Integrand size = 26, antiderivative size = 69 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2}+\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (10 a^3 (2 A+B x)+10 a^2 b x (3 A+2 B x)+5 a b^2 x^2 (4 A+3 B x)+b^3 x^3 (5 A+4 B x)\right )}{20 (a+b x)} \]
(x*Sqrt[(a + b*x)^2]*(10*a^3*(2*A + B*x) + 10*a^2*b*x*(3*A + 2*B*x) + 5*a* b^2*x^2*(4*A + 3*B*x) + b^3*x^3*(5*A + 4*B*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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1100 |
\(\displaystyle \frac {(A b-a B) \int \left (a^2+2 b x a+b^2 x^2\right )^{3/2}dx}{b}+\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
\(\Big \downarrow \) 1079 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} (A b-a B) \int \left (x b^2+a b\right )^3dx}{b^4 (a+b x)}+\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 b^2}+\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}\) |
((A*b - a*B)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^2) + (B*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*b^2)
3.18.25.3.1 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 = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30
method | result | size |
gosper | \(\frac {x \left (4 x^{4} B \,b^{3}+5 A \,b^{3} x^{3}+15 B a \,b^{2} x^{3}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+30 A \,a^{2} b x +10 a^{3} B x +20 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\) | \(90\) |
default | \(\frac {x \left (4 x^{4} B \,b^{3}+5 A \,b^{3} x^{3}+15 B a \,b^{2} x^{3}+20 A a \,b^{2} x^{2}+20 B \,a^{2} b \,x^{2}+30 A \,a^{2} b x +10 a^{3} B x +20 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\) | \(90\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, B \,b^{3} x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{3}+3 B a \,b^{2}\right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A a \,b^{2}+3 B b \,a^{2}\right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} A x}{b x +a}\) | \(153\) |
1/20*x*(4*B*b^3*x^4+5*A*b^3*x^3+15*B*a*b^2*x^3+20*A*a*b^2*x^2+20*B*a^2*b*x ^2+30*A*a^2*b*x+10*B*a^3*x+20*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, B b^{3} x^{5} + A a^{3} x + \frac {1}{4} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{4} + {\left (B a^{2} b + A a b^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2} \]
1/5*B*b^3*x^5 + A*a^3*x + 1/4*(3*B*a*b^2 + A*b^3)*x^4 + (B*a^2*b + A*a*b^2 )*x^3 + 1/2*(B*a^3 + 3*A*a^2*b)*x^2
Leaf count of result is larger than twice the leaf count of optimal. 745 vs. \(2 (65) = 130\).
Time = 0.69 (sec) , antiderivative size = 745, normalized size of antiderivative = 10.80 \[ \int (A+B 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 b^{2} x^{4}}{5} + \frac {x^{3} \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b^{2}} + \frac {x^{2} \cdot \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b^{2}} + \frac {x \left (6 A a^{2} b^{2} + 4 B a^{3} b - \frac {3 a^{2} \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b^{2}} - \frac {5 a \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b}\right )}{2 b^{2}} + \frac {4 A a^{3} b + B a^{4} - \frac {2 a^{2} \cdot \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (6 A a^{2} b^{2} + 4 B a^{3} b - \frac {3 a^{2} \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b^{2}} - \frac {5 a \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b}\right )}{2 b}}{b^{2}}\right ) + \frac {\left (\frac {a}{b} + x\right ) \left (A a^{4} - \frac {a^{2} \cdot \left (6 A a^{2} b^{2} + 4 B a^{3} b - \frac {3 a^{2} \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b^{2}} - \frac {5 a \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b}\right )}{2 b^{2}} - \frac {a \left (4 A a^{3} b + B a^{4} - \frac {2 a^{2} \cdot \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (6 A a^{2} b^{2} + 4 B a^{3} b - \frac {3 a^{2} \left (A b^{4} + \frac {11 B a b^{3}}{5}\right )}{4 b^{2}} - \frac {5 a \left (4 A a b^{3} + \frac {26 B a^{2} b^{2}}{5} - \frac {7 a \left (A b^{4} + \frac {11 B a b^{3}}{5}\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 {B \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{14 a b} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}} \cdot \left (2 A b - B a\right )}{10 b}}{a b} & \text {for}\: a b \neq 0 \\\left (A x + \frac {B 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*b**2*x**4/5 + x**3*(A*b**4 + 11*B*a*b**3/5)/(4*b**2) + x**2*(4*A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b **4 + 11*B*a*b**3/5)/(4*b))/(3*b**2) + x*(6*A*a**2*b**2 + 4*B*a**3*b - 3*a **2*(A*b**4 + 11*B*a*b**3/5)/(4*b**2) - 5*a*(4*A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b**4 + 11*B*a*b**3/5)/(4*b))/(3*b))/(2*b**2) + (4*A*a**3*b + B*a **4 - 2*a**2*(4*A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b**4 + 11*B*a*b**3/5) /(4*b))/(3*b**2) - 3*a*(6*A*a**2*b**2 + 4*B*a**3*b - 3*a**2*(A*b**4 + 11*B *a*b**3/5)/(4*b**2) - 5*a*(4*A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b**4 + 1 1*B*a*b**3/5)/(4*b))/(3*b))/(2*b))/b**2) + (a/b + x)*(A*a**4 - a**2*(6*A*a **2*b**2 + 4*B*a**3*b - 3*a**2*(A*b**4 + 11*B*a*b**3/5)/(4*b**2) - 5*a*(4* A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b**4 + 11*B*a*b**3/5)/(4*b))/(3*b))/( 2*b**2) - a*(4*A*a**3*b + B*a**4 - 2*a**2*(4*A*a*b**3 + 26*B*a**2*b**2/5 - 7*a*(A*b**4 + 11*B*a*b**3/5)/(4*b))/(3*b**2) - 3*a*(6*A*a**2*b**2 + 4*B*a **3*b - 3*a**2*(A*b**4 + 11*B*a*b**3/5)/(4*b**2) - 5*a*(4*A*a*b**3 + 26*B* a**2*b**2/5 - 7*a*(A*b**4 + 11*B*a*b**3/5)/(4*b))/(3*b))/(2*b))/b)*log(a/b + x)/sqrt(b**2*(a/b + x)**2), Ne(b**2, 0)), ((B*(a**2 + 2*a*b*x)**(7/2)/( 14*a*b) + (a**2 + 2*a*b*x)**(5/2)*(2*A*b - B*a)/(10*b))/(a*b), Ne(a*b, 0)) , ((A*x + B*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 = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.81 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{5 \, b^{2}} \]
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*x - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3 /2)*B*a*x/b - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^2/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a/b + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/b^2
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (61) = 122\).
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.09 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, B b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + B a^{2} b x^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, A a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + A a^{3} x \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (B a^{5} - 5 \, A a^{4} b\right )} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{2}} \]
1/5*B*b^3*x^5*sgn(b*x + a) + 3/4*B*a*b^2*x^4*sgn(b*x + a) + 1/4*A*b^3*x^4* sgn(b*x + a) + B*a^2*b*x^3*sgn(b*x + a) + A*a*b^2*x^3*sgn(b*x + a) + 1/2*B *a^3*x^2*sgn(b*x + a) + 3/2*A*a^2*b*x^2*sgn(b*x + a) + A*a^3*x*sgn(b*x + a ) - 1/20*(B*a^5 - 5*A*a^4*b)*sgn(b*x + a)/b^2
Time = 10.98 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61 \[ \int (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}\,\left (5\,A\,b-B\,a+4\,B\,b\,x\right )}{20\,b^2} \]