\(\int \frac {(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1083]

Optimal result

Integrand size = 32, antiderivative size = 41 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{2 c^2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, 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 = {656, 621} \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{2 c^2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[In]

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

[Out]

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

Rule 656

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx}{c} \\ & = -\frac {1}{2 c^2 e (d+e x) \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 = 28, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {d+e x}{2 c e \left (c (d+e x)^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {1}{2 c^{2} \left (e x +d \right ) \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(27\)
gosper	\(-\frac {\left (e x +d \right )^{3}}{2 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
default	\(-\frac {\left (e x +d \right )^{3}}{2 e \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
trager	\(\frac {\left (e x +2 d \right ) x \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{2 c^{3} d^{2} \left (e x +d \right )^{3}}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.68 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{2 \, {\left (c^{3} e^{4} x^{3} + 3 \, c^{3} d e^{3} x^{2} + 3 \, c^{3} d^{2} e^{2} x + c^{3} d^{3} e\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (37) = 74\).

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.02 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {c^{2} d^{2} e^{4}}{4 \, \left (c e^{2}\right )^{\frac {9}{2}} {\left (x + \frac {d}{e}\right )}^{4}} + \frac {2 \, c d e^{3}}{3 \, \left (c e^{2}\right )^{\frac {7}{2}} {\left (x + \frac {d}{e}\right )}^{3}} - \frac {2 \, d}{3 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e} - \frac {e^{2}}{2 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {d^{2}}{4 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{4}} \]

[In]

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

[Out]

-1/4*c^2*d^2*e^4/((c*e^2)^(9/2)*(x + d/e)^4) + 2/3*c*d*e^3/((c*e^2)^(7/2)*(x + d/e)^3) - 2/3*d/((c*e^2*x^2 + 2
*c*d*e*x + c*d^2)^(3/2)*c*e) - 1/2*e^2/((c*e^2)^(5/2)*(x + d/e)^2) + 1/4*d^2/((c*e^2)^(5/2)*(x + d/e)^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {1}{2 \, {\left (e x + d\right )}^{2} c^{\frac {5}{2}} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.91 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{2\,c^3\,e\,{\left (d+e\,x\right )}^3} \]

[In]

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