Integrand size = 26, antiderivative size = 69 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]
1/4*(-a*e+b*d)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/b^2+1/5*e*(b^2*x^2+2*a* b*x+a^2)^(5/2)/b^2
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (10 a^3 (2 d+e x)+10 a^2 b x (3 d+2 e x)+5 a b^2 x^2 (4 d+3 e x)+b^3 x^3 (5 d+4 e x)\right )}{20 (a+b x)} \]
(x*Sqrt[(a + b*x)^2]*(10*a^3*(2*d + e*x) + 10*a^2*b*x*(3*d + 2*e*x) + 5*a* b^2*x^2*(4*d + 3*e*x) + b^3*x^3*(5*d + 4*e*x)))/(20*(a + b*x))
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1100, 1079, 17}
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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x) \, dx\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {(b d-a e) \int \left (a^2+2 b x a+b^2 x^2\right )^{3/2}dx}{b}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \int \left (x b^2+a b\right )^3dx}{b^4 (a+b x)}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)}{4 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
((b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^2) + (e*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*b^2)
3.16.58.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c *x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(b/2 + c *x)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0]
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] && EqQ[b^2 - 4*a*c, 0]
Time = 3.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30
method | result | size |
gosper | \(\frac {x \left (4 b^{3} e \,x^{4}+15 a \,b^{2} e \,x^{3}+5 b^{3} d \,x^{3}+20 a^{2} b e \,x^{2}+20 a \,b^{2} d \,x^{2}+10 a^{3} e x +30 a^{2} b d x +20 a^{3} d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\) | \(90\) |
default | \(\frac {x \left (4 b^{3} e \,x^{4}+15 a \,b^{2} e \,x^{3}+5 b^{3} d \,x^{3}+20 a^{2} b e \,x^{2}+20 a \,b^{2} d \,x^{2}+10 a^{3} e x +30 a^{2} b d x +20 a^{3} d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\) | \(90\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} e \,x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a \,b^{2} e +b^{3} d \right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a^{2} b e +3 a \,b^{2} d \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e \,a^{3}+3 d \,a^{2} b \right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} d x}{b x +a}\) | \(153\) |
1/20*x*(4*b^3*e*x^4+15*a*b^2*e*x^3+5*b^3*d*x^3+20*a^2*b*e*x^2+20*a*b^2*d*x ^2+10*a^3*e*x+30*a^2*b*d*x+20*a^3*d)*((b*x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} e x^{5} + a^{3} d x + \frac {1}{4} \, {\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} + {\left (a b^{2} d + a^{2} b e\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b d + a^{3} e\right )} x^{2} \]
1/5*b^3*e*x^5 + a^3*d*x + 1/4*(b^3*d + 3*a*b^2*e)*x^4 + (a*b^2*d + a^2*b*e )*x^3 + 1/2*(3*a^2*b*d + a^3*e)*x^2
Leaf count of result is larger than twice the leaf count of optimal. 745 vs. \(2 (65) = 130\).
Time = 0.71 (sec) , antiderivative size = 745, normalized size of antiderivative = 10.80 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {b^{2} e x^{4}}{5} + \frac {x^{3} \cdot \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b^{2}} + \frac {x^{2} \cdot \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b^{2}} + \frac {x \left (4 a^{3} b e + 6 a^{2} b^{2} d - \frac {3 a^{2} \cdot \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b^{2}} - \frac {5 a \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b}\right )}{2 b^{2}} + \frac {a^{4} e + 4 a^{3} b d - \frac {2 a^{2} \cdot \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (4 a^{3} b e + 6 a^{2} b^{2} d - \frac {3 a^{2} \cdot \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b^{2}} - \frac {5 a \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b}\right )}{2 b}}{b^{2}}\right ) + \frac {\left (\frac {a}{b} + x\right ) \left (a^{4} d - \frac {a^{2} \cdot \left (4 a^{3} b e + 6 a^{2} b^{2} d - \frac {3 a^{2} \cdot \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b^{2}} - \frac {5 a \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b}\right )}{2 b^{2}} - \frac {a \left (a^{4} e + 4 a^{3} b d - \frac {2 a^{2} \cdot \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (4 a^{3} b e + 6 a^{2} b^{2} d - \frac {3 a^{2} \cdot \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b^{2}} - \frac {5 a \left (\frac {26 a^{2} b^{2} e}{5} + 4 a b^{3} d - \frac {7 a \left (\frac {11 a b^{3} e}{5} + b^{4} d\right )}{4 b}\right )}{3 b}\right )}{2 b}\right )}{b}\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}} \left (- a e + 2 b d\right )}{10 b} + \frac {e \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{14 a b}}{a b} & \text {for}\: a b \neq 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \left (a^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(b**2*e*x**4/5 + x**3*(11*a*b* *3*e/5 + b**4*d)/(4*b**2) + x**2*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11* a*b**3*e/5 + b**4*d)/(4*b))/(3*b**2) + x*(4*a**3*b*e + 6*a**2*b**2*d - 3*a **2*(11*a*b**3*e/5 + b**4*d)/(4*b**2) - 5*a*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3*e/5 + b**4*d)/(4*b))/(3*b))/(2*b**2) + (a**4*e + 4*a**3* b*d - 2*a**2*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3*e/5 + b**4*d) /(4*b))/(3*b**2) - 3*a*(4*a**3*b*e + 6*a**2*b**2*d - 3*a**2*(11*a*b**3*e/5 + b**4*d)/(4*b**2) - 5*a*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3* e/5 + b**4*d)/(4*b))/(3*b))/(2*b))/b**2) + (a/b + x)*(a**4*d - a**2*(4*a** 3*b*e + 6*a**2*b**2*d - 3*a**2*(11*a*b**3*e/5 + b**4*d)/(4*b**2) - 5*a*(26 *a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3*e/5 + b**4*d)/(4*b))/(3*b))/( 2*b**2) - a*(a**4*e + 4*a**3*b*d - 2*a**2*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3*e/5 + b**4*d)/(4*b))/(3*b**2) - 3*a*(4*a**3*b*e + 6*a**2*b **2*d - 3*a**2*(11*a*b**3*e/5 + b**4*d)/(4*b**2) - 5*a*(26*a**2*b**2*e/5 + 4*a*b**3*d - 7*a*(11*a*b**3*e/5 + b**4*d)/(4*b))/(3*b))/(2*b))/b)*log(a/b + x)/sqrt(b**2*(a/b + x)**2), Ne(b**2, 0)), (((a**2 + 2*a*b*x)**(5/2)*(-a *e + 2*b*d)/(10*b) + e*(a**2 + 2*a*b*x)**(7/2)/(14*a*b))/(a*b), Ne(a*b, 0) ), ((d*x + e*x**2/2)*(a**2)**(3/2), True))
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (61) = 122\).
Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.81 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e}{5 \, b^{2}} \]
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*d*x - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3 /2)*a*e*x/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d/b - 1/4*(b^2*x^2 + 2 *a*b*x + a^2)^(3/2)*a^2*e/b^2 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*e/b^2
Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (61) = 122\).
Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.10 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} e x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a b^{2} e x^{4} \mathrm {sgn}\left (b x + a\right ) + a b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} b e x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{3} e x^{2} \mathrm {sgn}\left (b x + a\right ) + a^{3} d x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (5 \, a^{4} b d - a^{5} e\right )} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{2}} \]
1/5*b^3*e*x^5*sgn(b*x + a) + 1/4*b^3*d*x^4*sgn(b*x + a) + 3/4*a*b^2*e*x^4* sgn(b*x + a) + a*b^2*d*x^3*sgn(b*x + a) + a^2*b*e*x^3*sgn(b*x + a) + 3/2*a ^2*b*d*x^2*sgn(b*x + a) + 1/2*a^3*e*x^2*sgn(b*x + a) + a^3*d*x*sgn(b*x + a ) + 1/20*(5*a^4*b*d - a^5*e)*sgn(b*x + a)/b^2
Time = 9.90 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a+b\,x\right )\,\left (5\,b\,d-a\,e+4\,b\,e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{20\,b^2} \]