^{3}+2 x^{2} b^{3} d \,e^{2}-15 x \,a^{2} b \,e^{3}+20 x a \,b^{2} d \,e^{2}-8 x \,b^{3} d^{2} e +5 a^{3} e^{3}-30 a^{2} b d \,e^{2}+40 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{5 \sqrt {e x +d}\, e^{4}}\) | \(116\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}-2 b^{3} d \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} b \,e^{2} \sqrt {e x +d}-12 a \,b^{2} d e \sqrt {e x +d}+6 b^{3} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(136\) |
default | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}-2 b^{3} d \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} b \,e^{2} \sqrt {e x +d}-12 a \,b^{2} d e \sqrt {e x +d}+6 b^{3} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(136\) |
2/5*((e^3*x^3-2*d*e^2*x^2+8*d^2*e*x+16*d^3)*b^3-40*e*a*(-1/8*e^2*x^2+1/2*d *e*x+d^2)*b^2+30*e^2*(1/2*e*x+d)*a^2*b-5*a^3*e^3)/(e*x+d)^(1/2)/e^4
Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - {\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{5} x + d e^{4}\right )}} \]
2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^ 3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2 + (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a ^2*b*e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)
Time = 2.31 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (3 a b^{2} e - 3 b^{3} d\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (3 a^{2} b e^{2} - 6 a b^{2} d e + 3 b^{3} d^{2}\right )}{e^{3}} - \frac {\left (a e - b d\right )^{3}}{e^{3} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {a^{3} b x + \frac {3 a^{2} b^{2} x^{2}}{2} + a b^{3} x^{3} + \frac {b^{4} x^{4}}{4}}{b} & \text {otherwise} \end {cases}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(b**3*(d + e*x)**(5/2)/(5*e**3) + (d + e*x)**(3/2)*(3*a*b**2* e - 3*b**3*d)/(3*e**3) + sqrt(d + e*x)*(3*a**2*b*e**2 - 6*a*b**2*d*e + 3*b **3*d**2)/e**3 - (a*e - b*d)**3/(e**3*sqrt(d + e*x)))/e, Ne(e, 0)), (Piece wise((a**3*x, Eq(b, 0)), ((a**3*b*x + 3*a**2*b**2*x**2/2 + a*b**3*x**3 + b **4*x**4/4)/b, True))/d**(3/2), True))
Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}} b^{3} - 5 \, {\left (b^{3} d - a b^{2} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} \sqrt {e x + d}}{e^{3}} + \frac {5 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}}{\sqrt {e x + d} e^{3}}\right )}}{5 \, e} \]
2/5*(((e*x + d)^(5/2)*b^3 - 5*(b^3*d - a*b^2*e)*(e*x + d)^(3/2) + 15*(b^3* d^2 - 2*a*b^2*d*e + a^2*b*e^2)*sqrt(e*x + d))/e^3 + 5*(b^3*d^3 - 3*a*b^2*d ^2*e + 3*a^2*b*d*e^2 - a^3*e^3)/(sqrt(e*x + d)*e^3))/e
Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {5}{2}} b^{3} e^{16} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d e^{16} + 15 \, \sqrt {e x + d} b^{3} d^{2} e^{16} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} e^{17} - 30 \, \sqrt {e x + d} a b^{2} d e^{17} + 15 \, \sqrt {e x + d} a^{2} b e^{18}\right )}}{5 \, e^{20}} \]
2*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)/(sqrt(e*x + d)*e^4) + 2/5*((e*x + d)^(5/2)*b^3*e^16 - 5*(e*x + d)^(3/2)*b^3*d*e^16 + 15*sqrt(e *x + d)*b^3*d^2*e^16 + 5*(e*x + d)^(3/2)*a*b^2*e^17 - 30*sqrt(e*x + d)*a*b ^2*d*e^17 + 15*sqrt(e*x + d)*a^2*b*e^18)/e^20
Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2\,b^3\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}-\frac {\left (6\,b^3\,d-6\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4}-\frac {2\,a^3\,e^3-6\,a^2\,b\,d\,e^2+6\,a\,b^2\,d^2\,e-2\,b^3\,d^3}{e^4\,\sqrt {d+e\,x}}+\frac {6\,b\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}}{e^4} \]