\(\int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\) [192]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \]

[Out]

1/5*(-e^2*x^2+d^2)^(3/2)/e^2/(e*x+d)^4-4/15*(-e^2*x^2+d^2)^(3/2)/d/e^2/(e*x+d)^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {807, 665} \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \]

[In]

Int[(x*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]

[Out]

(d^2 - e^2*x^2)^(3/2)/(5*e^2*(d + e*x)^4) - (4*(d^2 - e^2*x^2)^(3/2))/(15*d*e^2*(d + e*x)^3)

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}+\frac {4 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e} \\ & = \frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-d^2-3 d e x+4 e^2 x^2\right )}{15 d e^2 (d+e x)^3} \]

[In]

Integrate[(x*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-d^2 - 3*d*e*x + 4*e^2*x^2))/(15*d*e^2*(d + e*x)^3)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (4 e x +d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 \left (e x +d \right )^{3} d \,e^{2}}\)	\(42\)
trager	\(-\frac {\left (-4 e^{2} x^{2}+3 d e x +d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d \left (e x +d \right )^{3} e^{2}}\)	\(47\)
default	\(-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{5} d \left (x +\frac {d}{e}\right )^{3}}-\frac {d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{5}}\)	\(141\)

[In]

int(x*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/15*(-e*x+d)*(4*e*x+d)*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^3/d/e^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.59 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3} - {\left (4 \, e^{2} x^{2} - 3 \, d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{5} x^{3} + 3 \, d^{2} e^{4} x^{2} + 3 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} \]

[In]

integrate(x*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/15*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3 - (4*e^2*x^2 - 3*d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/(d*e^5*x^3
+ 3*d^2*e^4*x^2 + 3*d^3*e^3*x + d^4*e^2)

Sympy [F]

\[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate(x*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(x*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (56) = 112\).

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.95 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} + \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{3} x + d^{2} e^{2}\right )}} \]

[In]

integrate(x*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

2/5*sqrt(-e^2*x^2 + d^2)*d/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2) - 11/15*sqrt(-e^2*x^2 + d^2)/(e^4*x
^2 + 2*d*e^3*x + d^2*e^2) + 4/15*sqrt(-e^2*x^2 + d^2)/(d*e^3*x + d^2*e^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (56) = 112\).

Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.14 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + 1\right )}}{15 \, d e {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate(x*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

2/15*(5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 15*(
d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 1)/(d*e*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*a
bs(e))

Mupad [B] (verification not implemented)

Time = 11.79 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int \frac {x \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (d^2+3\,d\,e\,x-4\,e^2\,x^2\right )}{15\,d\,e^2\,{\left (d+e\,x\right )}^3} \]

[In]

int((x*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4,x)

[Out]

-((d^2 - e^2*x^2)^(1/2)*(d^2 - 4*e^2*x^2 + 3*d*e*x))/(15*d*e^2*(d + e*x)^3)