3.2.72 \(\int \frac {x^3}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [172]

Optimal. Leaf size=99 \[ \frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {866, 1649, 651} \begin {gather*} \frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(d - e*x)^2)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (4*d*(d - e*x))/(5*e^4*(d^2 - e^2*x^2)^(3/2)) + (5*d - 2*e*x
)/(5*d*e^4*Sqrt[d^2 - e^2*x^2])

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 866

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

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {x^3 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x) \left (-\frac {2 d^3}{e^3}+\frac {5 d^2 x}{e^2}-\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {-\frac {6 d^3}{e^3}+\frac {15 d^2 x}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 70, normalized size = 0.71 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (2 d^3+4 d^2 e x+d e^2 x^2-2 e^3 x^3\right )}{5 d e^4 (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(2*d^3 + 4*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3))/(5*d*e^4*(d - e*x)*(d + e*x)^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(87)=174\).
time = 0.06, size = 309, normalized size = 3.12

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (-2 e^{3} x^{3}+d \,e^{2} x^{2}+4 d^{2} e x +2 d^{3}\right )}{5 \left (e x +d \right ) d \,e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(65\)
trager	\(\frac {\left (-2 e^{3} x^{3}+d \,e^{2} x^{2}+4 d^{2} e x +2 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 d \,e^{4} \left (e x +d \right )^{3} \left (-e x +d \right )}\)	\(67\)
default	\(\frac {1}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {2 x}{d \,e^{3} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 d^{2} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{4}}-\frac {d^{3} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{5}}\)	\(309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 143, normalized size = 1.44 \begin {gather*} \frac {d^{2}}{5 \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{6} + 2 \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{5} + \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{4}\right )}} - \frac {2 \, x e^{\left (-3\right )}}{5 \, \sqrt {-x^{2} e^{2} + d^{2}} d} + \frac {e^{\left (-4\right )}}{\sqrt {-x^{2} e^{2} + d^{2}}} - \frac {4 \, d}{5 \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x e^{5} + \sqrt {-x^{2} e^{2} + d^{2}} d e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.23, size = 109, normalized size = 1.10 \begin {gather*} \frac {2 \, x^{4} e^{4} + 4 \, d x^{3} e^{3} - 4 \, d^{3} x e - 2 \, d^{4} + {\left (2 \, x^{3} e^{3} - d x^{2} e^{2} - 4 \, d^{2} x e - 2 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d x^{4} e^{8} + 2 \, d^{2} x^{3} e^{7} - 2 \, d^{4} x e^{5} - d^{5} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*x^4*e^4 + 4*d*x^3*e^3 - 4*d^3*x*e - 2*d^4 + (2*x^3*e^3 - d*x^2*e^2 - 4*d^2*x*e - 2*d^3)*sqrt(-x^2*e^2 +
d^2))/(d*x^4*e^8 + 2*d^2*x^3*e^7 - 2*d^4*x*e^5 - d^5*e^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 1.25, size = 171, normalized size = 1.73 \begin {gather*} -\frac {1}{40} \, {\left (\frac {16 i \, e^{\left (-3\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d} - \frac {5 \, e^{\left (-3\right )}}{d \sqrt {\frac {2 \, d}{x e + d} - 1} \mathrm {sgn}\left (\frac {1}{x e + d}\right )} - \frac {{\left (d^{4} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{12} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} - 5 \, d^{4} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{12} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4} + 15 \, d^{4} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{12} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{4}\right )} e^{\left (-15\right )}}{d^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{5}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(16*I*e^(-3)*sgn(1/(x*e + d))/d - 5*e^(-3)/(d*sqrt(2*d/(x*e + d) - 1)*sgn(1/(x*e + d))) - (d^4*(2*d/(x*e
+ d) - 1)^(5/2)*e^12*sgn(1/(x*e + d))^4 - 5*d^4*(2*d/(x*e + d) - 1)^(3/2)*e^12*sgn(1/(x*e + d))^4 + 15*d^4*sq
rt(2*d/(x*e + d) - 1)*e^12*sgn(1/(x*e + d))^4)*e^(-15)/(d^5*sgn(1/(x*e + d))^5))*e^(-1)

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Mupad [B]
time = 2.97, size = 66, normalized size = 0.67 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^3+4\,d^2\,e\,x+d\,e^2\,x^2-2\,e^3\,x^3\right )}{5\,d\,e^4\,{\left (d+e\,x\right )}^3\,\left (d-e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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