\(\int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\) [1069]

Optimal result

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

[Out]

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

[In]

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

[Out]

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

[In]

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

[Out]

-1/3*c/(e*(c*(d + e*x)^2)^(3/2))

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {1}{3 \left (e x +d \right )^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(24\)
pseudoelliptic	\(-\frac {1}{3 \left (e x +d \right )^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(24\)
gosper	\(-\frac {1}{3 \left (e x +d \right )^{2} e \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}\)	\(35\)
default	\(-\frac {1}{3 \left (e x +d \right )^{2} e \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}\)	\(35\)
trager	\(\frac {\left (x^{2} e^{2}+3 d e x +3 d^{2}\right ) x \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{3 d^{3} c \left (e x +d \right )^{4}}\)	\(57\)

[In]

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

[Out]

-1/3/(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. 73 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=-\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \, {\left (c e^{5} x^{4} + 4 \, c d e^{4} x^{3} + 6 \, c d^{2} e^{3} x^{2} + 4 \, c d^{3} e^{2} x + c d^{4} e\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {c \left (d + e x\right )^{2}} \left (d + e x\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=-\frac {1}{3 \, {\left (\sqrt {c} e^{4} x^{3} + 3 \, \sqrt {c} d e^{3} x^{2} + 3 \, \sqrt {c} d^{2} e^{2} x + \sqrt {c} d^{3} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=-\frac {1}{3 \, {\left (e x + d\right )}^{3} \sqrt {c} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.75 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=-\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{3\,c\,e\,{\left (d+e\,x\right )}^4} \]

[In]

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