3.92 \(\int \frac {x^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx\)

Optimal. Leaf size=147 \[ \frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5} \]

[Out]

3/8*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5+4/15*d^2*x^2*(-e^2*x^2+d^2)^(1/2)/e^3-1/4*d*x^3*(-e^2*x^2+d^2)^(1
/2)/e^2+1/5*x^4*(-e^2*x^2+d^2)^(1/2)/e+1/120*d^3*(-45*e*x+64*d)*(-e^2*x^2+d^2)^(1/2)/e^5

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Rubi [A]  time = 0.14, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*Sqrt[d^2 - e^2*x^2])/(d + e*x),x]

[Out]

(4*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (d*x^3*Sqrt[d^2 - e^2*x^2])/(4*e^2) + (x^4*Sqrt[d^2 - e^2*x^2])/(5*
e) + (d^3*(64*d - 45*e*x)*Sqrt[d^2 - e^2*x^2])/(120*e^5) + (3*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^5)

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^4 \sqrt {d^2-e^2 x^2}}{d+e x} \, dx &=\int \frac {x^4 (d-e x)}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x^3 \left (4 d^2 e-5 d e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {\int \frac {x^2 \left (15 d^3 e^2-16 d^2 e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{20 e^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {\int \frac {x \left (32 d^4 e^3-45 d^3 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{60 e^6}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {\left (3 d^5\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^4}\\ &=\frac {4 d^2 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{4 e^2}+\frac {x^4 \sqrt {d^2-e^2 x^2}}{5 e}+\frac {d^3 (64 d-45 e x) \sqrt {d^2-e^2 x^2}}{120 e^5}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^5}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 91, normalized size = 0.62 \[ \frac {45 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\sqrt {d^2-e^2 x^2} \left (64 d^4-45 d^3 e x+32 d^2 e^2 x^2-30 d e^3 x^3+24 e^4 x^4\right )}{120 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*Sqrt[d^2 - e^2*x^2])/(d + e*x),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(64*d^4 - 45*d^3*e*x + 32*d^2*e^2*x^2 - 30*d*e^3*x^3 + 24*e^4*x^4) + 45*d^5*ArcTan[(e*x)/
Sqrt[d^2 - e^2*x^2]])/(120*e^5)

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fricas [A]  time = 0.87, size = 95, normalized size = 0.65 \[ -\frac {90 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (24 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 32 \, d^{2} e^{2} x^{2} - 45 \, d^{3} e x + 64 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

-1/120*(90*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*e^4*x^4 - 30*d*e^3*x^3 + 32*d^2*e^2*x^2 - 45*d^
3*e*x + 64*d^4)*sqrt(-e^2*x^2 + d^2))/e^5

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giac [A]  time = 0.20, size = 77, normalized size = 0.52 \[ \frac {3}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) + \frac {1}{120} \, {\left (64 \, d^{4} e^{\left (-5\right )} - {\left (45 \, d^{3} e^{\left (-4\right )} - 2 \, {\left (16 \, d^{2} e^{\left (-3\right )} + 3 \, {\left (4 \, x e^{\left (-1\right )} - 5 \, d e^{\left (-2\right )}\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

3/8*d^5*arcsin(x*e/d)*e^(-5)*sgn(d) + 1/120*(64*d^4*e^(-5) - (45*d^3*e^(-4) - 2*(16*d^2*e^(-3) + 3*(4*x*e^(-1)
- 5*d*e^(-2))*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.02, size = 208, normalized size = 1.41 \[ \frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}}-\frac {5 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{4}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x}{8 e^{4}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4}}{e^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} x^{2}}{5 e^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d x}{4 e^{4}}-\frac {7 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2}}{15 e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x)

[Out]

-1/5/e^3*x^2*(-e^2*x^2+d^2)^(3/2)-7/15*d^2/e^5*(-e^2*x^2+d^2)^(3/2)+1/4*d/e^4*x*(-e^2*x^2+d^2)^(3/2)-5/8*d^3/e
^4*x*(-e^2*x^2+d^2)^(1/2)-5/8*d^5/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+d^4/e^5*(-(x+d/e)
^2*e^2+2*d*e*(x+d/e))^(1/2)+d^5/e^4/(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 = 0.99, size = 125, normalized size = 0.85 \[ \frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{5}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{8 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{5}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{4}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{15 \, e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

3/8*d^5*arcsin(e*x/d)/e^5 - 5/8*sqrt(-e^2*x^2 + d^2)*d^3*x/e^4 - 1/5*(-e^2*x^2 + d^2)^(3/2)*x^2/e^3 + sqrt(-e^
2*x^2 + d^2)*d^4/e^5 + 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e^4 - 7/15*(-e^2*x^2 + d^2)^(3/2)*d^2/e^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\sqrt {d^2-e^2\,x^2}}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d^2 - e^2*x^2)^(1/2))/(d + e*x),x)

[Out]

int((x^4*(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 \[ \int \frac {x^{4} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(-e**2*x**2+d**2)**(1/2)/(e*x+d),x)

[Out]

Integral(x**4*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x), x)

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