\(\int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx\) [1029]

Optimal result

Integrand size = 32, antiderivative size = 34 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 c^2 e} \]

[Out]

1/5*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)/c^2/e

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, 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 = {657, 643} \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 c^2 e} \]

[In]

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

[Out]

(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)/(5*c^2*e)

Rule 643

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

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/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 - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^4 \sqrt {c (d+e x)^2}}{5 e} \]

[In]

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

[Out]

((d + e*x)^4*Sqrt[c*(d + e*x)^2])/(5*e)

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\sqrt {c \left (e x +d \right )^{2}}\, \left (e x +d \right )^{4}}{5 e}\)	\(24\)
pseudoelliptic	\(\frac {\sqrt {c \left (e x +d \right )^{2}}\, \left (e x +d \right )^{4}}{5 e}\)	\(24\)
default	\(\frac {\left (e x +d \right )^{4} \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{5 e}\)	\(35\)
gosper	\(\frac {x \left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{5 e x +5 d}\)	\(73\)
trager	\(\frac {x \left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{5 e x +5 d}\)	\(73\)

[In]

int((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/5*(c*(e*x+d)^2)^(1/2)*(e*x+d)^4/e

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (30) = 60\).

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.18 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e^{4} x^{5} + 5 \, d e^{3} x^{4} + 10 \, d^{2} e^{2} x^{3} + 10 \, d^{3} e x^{2} + 5 \, d^{4} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{5 \, {\left (e x + d\right )}} \]

[In]

integrate((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*(e^4*x^5 + 5*d*e^3*x^4 + 10*d^2*e^2*x^3 + 10*d^3*e*x^2 + 5*d^4*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. 187 vs. \(2 (31) = 62\).

Time = 0.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 5.50 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\begin {cases} \frac {d^{4} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5 e} + \frac {4 d^{3} x \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {6 d^{2} e x^{2} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {4 d e^{2} x^{3} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {e^{3} x^{4} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\d^{3} x \sqrt {c d^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**3*(c*e**2*x**2+2*c*d*e*x+c*d**2)**(1/2),x)

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((e*x+d)^3*(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 = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e x + d\right )}^{5} \sqrt {c} \mathrm {sgn}\left (e x + d\right )}{5 \, e} \]

[In]

integrate((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="giac")

[Out]

1/5*(e*x + d)^5*sqrt(c)*sgn(e*x + d)/e

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 239, normalized size of antiderivative = 7.03 \[ \int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {d^3\,\left (\frac {x}{2}+\frac {d}{2\,e}\right )\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{4}+\frac {e\,x^2\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{5\,c}-\frac {23\,d^2\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (8\,e^2\,\left (d^2+e^2\,x^2\right )-12\,d^2\,e^2+4\,d\,e^3\,x\right )}{480\,e^3}+\frac {3\,d\,x\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{4\,c}-\frac {7\,d\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (c\,d^3+3\,e\,x\,\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )-4\,c\,d^2\,e\,x-5\,c\,d\,e^2\,x^2\right )}{60\,c\,e} \]

[In]

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

[Out]

(d^3*(x/2 + d/(2*e))*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2))/4 + (e*x^2*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2))/
(5*c) - (23*d^2*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)*(8*e^2*(d^2 + e^2*x^2) - 12*d^2*e^2 + 4*d*e^3*x))/(480*e
^3) + (3*d*x*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2))/(4*c) - (7*d*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)*(c*d^3
+ 3*e*x*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x) - 4*c*d^2*e*x - 5*c*d*e^2*x^2))/(60*c*e)