3.9.40 \(\int \frac {(d+e x)^2}{(d^2-e^2 x^2)^{5/2}} \, dx\) [840]

Optimal. Leaf size=53 \[ \frac {2 (d+e x)}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {x}{3 d^2 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {667, 197} \begin {gather*} \frac {x}{3 d^2 \sqrt {d^2-e^2 x^2}}+\frac {2 (d+e x)}{3 e \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2*(d + e*x))/(3*e*(d^2 - e^2*x^2)^(3/2)) + x/(3*d^2*Sqrt[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 667

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]

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 42, normalized size = 0.79 \begin {gather*} \frac {(2 d-e x) \sqrt {d^2-e^2 x^2}}{3 d^2 e (d-e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(d^2 - e^2*x^2)^(5/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(140\) vs. \(2(45)=90\).
time = 0.46, size = 141, normalized size = 2.66

method	result	size
trager	\(\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{2} \left (-e x +d \right )^{2} e}\)	\(39\)
gosper	\(\frac {\left (e x +d \right )^{3} \left (-e x +d \right ) \left (-e x +2 d \right )}{3 d^{2} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(44\)
default	\(e^{2} \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )+\frac {2 d}{3 e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(-e^2*x^2+d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 54, normalized size = 1.02 \begin {gather*} \frac {2 \, d e^{\left (-1\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {x}{3 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.42, size = 70, normalized size = 1.32 \begin {gather*} \frac {2 \, x^{2} e^{2} - 4 \, d x e + 2 \, d^{2} - \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e - 2 \, d\right )}}{3 \, {\left (d^{2} x^{2} e^{3} - 2 \, d^{3} x e^{2} + d^{4} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**2/(-(-d + e*x)*(d + e*x))**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (43) = 86\).
time = 2.93, size = 98, normalized size = 1.85 \begin {gather*} -\frac {2 \, {\left (\frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} - 2\right )} e^{\left (-1\right )}}{3 \, d^{2} {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.48, size = 38, normalized size = 0.72 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d-e\,x\right )}{3\,d^2\,e\,{\left (d-e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(d^2 - e^2*x^2)^(5/2),x)

[Out]

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

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