\(\int \frac {x^2 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [47]

Optimal result

Integrand size = 27, antiderivative size = 87 \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {7 (d+e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1649, 792, 197} \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}}+\frac {d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {7 (d+e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(d*(d + e*x)^2)/(5*e^3*(d^2 - e^2*x^2)^(5/2)) - (7*(d + e*x))/(15*e^3*(d^2 - e^2*x^2)^(3/2)) + x/(15*d^2*e^2*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 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[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*} \text {integral}& = \frac {d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {\left (\frac {2 d^2}{e^2}+\frac {5 d x}{e}\right ) (d+e x)}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = \frac {d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {7 (d+e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 e^2} \\ & = \frac {d (d+e x)^2}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {7 (d+e x)}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{15 d^2 e^2 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-4 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )}{15 d^2 e^3 (d-e x)^3 (d+e x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (-e^{3} x^{3}+2 d \,e^{2} x^{2}-8 d^{2} e x +4 d^{3}\right )}{15 d^{2} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(66\)
trager	\(-\frac {\left (-e^{3} x^{3}+2 d \,e^{2} x^{2}-8 d^{2} e x +4 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{2} e^{3} \left (-e x +d \right )^{3} \left (e x +d \right )}\)	\(68\)
default	\(e^{2} \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+2 d e \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )\)	\(282\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.34 \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {4 \, e^{4} x^{4} - 8 \, d e^{3} x^{3} + 8 \, d^{3} e x - 4 \, d^{4} + {\left (e^{3} x^{3} - 2 \, d e^{2} x^{2} + 8 \, d^{2} e x - 4 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{7} x^{4} - 2 \, d^{3} e^{6} x^{3} + 2 \, d^{5} e^{4} x - d^{6} e^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*(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 = 131, normalized size of antiderivative = 1.51 \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2 \, d x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {d^{2} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {4 \, d^{3}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} + \frac {x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{2}} \]

[In]

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

[Out]

1/2*x^3/(-e^2*x^2 + d^2)^(5/2) + 2/3*d*x^2/((-e^2*x^2 + d^2)^(5/2)*e) - 1/10*d^2*x/((-e^2*x^2 + d^2)^(5/2)*e^2
) - 4/15*d^3/((-e^2*x^2 + d^2)^(5/2)*e^3) + 1/30*x/((-e^2*x^2 + d^2)^(3/2)*e^2) + 1/15*x/(sqrt(-e^2*x^2 + d^2)
*d^2*e^2)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (4\,d^3-8\,d^2\,e\,x+2\,d\,e^2\,x^2-e^3\,x^3\right )}{15\,d^2\,e^3\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \]

[In]

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

[Out]

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