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

Optimal. Leaf size=86 \[ \frac {(d+e x)^3}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x}{5 d^2 e \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 86, 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, 653, 191} \[ \frac {(d+e x)^3}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x}{5 d^2 e \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)^3/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x))/(5*e^2*(d^2 - e^2*x^2)^(3/2)) - x/(5*d^2*e*Sqrt[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 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

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

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Mathematica [A]  time = 0.18, size = 55, normalized size = 0.64 \[ -\frac {(d+e x) \left (d^2-3 d e x+e^2 x^2\right )}{5 d^2 e^2 (d-e x)^2 \sqrt {d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*((d + e*x)*(d^2 - 3*d*e*x + e^2*x^2))/(d^2*e^2*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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fricas [A]  time = 0.90, size = 104, normalized size = 1.21 \[ -\frac {e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3} - {\left (e^{2} x^{2} - 3 \, d e x + d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{2} e^{5} x^{3} - 3 \, d^{3} e^{4} x^{2} + 3 \, d^{4} e^{3} x - d^{5} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 60, normalized size = 0.70 \[ \frac {{\left (d^{3} e^{\left (-2\right )} + {\left (x {\left (\frac {x^{2} e^{3}}{d^{2}} - 5 \, e\right )} - 5 \, d\right )} x^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(d^3*e^(-2) + (x*(x^2*e^3/d^2 - 5*e) - 5*d)*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 = 52, normalized size = 0.60 \[ -\frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (e^{2} x^{2}-3 d e x +d^{2}\right )}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d^{2} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 128, normalized size = 1.49 \[ \frac {e x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d^{2} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {d^{3}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.66, size = 46, normalized size = 0.53 \[ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (d^2-3\,d\,e\,x+e^2\,x^2\right )}{5\,d^2\,e^2\,{\left (d-e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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