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

Optimal. Leaf size=115 \[ -\frac{d^2 (32 d+21 e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}+\frac{7 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

(-2*d*x^2*Sqrt[d^2 - e^2*x^2])/(3*e) - (x^3*Sqrt[d^2 - e^2*x^2])/4 - (d^2*(32*d + 21*e*x)*Sqrt[d^2 - e^2*x^2])
/(24*e^3) + (7*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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Rubi [A]  time = 0.145128, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1809, 833, 780, 217, 203} \[ -\frac{d^2 (32 d+21 e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}+\frac{7 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x^2*Sqrt[d^2 - e^2*x^2])/(3*e) - (x^3*Sqrt[d^2 - e^2*x^2])/4 - (d^2*(32*d + 21*e*x)*Sqrt[d^2 - e^2*x^2])
/(24*e^3) + (7*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 833

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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{x^2 (d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx &=-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{x^2 \left (-7 d^2 e^2-8 d e^3 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}+\frac{\int \frac{x \left (16 d^3 e^3+21 d^2 e^4 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{12 e^4}\\ &=-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}-\frac{d^2 (32 d+21 e x) \sqrt{d^2-e^2 x^2}}{24 e^3}+\frac{\left (7 d^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}-\frac{d^2 (32 d+21 e x) \sqrt{d^2-e^2 x^2}}{24 e^3}+\frac{\left (7 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^2}\\ &=-\frac{2 d x^2 \sqrt{d^2-e^2 x^2}}{3 e}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2}-\frac{d^2 (32 d+21 e x) \sqrt{d^2-e^2 x^2}}{24 e^3}+\frac{7 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3}\\ \end{align*}

Mathematica [A]  time = 0.075886, size = 81, normalized size = 0.7 \[ \frac{21 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\sqrt{d^2-e^2 x^2} \left (21 d^2 e x+32 d^3+16 d e^2 x^2+6 e^3 x^3\right )}{24 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(32*d^3 + 21*d^2*e*x + 16*d*e^2*x^2 + 6*e^3*x^3)) + 21*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*
x^2]])/(24*e^3)

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Maple [A]  time = 0.057, size = 124, normalized size = 1.1 \begin{align*} -{\frac{{x}^{3}}{4}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{7\,{d}^{2}x}{8\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{7\,{d}^{4}}{8\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{2\,d{x}^{2}}{3\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{4\,{d}^{3}}{3\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.47184, size = 157, normalized size = 1.37 \begin{align*} -\frac{1}{4} \, \sqrt{-e^{2} x^{2} + d^{2}} x^{3} - \frac{2 \, \sqrt{-e^{2} x^{2} + d^{2}} d x^{2}}{3 \, e} + \frac{7 \, d^{4} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}} e^{2}} - \frac{7 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{2}} - \frac{4 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.80938, size = 181, normalized size = 1.57 \begin{align*} -\frac{42 \, d^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (6 \, e^{3} x^{3} + 16 \, d e^{2} x^{2} + 21 \, d^{2} e x + 32 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(42*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 + 16*d*e^2*x^2 + 21*d^2*e*x + 32*d^3)*sqr
t(-e^2*x^2 + d^2))/e^3

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Sympy [C]  time = 7.75333, size = 389, normalized size = 3.38 \begin{align*} d^{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 ) + 2 d e \left (\begin{cases} - \frac{2 d^{2} \sqrt{d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac{x^{2} \sqrt{d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text{for}\: e \neq 0 \\\frac{x^{4}}{4 \sqrt{d^{2}}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{3 i d^{4} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{8 e^{5}} + \frac{3 i d^{3} x}{8 e^{4} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{i d x^{3}}{8 e^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{i x^{5}}{4 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{3 d^{4} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{8 e^{5}} - \frac{3 d^{3} x}{8 e^{4} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{d x^{3}}{8 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{x^{5}}{4 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(x**2*(e*x+d)**2/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**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)) + 2*d*e*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(d**2 - e**2*x**2)/(3*e**2
), Ne(e, 0)), (x**4/(4*sqrt(d**2)), True)) + e**2*Piecewise((-3*I*d**4*acosh(e*x/d)/(8*e**5) + 3*I*d**3*x/(8*e
**4*sqrt(-1 + e**2*x**2/d**2)) - I*d*x**3/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - I*x**5/(4*d*sqrt(-1 + e**2*x**2
/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (3*d**4*asin(e*x/d)/(8*e**5) - 3*d**3*x/(8*e**4*sqrt(1 - e**2*x**2/d**
2)) + d*x**3/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.192, size = 85, normalized size = 0.74 \begin{align*} \frac{7}{8} \, d^{4} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{24} \,{\left (32 \, d^{3} e^{\left (-3\right )} +{\left (21 \, d^{2} e^{\left (-2\right )} + 2 \,{\left (8 \, d e^{\left (-1\right )} + 3 \, x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8*d^4*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/24*(32*d^3*e^(-3) + (21*d^2*e^(-2) + 2*(8*d*e^(-1) + 3*x)*x)*x)*sqrt(-
x^2*e^2 + d^2)