Integrand size = 31, antiderivative size = 41 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {(4 b d-a e+3 b e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{12 b^2} \] Output:
1/12*(3*b*e*x-a*e+4*b*d)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/b^2
Time = 1.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.56 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (6 a^2 (2 d+e x)+4 a b x (3 d+2 e x)+b^2 x^2 (4 d+3 e x)\right )}{12 (a+b x)} \] Input:
Integrate[(a + b*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(x*Sqrt[(a + b*x)^2]*(6*a^2*(2*d + e*x) + 4*a*b*x*(3*d + 2*e*x) + b^2*x^2* (4*d + 3*e*x)))/(12*(a + b*x))
Time = 0.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1187, 27, 49, 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.
\(\displaystyle \int (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x)^2 (d+e x)dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^2 (d+e x)dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {e (a+b x)^3}{b}+\frac {(b d-a e) (a+b x)^2}{b}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {(a+b x)^3 (b d-a e)}{3 b^2}+\frac {e (a+b x)^4}{4 b^2}\right )}{a+b x}\) |
Input:
Int[(a + b*x)*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((b*d - a*e)*(a + b*x)^3)/(3*b^2) + (e*(a + b*x)^4)/(4*b^2)))/(a + b*x)
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((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[(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, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.55 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{3} \left (-3 b e x +a e -4 b d \right )}{12 b^{2}}\) | \(32\) |
gosper | \(\frac {x \left (3 b^{2} e \,x^{3}+8 a \,x^{2} e b +4 b^{2} d \,x^{2}+6 a^{2} e x +12 a b d x +12 a^{2} d \right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\) | \(66\) |
orering | \(\frac {x \left (3 b^{2} e \,x^{3}+8 a \,x^{2} e b +4 b^{2} d \,x^{2}+6 a^{2} e x +12 a b d x +12 a^{2} d \right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} e \,x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (2 a b e +b^{2} d \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{2} e +2 d a b \right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{2} d x}{b x +a}\) | \(113\) |
Input:
int((b*x+a)*(e*x+d)*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/12*csgn(b*x+a)*(b*x+a)^3*(-3*b*e*x+a*e-4*b*d)/b^2
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac {1}{3} \, {\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b d + a^{2} e\right )} x^{2} \] Input:
integrate((b*x+a)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/4*b^2*e*x^4 + a^2*d*x + 1/3*(b^2*d + 2*a*b*e)*x^3 + 1/2*(2*a*b*d + a^2*e )*x^2
Time = 1.81 (sec) , antiderivative size = 393, normalized size of antiderivative = 9.59 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=a d \left (\begin {cases} \left (\frac {a}{2 b} + \frac {x}{2}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3 a b} & \text {for}\: a b \neq 0 \\x \sqrt {a^{2}} & \text {otherwise} \end {cases}\right ) + a e \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + b d \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + b e \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x+a)*(e*x+d)*((b*x+a)**2)**(1/2),x)
Output:
a*d*Piecewise(((a/(2*b) + x/2)*sqrt(a**2 + 2*a*b*x + b**2*x**2), Ne(b**2, 0)), ((a**2 + 2*a*b*x)**(3/2)/(3*a*b), Ne(a*b, 0)), (x*sqrt(a**2), True)) + a*e*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + a*x/(6 *b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2 *a*b*x)**(5/2)/5)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*sqrt(a**2)/2, True)) + b*d*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + a*x/(6* b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2* a*b*x)**(5/2)/5)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*sqrt(a**2)/2, True)) + b*e*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**3/(12*b**3) - a**2*x/( 12*b**2) + a*x**2/(12*b) + x**3/4), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x)* *(3/2)/3 - 2*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/( 4*a**3*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2)/3, True))
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (37) = 74\).
Time = 0.04 (sec) , antiderivative size = 251, normalized size of antiderivative = 6.12 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d x + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (b d + a e\right )} a x}{2 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e x}{4 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (b d + a e\right )} a^{2}}{2 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e}{12 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (b d + a e\right )}}{3 \, b^{2}} \] Input:
integrate((b*x+a)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d*x + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 )*a^2*e*x/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*d/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*e/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(b*d + a*e )*a*x/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*e*x/b - 1/2*sqrt(b^2*x^2 + 2 *a*b*x + a^2)*(b*d + a*e)*a^2/b^2 - 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a *e/b^2 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(b*d + a*e)/b^2
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (37) = 74\).
Time = 1.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.68 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b^{2} e x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, a b e x^{3} \mathrm {sgn}\left (b x + a\right ) + a b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{2} e x^{2} \mathrm {sgn}\left (b x + a\right ) + a^{2} d x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (4 \, a^{3} b d - a^{4} e\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{2}} \] Input:
integrate((b*x+a)*(e*x+d)*((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
1/4*b^2*e*x^4*sgn(b*x + a) + 1/3*b^2*d*x^3*sgn(b*x + a) + 2/3*a*b*e*x^3*sg n(b*x + a) + a*b*d*x^2*sgn(b*x + a) + 1/2*a^2*e*x^2*sgn(b*x + a) + a^2*d*x *sgn(b*x + a) + 1/12*(4*a^3*b*d - a^4*e)*sgn(b*x + a)/b^2
Time = 11.47 (sec) , antiderivative size = 219, normalized size of antiderivative = 5.34 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {d\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^3}+\frac {e\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b}-\frac {a^2\,e\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b}-\frac {5\,a\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^4}+\frac {a\,\left (a+b\,x\right )\,\left (3\,b\,d-a\,e+2\,b\,e\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,b^2} \] Input:
int(((a + b*x)^2)^(1/2)*(a + b*x)*(d + e*x),x)
Output:
(d*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b *x)^(1/2))/(24*b^3) + (e*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b) - (a^2*e *(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b) - (5*a*e*(8*b^2*(a ^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/( 96*b^4) + (a*(a + b*x)*(3*b*d - a*e + 2*b*e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^( 1/2))/(6*b^2)
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int (a+b x) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \left (3 b^{2} e \,x^{3}+8 a b e \,x^{2}+4 b^{2} d \,x^{2}+6 a^{2} e x +12 a b d x +12 a^{2} d \right )}{12} \] Input:
int((b*x+a)*(e*x+d)*((b*x+a)^2)^(1/2),x)
Output:
(x*(12*a**2*d + 6*a**2*e*x + 12*a*b*d*x + 8*a*b*e*x**2 + 4*b**2*d*x**2 + 3 *b**2*e*x**3))/12