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

Optimal. Leaf size=77 \[ \frac{2 x}{5 d^4 \sqrt{d^2-e^2 x^2}}+\frac{x}{5 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (d+e x)}{5 e \left (d^2-e^2 x^2\right )^{5/2}} \]

[Out]

(2*(d + e*x))/(5*e*(d^2 - e^2*x^2)^(5/2)) + x/(5*d^2*(d^2 - e^2*x^2)^(3/2)) + (2*x)/(5*d^4*Sqrt[d^2 - e^2*x^2]
)

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Rubi [A]  time = 0.0197757, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {653, 192, 191} \[ \frac{2 x}{5 d^4 \sqrt{d^2-e^2 x^2}}+\frac{x}{5 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (d+e x)}{5 e \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x))/(5*e*(d^2 - e^2*x^2)^(5/2)) + x/(5*d^2*(d^2 - e^2*x^2)^(3/2)) + (2*x)/(5*d^4*Sqrt[d^2 - e^2*x^2]
)

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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

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]

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{2 (d+e x)}{5 e \left (d^2-e^2 x^2\right )^{5/2}}+\frac{3}{5} \int \frac{1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac{2 (d+e x)}{5 e \left (d^2-e^2 x^2\right )^{5/2}}+\frac{x}{5 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 \int \frac{1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{5 d^2}\\ &=\frac{2 (d+e x)}{5 e \left (d^2-e^2 x^2\right )^{5/2}}+\frac{x}{5 d^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 x}{5 d^4 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0441357, size = 63, normalized size = 0.82 \[ \frac{d^2 e x+2 d^3-4 d e^2 x^2+2 e^3 x^3}{5 d^4 e (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^3 + d^2*e*x - 4*d*e^2*x^2 + 2*e^3*x^3)/(5*d^4*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.049, size = 65, normalized size = 0.8 \begin{align*}{\frac{ \left ( -ex+d \right ) \left ( ex+d \right ) ^{3} \left ( 2\,{e}^{3}{x}^{3}-4\,{e}^{2}{x}^{2}d+x{d}^{2}e+2\,{d}^{3} \right ) }{5\,{d}^{4}e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.987325, size = 105, normalized size = 1.36 \begin{align*} \frac{2 \, x}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{2 \, d}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{x}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d^{2}} + \frac{2 \, x}{5 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93368, size = 230, normalized size = 2.99 \begin{align*} \frac{2 \, e^{4} x^{4} - 4 \, d e^{3} x^{3} + 4 \, d^{3} e x - 2 \, d^{4} -{\left (2 \, e^{3} x^{3} - 4 \, d e^{2} x^{2} + d^{2} e x + 2 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (d^{4} e^{5} x^{4} - 2 \, d^{5} e^{4} x^{3} + 2 \, d^{7} e^{2} x - d^{8} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17718, size = 82, normalized size = 1.06 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left (x^{2}{\left (\frac{2 \, x^{2} e^{4}}{d^{4}} - \frac{5 \, e^{2}}{d^{2}}\right )} + 5\right )} x + 2 \, d e^{\left (-1\right )}\right )}}{5 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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