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

Optimal. Leaf size=83 \[ -\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1809, 780, 217, 203} \begin {gather*} -\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x \left (-5 d^2 e^2-6 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 e^2}\\ &=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e}\\ &=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e}\\ &=-\frac {1}{3} x^2 \sqrt {d^2-e^2 x^2}-\frac {d (5 d+3 e x) \sqrt {d^2-e^2 x^2}}{3 e^2}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.83 \begin {gather*} \frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (5 d^2+3 d e x+e^2 x^2\right )}{3 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 89, normalized size = 1.07 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-5 d^2-3 d e x-e^2 x^2\right )}{3 e^2}+\frac {d^3 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 71, normalized size = 0.86 \begin {gather*} -\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (e^{2} x^{2} + 3 \, d e x + 5 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (e^2*x^2 + 3*d*e*x + 5*d^2)*sqrt(-e^2*x^2 + d^2))/e^2

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giac [A]  time = 0.25, size = 49, normalized size = 0.59 \begin {gather*} d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\relax (d) - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (5 \, d^{2} e^{\left (-2\right )} + {\left (3 \, d e^{\left (-1\right )} + x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 98, normalized size = 1.18 \begin {gather*} \frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{2}}{3}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d x}{e}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}}{3 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^2*(-e^2*x^2+d^2)^(1/2)-5/3*d^2/e^2*(-e^2*x^2+d^2)^(1/2)-d*x*(-e^2*x^2+d^2)^(1/2)/e+d^3/e/(e^2)^(1/2)*ar
ctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)

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maxima [A]  time = 0.97, size = 77, normalized size = 0.93 \begin {gather*} -\frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{2} + \frac {d^{3} \arcsin \left (\frac {e x}{d}\right )}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{e} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.62, size = 218, normalized size = 2.63 \begin {gather*} d^{2} \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 ) + 2 d e \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 ) + e^{2} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e**2, True)) + 2*d*e*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)) + e**2*Pi
ecewise((-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*s
qrt(d**2)), True))

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