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

Optimal. Leaf size=97 \[ \frac {\sqrt {d^2-e^2 x^2}}{5 e^2 (d+e x)^3}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e^2 (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2}}{5 d^2 e^2 (d+e x)} \]

[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)

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Rubi [A]
time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {807, 673, 665} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {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*}

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Mathematica [A]
time = 0.28, size = 49, normalized size = 0.51 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(239\) vs. \(2(85)=170\).
time = 0.07, size = 240, normalized size = 2.47

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	\(-\frac {d \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{4}}+\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{e^{3}}\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 117, normalized size = 1.21 \begin {gather*} \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{3} e^{5} + 3 \, d x^{2} e^{4} + 3 \, d^{2} x e^{3} + d^{3} e^{2}\right )}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d x^{2} e^{4} + 2 \, d^{2} x e^{3} + d^{3} e^{2}\right )}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d^{2} x e^{3} + d^{3} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.85, size = 94, normalized size = 0.97 \begin {gather*} -\frac {x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3} + {\left (x^{2} e^{2} + 3 \, d x e + d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d^{2} x^{3} e^{5} + 3 \, d^{3} x^{2} e^{4} + 3 \, d^{4} x e^{3} + d^{5} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 1.90, size = 128, normalized size = 1.32 \begin {gather*} \frac {2 \, {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{\left (-6\right )}}{x^{3}} + 1\right )} e^{\left (-2\right )}}{5 \, d^{2} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Mupad [B]
time = 2.59, size = 45, normalized size = 0.46 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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