3.48 \(\int \frac {x (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\)

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 = 0.03, 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 = {789, 639, 191} \[ \frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

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 191

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 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(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 789

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 = 0.05, size = 62, normalized size = 0.70 \[ \frac {d^3-2 d^2 e x+8 d e^2 x^2-4 e^3 x^3}{15 d^3 e^2 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)^2*Sqrt[d^2 - e^2*x^2])

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fricas [A]  time = 0.86, size = 117, normalized size = 1.31 \[ \frac {e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4} + {\left (4 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 2 \, d^{2} e x - d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{6} x^{4} - 2 \, d^{4} e^{5} x^{3} + 2 \, d^{6} e^{3} x - d^{7} e^{2}\right )}} \]

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

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giac [A]  time = 0.28, size = 64, normalized size = 0.72 \[ \frac {{\left ({\left (2 \, x {\left (\frac {2 \, x^{2} e^{3}}{d^{3}} - \frac {5 \, e}{d}\right )} - 5\right )} x^{2} - d^{2} e^{\left (-2\right )}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

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]

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

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maple [A]  time = 0.01, size = 64, normalized size = 0.72 \[ \frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (-4 e^{3} x^{3}+8 d \,e^{2} x^{2}-2 d^{2} e x +d^{3}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d^{3} e^{2}} \]

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)

[Out]

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

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maxima [A]  time = 0.44, size = 109, normalized size = 1.22 \[ \frac {x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {2 \, x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e} - \frac {4 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e} \]

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]

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

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

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

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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