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

Optimal. Leaf size=64 \[ \frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 e^2 (d+e x)^4}-\frac {4 \left (d^2-e^2 x^2\right )^{3/2}}{15 d e^2 (d+e x)^3} \]

[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

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Rubi [A]
time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {807, 665} \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[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*} \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 \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*}

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Mathematica [A]
time = 0.27, size = 52, normalized size = 0.81 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(140\) vs. \(2(56)=112\).
time = 0.06, size = 141, normalized size = 2.20

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	\(-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{5} d \left (x +\frac {d}{e}\right )^{3}}-\frac {d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{5}}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (54) = 108\).
time = 0.42, size = 113, normalized size = 1.77 \begin {gather*} \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{5 \, {\left (x^{3} e^{5} + 3 \, d x^{2} e^{4} + 3 \, d^{2} x e^{3} + d^{3} e^{2}\right )}} - \frac {11 \, \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}} + \frac {4 \, \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d x e^{3} + d^{2} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.83, size = 96, normalized size = 1.50 \begin {gather*} -\frac {x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3} - {\left (4 \, x^{2} e^{2} - 3 \, d x e - d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d x^{3} e^{5} + 3 \, d^{2} x^{2} e^{4} + 3 \, d^{3} x e^{3} + d^{4} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3 - (4*x^2*e^2 - 3*d*x*e - d^2)*sqrt(-x^2*e^2 + d^2))/(d*x^3*e^5
+ 3*d^2*x^2*e^4 + 3*d^3*x*e^3 + d^4*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 )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).
time = 1.20, size = 128, normalized size = 2.00 \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 {15 \, {\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 )}}{15 \, d {\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^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

2/15*(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 + 15*(d*e + sq
rt(-x^2*e^2 + d^2)*e)^3*e^(-6)/x^3 + 1)*e^(-2)/(d*((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^5)

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Mupad [B]
time = 2.90, size = 46, normalized size = 0.72 \begin {gather*} -\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} \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)^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)

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