\(\int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx\) [1030]

Optimal result

Integrand size = 32, antiderivative size = 39 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {656, 623} \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e} \]

[In]

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

[Out]

((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2))/(4*c*e)

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 656

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 c e} \]

[In]

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

[Out]

((d + e*x)*(c*(d + e*x)^2)^(3/2))/(4*c*e)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\left (e x +d \right )^{3} \sqrt {c \left (e x +d \right )^{2}}}{4 e}\)	\(24\)
default	\(\frac {\left (e x +d \right )^{3} \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 e}\)	\(35\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 e x +4 d}\)	\(62\)
trager	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 e x +4 d}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.62 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \, {\left (e x + d\right )}} \]

[In]

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

[Out]

1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(e*x + d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (36) = 72\).

Time = 0.80 (sec) , antiderivative size = 160, normalized size of antiderivative = 4.10 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\begin {cases} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} \left (\frac {d^{3}}{4 e} + \frac {3 d^{2} x}{4} + \frac {3 d e x^{2}}{4} + \frac {e^{2} x^{3}}{4}\right ) & \text {for}\: c e^{2} \neq 0 \\\frac {\frac {d^{2} \left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{12} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {5}{2}}}{10 c} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {7}{2}}}{28 c^{2} d^{2}}}{c d e} & \text {for}\: c d e \neq 0 \\\sqrt {c d^{2}} \left (\begin {cases} d^{2} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{3}}{3 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)*(d**3/(4*e) + 3*d**2*x/4 + 3*d*e*x**2/4 + e**2*x**3/4), Ne(c
*e**2, 0)), ((d**2*(c*d**2 + 2*c*d*e*x)**(3/2)/12 + (c*d**2 + 2*c*d*e*x)**(5/2)/(10*c) + (c*d**2 + 2*c*d*e*x)*
*(7/2)/(28*c**2*d**2))/(c*d*e), Ne(c*d*e, 0)), (sqrt(c*d**2)*Piecewise((d**2*x, Eq(e, 0)), ((d + e*x)**3/(3*e)
, True)), True))

Maxima [F(-2)]

Exception generated. \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e x + d\right )}^{4} \sqrt {c} \mathrm {sgn}\left (e x + d\right )}{4 \, e} \]

[In]

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

[Out]

1/4*(e*x + d)^4*sqrt(c)*sgn(e*x + d)/e

Mupad [B] (verification not implemented)

Time = 10.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.95 \[ \int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (c\,d^3+e\,x\,\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )+2\,c\,d^2\,e\,x+c\,d\,e^2\,x^2\right )}{4\,c\,e} \]

[In]

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

[Out]

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