### 3.829 $$\int \frac{(d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx$$

Optimal. Leaf size=74 $-\frac{2 d \sqrt{d^2-e^2 x^2}}{e}-\frac{1}{2} x \sqrt{d^2-e^2 x^2}+\frac{3 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

[Out]

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

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Rubi [A]  time = 0.0267756, antiderivative size = 83, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {671, 641, 217, 203} $-\frac{3 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{(d+e x) \sqrt{d^2-e^2 x^2}}{2 e}+\frac{3 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 671

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

Rule 641

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0366859, size = 58, normalized size = 0.78 $\frac{3 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-(4 d+e x) \sqrt{d^2-e^2 x^2}}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 71, normalized size = 1. \begin{align*} -{\frac{x}{2}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}+{\frac{3\,{d}^{2}}{2}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-2\,{\frac{d\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}{e}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.73513, size = 85, normalized size = 1.15 \begin{align*} \frac{3 \, d^{2} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}}} - \frac{1}{2} \, \sqrt{-e^{2} x^{2} + d^{2}} x - \frac{2 \, \sqrt{-e^{2} x^{2} + d^{2}} d}{e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15827, size = 126, normalized size = 1.7 \begin{align*} -\frac{6 \, d^{2} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + \sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + 4 \, d\right )}}{2 \, e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(6*d^2*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + sqrt(-e^2*x^2 + d^2)*(e*x + 4*d))/e

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Sympy [A]  time = 3.7291, size = 270, normalized size = 3.65 \begin{align*} d^{2} \left (\begin{cases} \frac{\sqrt{\frac{d^{2}}{e^{2}}} \operatorname{asin}{\left (x \sqrt{\frac{e^{2}}{d^{2}}} \right )}}{\sqrt{d^{2}}} & \text{for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac{\sqrt{- \frac{d^{2}}{e^{2}}} \operatorname{asinh}{\left (x \sqrt{- \frac{e^{2}}{d^{2}}} \right )}}{\sqrt{d^{2}}} & \text{for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac{\sqrt{\frac{d^{2}}{e^{2}}} \operatorname{acosh}{\left (x \sqrt{\frac{e^{2}}{d^{2}}} \right )}}{\sqrt{- d^{2}}} & \text{for}\: d^{2} < 0 \wedge e^{2} < 0 \end{cases}\right ) + 2 d e \left (\begin{cases} \frac{x^{2}}{2 \sqrt{d^{2}}} & \text{for}\: e^{2} = 0 \\- \frac{\sqrt{d^{2} - e^{2} x^{2}}}{e^{2}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{i d x \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e^{3}} - \frac{d x}{2 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{x^{3}}{2 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\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)**(1/2),x)

[Out]

d**2*Piecewise((sqrt(d**2/e**2)*asin(x*sqrt(e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 > 0)), (sqrt(-d**2/e**2
)*asinh(x*sqrt(-e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 < 0)), (sqrt(d**2/e**2)*acosh(x*sqrt(e**2/d**2))/sq
rt(-d**2), (d**2 < 0) & (e**2 < 0))) + 2*d*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*
x**2)/e**2, True)) + e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e**3) - I*d*x*sqrt(-1 + e**2*x**2/d**2)/(2*e**2),
Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2*d
*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.33922, size = 54, normalized size = 0.73 \begin{align*} \frac{3}{2} \, d^{2} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{2} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left (4 \, d e^{\left (-1\right )} + x\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)^(1/2),x, algorithm="giac")

[Out]

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