\(\int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [183]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e^2 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2}}{5 d^2 e^2 (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {807, 673, 665} \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{5 d^2 e^2 (d+e x)}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e^2 (d+e x)^2}+\frac {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3} \]

[In]

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

[Out]

Sqrt[d^2 - e^2*x^2]/(5*e^2*(d + e*x)^3) - Sqrt[d^2 - e^2*x^2]/(5*d*e^2*(d + e*x)^2) - Sqrt[d^2 - e^2*x^2]/(5*d
^2*e^2*(d + e*x))

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 673

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*(m + p + 1))), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^
p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[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 {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3}+\frac {3 \int \frac {1}{(d+e x)^2 \sqrt {d^2-e^2 x^2}} \, dx}{5 e} \\ & = \frac {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e^2 (d+e x)^2}+\frac {\int \frac {1}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx}{5 d e} \\ & = \frac {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e^2 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2}}{5 d^2 e^2 (d+e x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.51 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (d^2+3 d e x+e^2 x^2\right )}{5 d^2 e^2 (d+e x)^3} \]

[In]

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

[Out]

-1/5*(Sqrt[d^2 - e^2*x^2]*(d^2 + 3*d*e*x + e^2*x^2))/(d^2*e^2*(d + e*x)^3)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.47

method	result	size
trager	\(-\frac {\left (e^{2} x^{2}+3 d e x +d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 d^{2} \left (e x +d \right )^{3} e^{2}}\)	\(46\)
gosper	\(-\frac {\left (-e x +d \right ) \left (e^{2} x^{2}+3 d e x +d^{2}\right )}{5 \left (e x +d \right )^{2} d^{2} e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}\)	\(52\)
default	\(\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{e^{3}}-\frac {d \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{4}}\)	\(240\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3} + {\left (e^{2} x^{2} + 3 \, d e x + d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{2} e^{5} x^{3} + 3 \, d^{3} e^{4} x^{2} + 3 \, d^{4} e^{3} x + d^{5} e^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.33 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d e^{4} x^{2} + 2 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{2} e^{3} x + d^{3} e^{2}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.41 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + 1\right )}}{5 \, d^{2} e {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

2/5*(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) + 5*(d*
e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 1)/(d^2*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 = 11.83 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.46 \[ \int \frac {x}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (d^2+3\,d\,e\,x+e^2\,x^2\right )}{5\,d^2\,e^2\,{\left (d+e\,x\right )}^3} \]

[In]

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

[Out]

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