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

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

[Out]

7/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-2/3*d*x^2*(-e^2*x^2+d^2)^(1/2)/e-1/4*x^3*(-e^2*x^2+d^2)^(1/2)-1/2
4*d^2*(21*e*x+32*d)*(-e^2*x^2+d^2)^(1/2)/e^3

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Rubi [A]  time = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.07, size = 81, normalized size = 0.70 \[ \frac {21 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (32 d^3+21 d^2 e x+16 d e^2 x^2+6 e^3 x^3\right )}{24 e^3} \]

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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fricas [A]  time = 0.80, size = 83, normalized size = 0.72 \[ -\frac {42 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} + 16 \, d e^{2} x^{2} + 21 \, d^{2} e x + 32 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e^{3}} \]

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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giac [A]  time = 0.25, size = 63, normalized size = 0.55 \[ \frac {7}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{24} \, {\left (32 \, d^{3} e^{\left (-3\right )} + {\left (21 \, d^{2} e^{\left (-2\right )} + 2 \, {\left (8 \, d e^{\left (-1\right )} + 3 \, x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

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)

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maple [A]  time = 0.01, size = 124, normalized size = 1.08 \[ \frac {7 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{3}}{4}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,x^{2}}{3 e}-\frac {7 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x}{8 e^{2}}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}}{3 e^{3}} \]

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)/(-
e^2*x^2+d^2)^(1/2)*x)-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 = 0.97, size = 103, normalized size = 0.90 \[ -\frac {1}{4} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{3} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d x^{2}}{3 \, e} + \frac {7 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{3}} - \frac {7 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{2}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e^{3}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 9.33, size = 386, normalized size = 3.36 \[ d^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {i d x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x}{2 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{3}}{2 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {3 i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{5}} + \frac {3 i d^{3} x}{8 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{3}}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{5}} - \frac {3 d^{3} x}{8 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{3}}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

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/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*s
qrt(-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/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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