\(\int \frac {1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx\) [1076]

Optimal result

Integrand size = 32, antiderivative size = 31 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, 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} \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[In]

Int[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

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

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}& = c \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx \\ & = -\frac {1}{3 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 e \left (c (d+e x)^2\right )^{3/2}} \]

[In]

Integrate[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

-1/3*1/(e*(c*(d + e*x)^2)^(3/2))

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {1}{3 c \left (e x +d \right )^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(27\)
pseudoelliptic	\(-\frac {1}{3 c \left (e x +d \right )^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(27\)
gosper	\(-\frac {1}{3 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {3}{2}}}\)	\(28\)
default	\(-\frac {1}{3 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {3}{2}}}\)	\(28\)
trager	\(\frac {\left (x^{2} e^{2}+3 d e x +3 d^{2}\right ) x \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{3 c^{2} d^{3} \left (e x +d \right )^{4}}\)	\(57\)

[In]

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

[Out]

-1/3/c/(e*x+d)^2/(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. 83 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.68 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \, {\left (c^{2} e^{5} x^{4} + 4 \, c^{2} d e^{4} x^{3} + 6 \, c^{2} d^{2} e^{3} x^{2} + 4 \, c^{2} d^{3} e^{2} x + c^{2} d^{4} e\right )}} \]

[In]

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

[Out]

-1/3*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(c^2*e^5*x^4 + 4*c^2*d*e^4*x^3 + 6*c^2*d^2*e^3*x^2 + 4*c^2*d^3*e^2*x
+ c^2*d^4*e)

Sympy [A] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=\begin {cases} - \frac {1}{3 e \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {3}{2}}} & \text {for}\: e \neq 0 \\\frac {x}{d \left (c d^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-1/(3*e*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(3/2)), Ne(e, 0)), (x/(d*(c*d**2)**(3/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 \, {\left (c^{\frac {3}{2}} e^{4} x^{3} + 3 \, c^{\frac {3}{2}} d e^{3} x^{2} + 3 \, c^{\frac {3}{2}} d^{2} e^{2} x + c^{\frac {3}{2}} d^{3} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 \, {\left (e x + d\right )}^{3} c^{\frac {3}{2}} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.78 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{3\,c^2\,e\,{\left (d+e\,x\right )}^4} \]

[In]

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

[Out]

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