3.102 \(\int \frac {x^2 (d^2-e^2 x^2)^{3/2}}{d+e x} \, dx\)

Optimal. Leaf size=113 \[ \frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}+\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2} \]

[Out]

1/12*d*(-3*e*x+4*d)*(-e^2*x^2+d^2)^(3/2)/e^3-1/5*(-e^2*x^2+d^2)^(5/2)/e^3+1/8*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1
/2))/e^3+1/8*d^3*x*(-e^2*x^2+d^2)^(1/2)/e^2

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Rubi [A]  time = 0.13, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1639, 12, 785, 780, 195, 217, 203} \[ \frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}+\frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d^2 - e^2*x^2)^(3/2))/(d + e*x),x]

[Out]

(d^3*x*Sqrt[d^2 - e^2*x^2])/(8*e^2) + (d*(4*d - 3*e*x)*(d^2 - e^2*x^2)^(3/2))/(12*e^3) - (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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 0]

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {\int \frac {5 d e^3 x \left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx}{5 e^4}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {d \int \frac {x \left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx}{e}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}-\frac {\int x \left (d^2 e-d e^2 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{e^2}\\ &=\frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^3 \int \sqrt {d^2-e^2 x^2} \, dx}{4 e^2}\\ &=\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}+\frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\frac {d^5 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}+\frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e^3}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e^3}+\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 )}{8 e^2}\\ &=\frac {d^3 x \sqrt {d^2-e^2 x^2}}{8 e^2}+\frac {d (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 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}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 112, normalized size = 0.99 \[ \frac {\sqrt {d^2-e^2 x^2} \left (15 d^4 \sin ^{-1}\left (\frac {e x}{d}\right )+\sqrt {1-\frac {e^2 x^2}{d^2}} \left (16 d^4-15 d^3 e x+8 d^2 e^2 x^2+30 d e^3 x^3-24 e^4 x^4\right )\right )}{120 e^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d^2 - e^2*x^2)^(3/2))/(d + e*x),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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fricas [A]  time = 0.88, size = 94, normalized size = 0.83 \[ -\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (24 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} - 8 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x - 16 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(3/2)/(e*x+d),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 [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(3/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(4*d^5*exp(2)^3-4*d^5*exp(1)^4*exp(2
))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+e
xp(2)^2)/exp(1)^6/exp(1)+1/8*d^5*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)+2*((((-192*exp(1)^7*1/1920/exp(
1)^6*x+240*exp(1)^6*d*1/1920/exp(1)^6)*x+64*exp(1)^5*d^2*1/1920/exp(1)^6)*x-120*exp(1)^4*d^3*1/1920/exp(1)^6)*
x+128*exp(1)^3*d^4*1/1920/exp(1)^6)*sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.02, size = 222, normalized size = 1.96 \[ \frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}-\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x}{8 e^{2}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3} x}{2 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d x}{4 e^{2}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{2}}{3 e^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-e^2*x^2+d^2)^(3/2)/(e*x+d),x)

[Out]

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

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maxima [C]  time = 1.01, size = 174, normalized size = 1.54 \[ -\frac {i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{2 \, e^{3}} - \frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{3}} + \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{2 \, e^{2}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{2}} + \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{3 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(3/2)/(e*x+d),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d^2 - e^2*x^2)^(3/2))/(d + e*x),x)

[Out]

int((x^2*(d^2 - e^2*x^2)^(3/2))/(d + e*x), x)

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sympy [C]  time = 7.85, size = 279, normalized size = 2.47 \[ d \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - e \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-e**2*x**2+d**2)**(3/2)/(e*x+d),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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