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

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

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Rubi [A]  time = 0.08, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {797, 641, 195, 217, 203} \begin {gather*} \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 {gather*}

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 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.12, size = 112, normalized size = 0.85 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (15 d^4 \sin ^{-1}\left (\frac {e x}{d}\right )+\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-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\right )\right )}{120 e^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.89, size = 114, normalized size = 0.86 \begin {gather*} \frac {d^5 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^4}+\frac {\sqrt {d^2-e^2 x^2} \left (-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\right )}{120 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^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))/(120*e^3) + (d^5*Sqrt
[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e^4)

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fricas [A]  time = 0.40, size = 95, normalized size = 0.72 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.26, size = 74, normalized size = 0.56 \begin {gather*} \frac {1}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \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 {gather*}

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)

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maple [A]  time = 0.06, size = 125, normalized size = 0.95 \begin {gather*} \frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{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 {gather*}

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 = 0.98, size = 104, normalized size = 0.79 \begin {gather*} \frac {d^{5} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{3}} + \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 {gather*}

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*x/d)/e^3 + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.48, size = 279, normalized size = 2.11 \begin {gather*} 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}\: \left |{\frac {e^{2} x^{2}}{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 {gather*}

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/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*sqrt(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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