3.86 \(\int \frac {x^2 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=93 \[ \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1635, 789, 637} \[ \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d*(d + e*x)^3)/(5*e^3*(d^2 - e^2*x^2)^(5/2)) - (8*(d + e*x)^2)/(15*e^3*(d^2 - e^2*x^2)^(3/2)) + (7*(d + e*x))
/(15*d*e^3*Sqrt[d^2 - e^2*x^2])

Rule 637

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 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]

Rule 1635

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^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {\left (\frac {3 d^2}{e^2}+\frac {5 d x}{e}\right ) (d+e x)^2}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 \int \frac {d+e x}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 e^2}\\ &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 58, normalized size = 0.62 \[ \frac {(d+e x) \left (2 d^2-6 d e x+7 e^2 x^2\right )}{15 d e^3 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(2*d^2 - 6*d*e*x + 7*e^2*x^2))/(15*d*e^3*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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fricas [A]  time = 1.07, size = 106, normalized size = 1.14 \[ \frac {2 \, e^{3} x^{3} - 6 \, d e^{2} x^{2} + 6 \, d^{2} e x - 2 \, d^{3} - {\left (7 \, e^{2} x^{2} - 6 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d e^{6} x^{3} - 3 \, d^{2} e^{5} x^{2} + 3 \, d^{3} e^{4} x - d^{4} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 72, normalized size = 0.77 \[ -\frac {{\left (2 \, d^{4} e^{\left (-3\right )} - {\left (5 \, d^{2} e^{\left (-1\right )} - {\left (x {\left (\frac {7 \, x e^{2}}{d} + 15 \, e\right )} + 5 \, d\right )} x\right )} x^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*(2*d^4*e^(-3) - (5*d^2*e^(-1) - (x*(7*x*e^2/d + 15*e) + 5*d)*x)*x^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 = 55, normalized size = 0.59 \[ \frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (7 e^{2} x^{2}-6 d e x +2 d^{2}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d \,e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 154, normalized size = 1.66 \[ \frac {e x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {d^{2} x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {7 \, d^{3} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} + \frac {2 \, d^{4}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {7 \, d x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} + \frac {7 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

e*x^4/(-e^2*x^2 + d^2)^(5/2) + 3/2*d*x^3/(-e^2*x^2 + d^2)^(5/2) - 1/3*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e) - 7/1
0*d^3*x/((-e^2*x^2 + d^2)^(5/2)*e^2) + 2/15*d^4/((-e^2*x^2 + d^2)^(5/2)*e^3) + 7/30*d*x/((-e^2*x^2 + d^2)^(3/2
)*e^2) + 7/15*x/(sqrt(-e^2*x^2 + d^2)*d*e^2)

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mupad [B]  time = 2.69, size = 49, normalized size = 0.53 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2-6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d-e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**2*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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