3.1.48 \(\int \frac {x (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [48]

Optimal. Leaf size=89 \[ \frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 89, 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 = {803, 653, 197} \begin {gather*} \frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d + e*x)^2/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x))/(15*d*e^2*(d^2 - e^2*x^2)^(3/2)) - (4*x)/(15*d^3*e*S
qrt[d^2 - e^2*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 653

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)/(2*a*c*(p + 1)))*(a + c*x
^2)^(p + 1), x] + Dist[d*((2*p + 3)/(2*a*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 803

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*(p + 1))), x] - Dist[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 \int \frac {d+e x}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 e}\\ &=\frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d e}\\ &=\frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 69, normalized size = 0.78 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (d^3-2 d^2 e x+8 d e^2 x^2-4 e^3 x^3\right )}{15 d^3 e^2 (d-e x)^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(d^3 - 2*d^2*e*x + 8*d*e^2*x^2 - 4*e^3*x^3))/(15*d^3*e^2*(d - e*x)^3*(d + e*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(172\) vs. \(2(77)=154\).
time = 0.06, size = 173, normalized size = 1.94

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (-4 e^{3} x^{3}+8 d \,e^{2} x^{2}-2 d^{2} e x +d^{3}\right )}{15 d^{3} e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(64\)
trager	\(\frac {\left (-4 e^{3} x^{3}+8 d \,e^{2} x^{2}-2 d^{2} e x +d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} \left (-e x +d \right )^{3} e^{2} \left (e x +d \right )}\)	\(66\)
default	\(e^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+2 e d \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+\frac {d^{2}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 100, normalized size = 1.12 \begin {gather*} \frac {2 \, d x e^{\left (-1\right )}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d^{2} e^{\left (-2\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {x^{2}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {2 \, x e^{\left (-1\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d} - \frac {4 \, x e^{\left (-1\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.35, size = 110, normalized size = 1.24 \begin {gather*} \frac {x^{4} e^{4} - 2 \, d x^{3} e^{3} + 2 \, d^{3} x e - d^{4} + {\left (4 \, x^{3} e^{3} - 8 \, d x^{2} e^{2} + 2 \, d^{2} x e - d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{3} x^{4} e^{6} - 2 \, d^{4} x^{3} e^{5} + 2 \, d^{6} x e^{3} - d^{7} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(x^4*e^4 - 2*d*x^3*e^3 + 2*d^3*x*e - d^4 + (4*x^3*e^3 - 8*d*x^2*e^2 + 2*d^2*x*e - d^3)*sqrt(-x^2*e^2 + d^
2))/(d^3*x^4*e^6 - 2*d^4*x^3*e^5 + 2*d^6*x*e^3 - d^7*e^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((x*e + d)^2*x/(-x^2*e^2 + d^2)^(7/2), x)

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Mupad [B]
time = 2.86, size = 65, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (d^3-2\,d^2\,e\,x+8\,d\,e^2\,x^2-4\,e^3\,x^3\right )}{15\,d^3\,e^2\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(d^3 - 4*e^3*x^3 + 8*d*e^2*x^2 - 2*d^2*e*x))/(15*d^3*e^2*(d + e*x)*(d - e*x)^3)

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