\(\int \frac {1}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1085]

Optimal result

Integrand size = 24, antiderivative size = 41 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 c e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 41, 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 = {621} \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 c e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{4 c e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {d+e x}{4 e \left (c (d+e x)^2\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {1}{4 c^{2} \left (e x +d \right )^{3} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(27\)
gosper	\(-\frac {e x +d}{4 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(33\)
default	\(-\frac {e x +d}{4 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(33\)
trager	\(\frac {\left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) x \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 c^{3} d^{4} \left (e x +d \right )^{5}}\)	\(68\)

[In]

int(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/c^2/(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. 97 vs. \(2 (37) = 74\).

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.37 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \, {\left (c^{3} e^{6} x^{5} + 5 \, c^{3} d e^{5} x^{4} + 10 \, c^{3} d^{2} e^{4} x^{3} + 10 \, c^{3} d^{3} e^{3} x^{2} + 5 \, c^{3} d^{4} e^{2} x + c^{3} d^{5} e\right )}} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(c^3*e^6*x^5 + 5*c^3*d*e^5*x^4 + 10*c^3*d^2*e^4*x^3 + 10*c^3*d^3*e^3*
x^2 + 5*c^3*d^4*e^2*x + c^3*d^5*e)

Sympy [F]

\[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{4}} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-1/4/((c*e^2)^(5/2)*(x + d/e)^4)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 \, {\left (e x + d\right )}^{4} c^{\frac {5}{2}} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

integrate(1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.81 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{4\,c^3\,e\,{\left (d+e\,x\right )}^5} \]

[In]

int(1/(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)/(4*c^3*e*(d + e*x)^5)