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

Optimal. Leaf size=132 \[ \frac{d^3 x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

(d^3*x*Sqrt[d^2 - e^2*x^2])/(8*e^2) - (d^2*(d^2 - e^2*x^2)^(3/2))/(3*e^3) - (d*x*(d^2 - e^2*x^2)^(3/2))/(4*e^2
) + (d^2 - e^2*x^2)^(5/2)/(5*e^3) + (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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Rubi [A]  time = 0.0757274, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {797, 641, 195, 217, 203} \[ \frac{d^3 x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*x*Sqrt[d^2 - e^2*x^2])/(8*e^2) - (d^2*(d^2 - e^2*x^2)^(3/2))/(3*e^3) - (d*x*(d^2 - e^2*x^2)^(3/2))/(4*e^2
) + (d^2 - e^2*x^2)^(5/2)/(5*e^3) + (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

Rule 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 x^2 (d+e x) \sqrt{d^2-e^2 x^2} \, dx &=-\frac{\int (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}+\frac{d^2 \int (d+e x) \sqrt{d^2-e^2 x^2} \, dx}{e^2}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac{d \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}+\frac{d^3 \int \sqrt{d^2-e^2 x^2} \, dx}{e^2}\\ &=\frac{d^3 x \sqrt{d^2-e^2 x^2}}{2 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac{\left (3 d^3\right ) \int \sqrt{d^2-e^2 x^2} \, dx}{4 e^2}+\frac{d^5 \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^2}\\ &=\frac{d^3 x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac{\left (3 d^5\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}+\frac{d^5 \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^2}\\ &=\frac{d^3 x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^3}-\frac{\left (3 d^5\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{d^3 x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3}\\ \end{align*}

Mathematica [A]  time = 0.136492, size = 112, normalized size = 0.85 \[ \frac{\sqrt{d^2-e^2 x^2} \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (-8 d^2 e^2 x^2-15 d^3 e x-16 d^4+30 d e^3 x^3+24 e^4 x^4\right )+15 d^4 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{120 e^3 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(Sqrt[1 - (e^2*x^2)/d^2]*(-16*d^4 - 15*d^3*e*x - 8*d^2*e^2*x^2 + 30*d*e^3*x^3 + 24*e^4*x^
4) + 15*d^4*ArcSin[(e*x)/d]))/(120*e^3*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [A]  time = 0.058, size = 125, normalized size = 1. \begin{align*} -{\frac{{x}^{2}}{5\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{2\,{d}^{2}}{15\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{dx}{4\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{d}^{3}x}{8\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{d}^{5}}{8\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x^2*(-e^2*x^2+d^2)^(3/2)/e-2/15*d^2*(-e^2*x^2+d^2)^(3/2)/e^3-1/4*d*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/8*d^3*x*(
-e^2*x^2+d^2)^(1/2)/e^2+1/8*d^5/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [A]  time = 1.60867, size = 158, normalized size = 1.2 \begin{align*} \frac{d^{5} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}} e^{2}} + \frac{\sqrt{-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} x^{2}}{5 \, e} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d x}{4 \, e^{2}} - \frac{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d^{2}}{15 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*d^5*arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^2) + 1/8*sqrt(-e^2*x^2 + d^2)*d^3*x/e^2 - 1/5*(-e^2*x^2 + d^2
)^(3/2)*x^2/e - 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*d^2/e^3

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Fricas [A]  time = 1.8215, size = 205, normalized size = 1.55 \begin{align*} -\frac{30 \, d^{5} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (24 \, e^{4} x^{4} + 30 \, d e^{3} x^{3} - 8 \, d^{2} e^{2} x^{2} - 15 \, d^{3} e x - 16 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*e^4*x^4 + 30*d*e^3*x^3 - 8*d^2*e^2*x^2 - 15*d^3
*e*x - 16*d^4)*sqrt(-e^2*x^2 + d^2))/e^3

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Sympy [C]  time = 6.22928, size = 280, normalized size = 2.12 \begin{align*} d \left (\begin{cases} - \frac{i d^{4} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{8 e^{3}} + \frac{i d^{3} x}{8 e^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{3 i d x^{3}}{8 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} 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{d^{4} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{8 e^{3}} - \frac{d^{3} x}{8 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{3 d x^{3}}{8 \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} - \frac{e^{2} x^{5}}{4 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) + e \left (\begin{cases} - \frac{2 d^{4} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac{d^{2} x^{2} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac{x^{4} \sqrt{d^{2} - e^{2} x^{2}}}{5} & \text{for}\: e \neq 0 \\\frac{x^{4} \sqrt{d^{2}}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(
-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*asin
(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**
5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + e*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*s
qrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True))

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Giac [A]  time = 1.26262, size = 100, normalized size = 0.76 \begin{align*} \frac{1}{8} \, d^{5} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{120} \,{\left (16 \, d^{4} e^{\left (-3\right )} +{\left (15 \, d^{3} e^{\left (-2\right )} + 2 \,{\left (4 \, d^{2} e^{\left (-1\right )} - 3 \,{\left (4 \, x e + 5 \, d\right )} 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)*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/8*d^5*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/120*(16*d^4*e^(-3) + (15*d^3*e^(-2) + 2*(4*d^2*e^(-1) - 3*(4*x*e + 5*d
)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)