\(\int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1545]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {(b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x)}+\frac {b (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^2 (a+b x)} \]

[Out]

-1/3*(-a*e+b*d)*(e*x+d)^3*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+1/4*b*(e*x+d)^4*((b*x+a)^2)^(1/2)/e^2/(b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4}{4 e^2 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}{3 e^2 (a+b x)} \]

[In]

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

[Out]

-1/3*((b*d - a*e)*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)) + (b*(d + e*x)^4*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(4*e^2*(a + b*x))

Rule 45

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] && NeQ[b*c - a*d, 0] && 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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right ) (d+e x)^2 \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b (b d-a e) (d+e x)^2}{e}+\frac {b^2 (d+e x)^3}{e}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {(b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2 (a+b x)}+\frac {b (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^2 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.25 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \left (4 a \left (3 d^2+3 d e x+e^2 x^2\right )+b x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{12 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

(x*(4*a*(3*d^2 + 3*d*e*x + e^2*x^2) + b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2))*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[
(a + b*x)^2])))/(12*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72

method	result	size
gosper	\(\frac {x \left (3 e^{2} b \,x^{3}+4 x^{2} e^{2} a +8 b d e \,x^{2}+12 a d e x +6 b \,d^{2} x +12 a \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\)	\(66\)
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (3 x^{2} b^{2} e^{2}-2 x a b \,e^{2}+8 b^{2} d e x +a^{2} e^{2}-4 a b d e +6 b^{2} d^{2}\right )}{12 b^{3}}\)	\(68\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{2} b \,x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{2} a +2 b d e \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (2 a d e +b \,d^{2}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, x a \,d^{2}}{b x +a}\)	\(113\)

[In]

int((e*x+d)^2*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12*x*(3*b*e^2*x^3+4*a*e^2*x^2+8*b*d*e*x^2+12*a*d*e*x+6*b*d^2*x+12*a*d^2)*((b*x+a)^2)^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.52 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b e^{2} x^{4} + a d^{2} x + \frac {1}{3} \, {\left (2 \, b d e + a e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b d^{2} + 2 \, a d e\right )} x^{2} \]

[In]

integrate((e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/4*b*e^2*x^4 + a*d^2*x + 1/3*(2*b*d*e + a*e^2)*x^3 + 1/2*(b*d^2 + 2*a*d*e)*x^2

Sympy [A] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.20 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=d^{2} \left (\begin {cases} \left (\frac {a}{2 b} + \frac {x}{2}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3 a b} & \text {for}\: a b \neq 0 \\x \sqrt {a^{2}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((e*x+d)**2*((b*x+a)**2)**(1/2),x)

[Out]

d**2*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)) + 2*d*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)) + e**2*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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (66) = 132\).

Time = 0.19 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.66 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{2} x - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d e x}{b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e^{2} x}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{2}}{2 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d e}{b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e^{2}}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{2} x}{4 \, b^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d e}{3 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{2}}{12 \, b^{3}} \]

[In]

integrate((e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*d^2*x - sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d*e*x/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x
+ a^2)*a^2*e^2*x/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d^2/b - sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*d*e/b^2
+ 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*e^2/b^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*e^2*x/b^2 + 2/3*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*d*e/b^2 - 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*e^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b e^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, b d e x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a e^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + a d e x^{2} \mathrm {sgn}\left (b x + a\right ) + a d^{2} x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (6 \, a^{2} b^{2} d^{2} - 4 \, a^{3} b d e + a^{4} e^{2}\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{3}} \]

[In]

integrate((e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/4*b*e^2*x^4*sgn(b*x + a) + 2/3*b*d*e*x^3*sgn(b*x + a) + 1/3*a*e^2*x^3*sgn(b*x + a) + 1/2*b*d^2*x^2*sgn(b*x +
a) + a*d*e*x^2*sgn(b*x + a) + a*d^2*x*sgn(b*x + a) + 1/12*(6*a^2*b^2*d^2 - 4*a^3*b*d*e + a^4*e^2)*sgn(b*x + a
)/b^3

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.34 \[ \int (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=d^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}+\frac {d\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{12\,b^4}-\frac {a^2\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2}-\frac {5\,a\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{96\,b^5} \]

[In]

int(((a + b*x)^2)^(1/2)*(d + e*x)^2,x)

[Out]

d^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (e^2*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) + (d*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))/(12*b^4) - (a^2*e^2*(x/2 + a/
(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2) - (5*a*e^2*(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^5)