3.1.93 \(\int \frac {x^3 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\) [93]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 = {864, 847, 794, 223, 209} \begin {gather*} -\frac {3 d^4 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(d*x^2*Sqrt[d^2 - e^2*x^2])/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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 794

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 847

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 864

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]))

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 ) \text {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*}

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Mathematica [A]
time = 0.16, size = 100, normalized size = 0.85 \begin {gather*} \frac {e \sqrt {d^2-e^2 x^2} \left (-16 d^3+9 d^2 e x-8 d e^2 x^2+6 e^3 x^3\right )-9 d^4 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{24 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*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*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x)
+ Sqrt[d^2 - e^2*x^2]])/(24*e^5)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(242\) vs. \(2(102)=204\).
time = 0.08, size = 243, normalized size = 2.06

method	result	size
risch	\(-\frac {\left (-6 e^{3} x^{3}+8 d \,e^{2} x^{2}-9 d^{2} e x +16 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e^{4}}-\frac {3 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{3} \sqrt {e^{2}}}\)	\(86\)
default	\(\frac {-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 e^{2}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4 e^{2}}}{e}+\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 e^{4}}+\frac {d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{e^{3}}-\frac {d^{3} \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{4}}\)	\(243\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 93, normalized size = 0.79 \begin {gather*} -\frac {3}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} + \frac {5}{8} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} x e^{\left (-3\right )} - \sqrt {-x^{2} e^{2} + d^{2}} d^{3} e^{\left (-4\right )} - \frac {1}{4} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} x e^{\left (-3\right )} + \frac {1}{3} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{\left (-4\right )} \end {gather*}

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]

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

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Fricas [A]
time = 2.47, size = 78, normalized size = 0.66 \begin {gather*} \frac {1}{24} \, {\left (18 \, d^{4} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (6 \, x^{3} e^{3} - 8 \, d x^{2} e^{2} + 9 \, d^{2} x e - 16 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-4\right )} \end {gather*}

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(-x^2*e^2 + d^2))*e^(-1)/x) + (6*x^3*e^3 - 8*d*x^2*e^2 + 9*d^2*x*e - 16*d^3)*sqr
t(-x^2*e^2 + d^2))*e^(-4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \end {gather*}

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 = 0.66, size = 66, normalized size = 0.56 \begin {gather*} -\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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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