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

Optimal. Leaf size=103 \[ -\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2*Sqrt[d^2 - e^2*x^2])/e^3) - (d*x*Sqrt[d^2 - e^2*x^2])/(2*e^2) + (d^2 - e^2*x^2)^(3/2)/(3*e^3) + (d^3*Ar
cTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*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 \frac {x^2 (d+e x)}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {\int (d+e x) \sqrt {d^2-e^2 x^2} \, dx}{e^2}+\frac {d^2 \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2}\\ &=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d \int \sqrt {d^2-e^2 x^2} \, dx}{e^2}+\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2}\\ &=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}-\frac {d^3 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 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^2}\\ &=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}-\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 )}{2 e^2}\\ &=-\frac {d^2 \sqrt {d^2-e^2 x^2}}{e^3}-\frac {d x \sqrt {d^2-e^2 x^2}}{2 e^2}+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 70, normalized size = 0.68 \[ \frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (4 d^2+3 d e x+2 e^2 x^2\right )}{6 e^3} \]

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

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fricas [A]  time = 0.86, size = 72, normalized size = 0.70 \[ -\frac {6 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{2} x^{2} + 3 \, d e x + 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

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]

-1/6*(6*d^3*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^2*x^2 + 3*d*e*x + 4*d^2)*sqrt(-e^2*x^2 + d^2))/e^
3

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giac [A]  time = 0.24, size = 54, normalized size = 0.52 \[ \frac {1}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (4 \, d^{2} e^{\left (-3\right )} + {\left (2 \, x e^{\left (-1\right )} + 3 \, d e^{\left (-2\right )}\right )} x\right )} \]

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]

1/2*d^3*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/6*sqrt(-x^2*e^2 + d^2)*(4*d^2*e^(-3) + (2*x*e^(-1) + 3*d*e^(-2))*x)

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maple [A]  time = 0.02, size = 102, normalized size = 0.99 \[ \frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{2}}{3 e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d x}{2 e^{2}}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}}{3 e^{3}} \]

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]

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

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maxima [A]  time = 0.99, size = 81, normalized size = 0.79 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e} + \frac {d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x}{2 \, e^{2}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{3}} \]

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]

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

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mupad [B]  time = 3.14, size = 112, normalized size = 1.09 \[ \left \{\begin {array}{cl} \frac {d\,x^3}{3\,\sqrt {d^2}} & \text {\ if\ \ }e=0\\ -\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+e^2\,x^2\right )}{3\,e^3}-\frac {d^3\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}}-\frac {d\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e^2} & \text {\ if\ \ }e\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(e == 0, (d*x^3)/(3*(d^2)^(1/2)), e ~= 0, - ((d^2 - e^2*x^2)^(1/2)*(2*d^2 + e^2*x^2))/(3*e^3) - (d^3*
log(2*x*(-e^2)^(1/2) + 2*(d^2 - e^2*x^2)^(1/2)))/(2*(-e^2)^(3/2)) - (d*x*(d^2 - e^2*x^2)^(1/2))/(2*e^2))

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sympy [C]  time = 5.25, size = 177, normalized size = 1.72 \[ d \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {i d x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x}{2 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{3}}{2 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \]

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**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)), T
rue)) + 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))

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