\(\int (c d^2+2 c d e x+c e^2 x^2)^{3/2} \, dx\) [1042]

Optimal result

Integrand size = 24, antiderivative size = 36 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {623} \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {(d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {c \left (e x +d \right )^{3} \sqrt {c \left (e x +d \right )^{2}}}{4 e}\)	\(25\)
default	\(\frac {\left (e x +d \right ) \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {3}{2}}}{4 e}\)	\(33\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {3}{2}}}{4 \left (e x +d \right )^{3}}\)	\(62\)
trager	\(\frac {c x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 e x +4 d}\)	\(63\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (32) = 64\).

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.86 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {{\left (c e^{3} x^{4} + 4 \, c d e^{2} x^{3} + 6 \, c d^{2} e x^{2} + 4 \, c d^{3} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \, {\left (e x + d\right )}} \]

[In]

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

[Out]

1/4*(c*e^3*x^4 + 4*c*d*e^2*x^3 + 6*c*d^2*e*x^2 + 4*c*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. 76 vs. \(2 (34) = 68\).

Time = 0.93 (sec) , antiderivative size = 366, normalized size of antiderivative = 10.17 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=c d^{2} \left (\begin {cases} \left (\frac {d}{2 e} + \frac {x}{2}\right ) \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} & \text {for}\: c e^{2} \neq 0 \\\frac {\left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{3 c d e} & \text {for}\: c d e \neq 0 \\x \sqrt {c d^{2}} & \text {otherwise} \end {cases}\right ) + 2 c d e \left (\begin {cases} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} \left (- \frac {d^{2}}{6 e^{2}} + \frac {d x}{6 e} + \frac {x^{2}}{3}\right ) & \text {for}\: c e^{2} \neq 0 \\\frac {- \frac {c d^{2} \left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{3} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {5}{2}}}{5}}{2 c^{2} d^{2} e^{2}} & \text {for}\: c d e \neq 0 \\\frac {x^{2} \sqrt {c d^{2}}}{2} & \text {otherwise} \end {cases}\right ) + c e^{2} \left (\begin {cases} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} \left (\frac {d^{3}}{12 e^{3}} - \frac {d^{2} x}{12 e^{2}} + \frac {d x^{2}}{12 e} + \frac {x^{3}}{4}\right ) & \text {for}\: c e^{2} \neq 0 \\\frac {\frac {c^{2} d^{4} \left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{3} - \frac {2 c d^{2} \left (c d^{2} + 2 c d e x\right )^{\frac {5}{2}}}{5} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {7}{2}}}{7}}{4 c^{3} d^{3} e^{3}} & \text {for}\: c d e \neq 0 \\\frac {x^{3} \sqrt {c d^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

c*d**2*Piecewise(((d/(2*e) + x/2)*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2), Ne(c*e**2, 0)), ((c*d**2 + 2*c*d*e*x
)**(3/2)/(3*c*d*e), Ne(c*d*e, 0)), (x*sqrt(c*d**2), True)) + 2*c*d*e*Piecewise((sqrt(c*d**2 + 2*c*d*e*x + c*e*
*2*x**2)*(-d**2/(6*e**2) + d*x/(6*e) + x**2/3), Ne(c*e**2, 0)), ((-c*d**2*(c*d**2 + 2*c*d*e*x)**(3/2)/3 + (c*d
**2 + 2*c*d*e*x)**(5/2)/5)/(2*c**2*d**2*e**2), Ne(c*d*e, 0)), (x**2*sqrt(c*d**2)/2, True)) + c*e**2*Piecewise(
(sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)*(d**3/(12*e**3) - d**2*x/(12*e**2) + d*x**2/(12*e) + x**3/4), Ne(c*e**
2, 0)), ((c**2*d**4*(c*d**2 + 2*c*d*e*x)**(3/2)/3 - 2*c*d**2*(c*d**2 + 2*c*d*e*x)**(5/2)/5 + (c*d**2 + 2*c*d*e
*x)**(7/2)/7)/(4*c**3*d**3*e**3), Ne(c*d*e, 0)), (x**3*sqrt(c*d**2)/3, True))

Maxima [F(-2)]

Exception generated. \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),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 = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.75 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (2 \, {\left (e x^{2} + 2 \, d x\right )} c d^{2} \mathrm {sgn}\left (e x + d\right ) + \frac {c d^{4} \mathrm {sgn}\left (e x + d\right )}{e} + {\left (e x^{2} + 2 \, d x\right )}^{2} c e \mathrm {sgn}\left (e x + d\right )\right )} \sqrt {c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx=\frac {\left (x\,e^2+d\,e\right )\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{4\,e^2} \]

[In]

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

[Out]

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