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

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 89 \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {803, 653, 197} \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {(d+e x)^2}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{15 d e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {4 x}{15 d^3 e \sqrt {d^2-e^2 x^2}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (d^3-2 d^2 e x+8 d e^2 x^2-4 e^3 x^3\right )}{15 d^3 e^2 (d-e x)^3 (d+e x)} \]

[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))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72

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 \(e^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+\frac {d^{2}}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+2 d e \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\) \(173\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.31 \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4} + {\left (4 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 2 \, d^{2} e x - d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{6} x^{4} - 2 \, d^{4} e^{5} x^{3} + 2 \, d^{6} e^{3} x - d^{7} e^{2}\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.22 \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {2 \, x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e} - \frac {4 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e} \]

[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)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.73 \[ \int \frac {x (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (d^3-2\,d^2\,e\,x+8\,d\,e^2\,x^2-4\,e^3\,x^3\right )}{15\,d^3\,e^2\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \]

[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)