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

Optimal. Leaf size=118 \[ -\frac{d^2 (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{24 e^4}-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{3 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^4} \]

[Out]

-(d*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^2) + (x^3*Sqrt[d^2 - e^2*x^2])/(4*e) - (d^2*(16*d - 9*e*x)*Sqrt[d^2 - e^2*x^
2])/(24*e^4) - (3*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^4)

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Rubi [A]  time = 0.0989571, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {850, 833, 780, 217, 203} \[ -\frac{d^2 (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{24 e^4}-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{3 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^2) + (x^3*Sqrt[d^2 - e^2*x^2])/(4*e) - (d^2*(16*d - 9*e*x)*Sqrt[d^2 - e^2*x^
2])/(24*e^4) - (3*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^4)

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{x^3 \sqrt{d^2-e^2 x^2}}{d+e x} \, dx &=\int \frac{x^3 (d-e x)}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{\int \frac{x^2 \left (3 d^2 e-4 d e^2 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}+\frac{\int \frac{x \left (8 d^3 e^2-9 d^2 e^3 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{12 e^4}\\ &=-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{d^2 (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{24 e^4}-\frac{\left (3 d^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^3}\\ &=-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{d^2 (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{24 e^4}-\frac{\left (3 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^3}\\ &=-\frac{d x^2 \sqrt{d^2-e^2 x^2}}{3 e^2}+\frac{x^3 \sqrt{d^2-e^2 x^2}}{4 e}-\frac{d^2 (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{24 e^4}-\frac{3 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^4}\\ \end{align*}

Mathematica [A]  time = 0.115075, size = 80, normalized size = 0.68 \[ \frac{\sqrt{d^2-e^2 x^2} \left (9 d^2 e x-16 d^3-8 d e^2 x^2+6 e^3 x^3\right )-9 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{24 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 + 9*d^2*e*x - 8*d*e^2*x^2 + 6*e^3*x^3) - 9*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]
)/(24*e^4)

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Maple [A]  time = 0.064, size = 185, normalized size = 1.6 \begin{align*} -{\frac{x}{4\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{d}^{2}x}{8\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{5\,{d}^{4}}{8\,{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{d}{3\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{3}}{{e}^{4}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{{d}^{4}}{{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54405, size = 177, normalized size = 1.5 \begin{align*} \frac{18 \, d^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} + 9 \, d^{2} e x - 16 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

1/24*(18*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 - 8*d*e^2*x^2 + 9*d^2*e*x - 16*d^3)*sqrt(-
e^2*x^2 + d^2))/e^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(1/2)/(e*x+d),x)

[Out]

Integral(x**3*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x), x)

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Giac [A]  time = 1.15272, size = 89, normalized size = 0.75 \begin{align*} -\frac{3}{8} \, d^{4} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-4\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{24} \,{\left (16 \, d^{3} e^{\left (-4\right )} -{\left (9 \, d^{2} e^{\left (-3\right )} + 2 \,{\left (3 \, x e^{\left (-1\right )} - 4 \, d e^{\left (-2\right )}\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

-3/8*d^4*arcsin(x*e/d)*e^(-4)*sgn(d) - 1/24*(16*d^3*e^(-4) - (9*d^2*e^(-3) + 2*(3*x*e^(-1) - 4*d*e^(-2))*x)*x)
*sqrt(-x^2*e^2 + d^2)