\(\int \frac {(d+e x)^3}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1082]

Optimal result

Integrand size = 32, antiderivative size = 32 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, 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} \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[In]

Int[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-(1/(c^2*e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]))

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 \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx}{c} \\ & = -\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{c^2 e \sqrt {c (d+e x)^2}} \]

[In]

Integrate[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-(1/(c^2*e*Sqrt[c*(d + e*x)^2]))

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {1}{c^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(20\)
pseudoelliptic	\(-\frac {1}{c^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(20\)
gosper	\(-\frac {\left (e x +d \right )^{4}}{e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
default	\(-\frac {\left (e x +d \right )^{4}}{e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
trager	\(\frac {x \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{c^{3} d \left (e x +d \right )^{2}}\)	\(38\)

[In]

int((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{c^{3} e^{3} x^{2} + 2 \, c^{3} d e^{2} x + c^{3} d^{2} e} \]

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (32) = 64\).

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.19 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {\sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{c^{3} d^{2} e + 2 c^{3} d e^{2} x + c^{3} e^{3} x^{2}} & \text {for}\: e \neq 0 \\\frac {d^{3} x}{\left (c d^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**3/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (30) = 60\).

Time = 0.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 4.97 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {c^{2} d^{3} e^{4}}{\left (c e^{2}\right )^{\frac {9}{2}} {\left (x + \frac {d}{e}\right )}^{4}} - \frac {e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} + \frac {8 \, c d^{2} e^{3}}{3 \, \left (c e^{2}\right )^{\frac {7}{2}} {\left (x + \frac {d}{e}\right )}^{3}} - \frac {5 \, d^{2}}{3 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e} - \frac {2 \, d e^{2}}{\left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {d^{3}}{\left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{4}} \]

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-c^2*d^3*e^4/((c*e^2)^(9/2)*(x + d/e)^4) - e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 8/3*c*d^2*e^3/((c
*e^2)^(7/2)*(x + d/e)^3) - 5/3*d^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 2*d*e^2/((c*e^2)^(5/2)*(x + d
/e)^2) + d^3/((c*e^2)^(5/2)*(x + d/e)^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{{\left (e x + d\right )} c^{\frac {5}{2}} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

-1/((e*x + d)*c^(5/2)*e*sgn(e*x + d))

Mupad [B] (verification not implemented)

Time = 10.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{c^3\,e\,{\left (d+e\,x\right )}^2} \]

[In]

int((d + e*x)^3/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(5/2),x)

[Out]

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