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

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

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Rubi [A]  time = 0.10, 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 = {850, 833, 780, 217, 203} \begin {gather*} -\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} \end {gather*}

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 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 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 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]))

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*}

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Mathematica [A]  time = 0.10, size = 80, normalized size = 0.68 \begin {gather*} \frac {\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 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{24 e^4} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.27, size = 103, normalized size = 0.87 \begin {gather*} \frac {\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 )}{24 e^4}-\frac {3 d^4 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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))/(24*e^4) - (3*d^4*Sqrt[-e^2]*Log[-(Sqrt[
-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e^5)

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fricas [A]  time = 0.40, size = 83, normalized size = 0.70 \begin {gather*} \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 {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(-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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giac [A]  time = 0.21, 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}\relax (d) - \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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maple [A]  time = 0.01, size = 185, normalized size = 1.57 \begin {gather*} -\frac {d^{4} \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 )}{\sqrt {e^{2}}\, e^{3}}+\frac {5 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{3}}+\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x}{8 e^{3}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3}}{e^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} x}{4 e^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}{3 e^{4}} \end {gather*}

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)/
(-e^2*x^2+d^2)^(1/2)*x)+1/3*d/e^4*(-e^2*x^2+d^2)^(3/2)-d^3/e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-d^4/e^3/(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 [A]  time = 0.99, size = 101, normalized size = 0.86 \begin {gather*} -\frac {3 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{4}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x}{4 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{3 \, e^{4}} \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(e*x/d)/e^4 + 5/8*sqrt(-e^2*x^2 + d^2)*d^2*x/e^3 - sqrt(-e^2*x^2 + d^2)*d^3/e^4 - 1/4*(-e^2*x^2
+ d^2)^(3/2)*x/e^3 + 1/3*(-e^2*x^2 + d^2)^(3/2)*d/e^4

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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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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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