\(\int (d+e x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [1558]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(b d-a e) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {654, 623} \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)}{4 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

[In]

Int[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((b*d - a*e)*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(4*b^2) + (e*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(5*b^2)

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 654

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}+\frac {\left (2 b^2 d-2 a b e\right ) \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx}{2 b^2} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.20 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (10 a^3 (2 d+e x)+10 a^2 b x (3 d+2 e x)+5 a b^2 x^2 (4 d+3 e x)+b^3 x^3 (5 d+4 e x)\right )}{20 (a+b x)} \]

[In]

Integrate[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(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))

Maple [A] (verified)

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 \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} e \,x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a \,b^{2} e +b^{3} d \right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a^{2} b e +3 a \,b^{2} d \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e \,a^{3}+3 d \,a^{2} b \right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} d x}{b x +a}\) \(153\)

[In]

int((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} e x^{5} + a^{3} d x + \frac {1}{4} \, {\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} + {\left (a b^{2} d + a^{2} b e\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b d + a^{3} e\right )} x^{2} \]

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

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} \]

[In]

integrate((e*x+d)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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))

Maxima [B] (verification not implemented)

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 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e}{5 \, b^{2}} \]

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

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 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} e x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a b^{2} e x^{4} \mathrm {sgn}\left (b x + a\right ) + a b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} b e x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{3} e x^{2} \mathrm {sgn}\left (b x + a\right ) + a^{3} d x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (5 \, a^{4} b d - a^{5} e\right )} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{2}} \]

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.61 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a+b\,x\right )\,\left (5\,b\,d-a\,e+4\,b\,e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{20\,b^2} \]

[In]

int((d + e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

((a + b*x)*(5*b*d - a*e + 4*b*e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(20*b^2)