3.2.19 \(\int \frac {x^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [119]

Optimal. Leaf size=118 \[ \frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \]

[Out]

-3/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5+x^3*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(1/2)-4/3*x^2*(-e^2*x^2+d^2)^(1/
2)/e^3-1/6*d*(-9*e*x+16*d)*(-e^2*x^2+d^2)^(1/2)/e^5

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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 = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {864, 833, 847, 794, 223, 209} \begin {gather*} -\frac {3 d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}+\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) - (4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^3) - (d*(16*d - 9*e*x)*Sqrt[d^2 -
e^2*x^2])/(6*e^5) - (3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^5)

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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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^4}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^4 (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x^2 \left (3 d^3-4 d^2 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}+\frac {\int \frac {x \left (8 d^4 e-9 d^3 e^2 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{3 d^2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {\left (3 d^3\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}\\ &=\frac {x^3 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {4 x^2 \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {d (16 d-9 e x) \sqrt {d^2-e^2 x^2}}{6 e^5}-\frac {3 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 109, normalized size = 0.92 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-16 d^3-7 d^2 e x+d e^2 x^2-2 e^3 x^3\right )}{6 e^5 (d+e x)}+\frac {3 d^3 \left (-e^2\right )^{3/2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 7*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3))/(6*e^5*(d + e*x)) + (3*d^3*(-e^2)^(3/2)*Lo
g[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^8)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(104)=208\).
time = 0.07, size = 215, normalized size = 1.82

method	result	size
risch	\(-\frac {\left (2 e^{2} x^{2}-3 d e x +10 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e^{5}}-\frac {3 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{4} \sqrt {e^{2}}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}\)	\(120\)
default	\(\frac {-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}}{e}-\frac {d \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{e^{2}}-\frac {d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(-1/3*x^2/e^2*(-e^2*x^2+d^2)^(1/2)-2/3*d^2/e^4*(-e^2*x^2+d^2)^(1/2))-d/e^2*(-1/2*x/e^2*(-e^2*x^2+d^2)^(1/2
)+1/2*d^2/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))-d^2/e^5*(-e^2*x^2+d^2)^(1/2)-d^3/e^4/(e^
2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/e^6*d^3/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

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Maxima [A]
time = 0.48, size = 104, normalized size = 0.88 \begin {gather*} -\frac {3}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{\left (-3\right )} + \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{\left (-4\right )} - \frac {5}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-5\right )} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} d^{3}}{x e^{6} + d e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*d^3*arcsin(x*e/d)*e^(-5) - 1/3*sqrt(-x^2*e^2 + d^2)*x^2*e^(-3) + 1/2*sqrt(-x^2*e^2 + d^2)*d*x*e^(-4) - 5/
3*sqrt(-x^2*e^2 + d^2)*d^2*e^(-5) - sqrt(-x^2*e^2 + d^2)*d^3/(x*e^6 + d*e^5)

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Fricas [A]
time = 2.03, size = 108, normalized size = 0.92 \begin {gather*} -\frac {16 \, d^{3} x e + 16 \, d^{4} - 18 \, {\left (d^{3} x e + d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (2 \, x^{3} e^{3} - d x^{2} e^{2} + 7 \, d^{2} x e + 16 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{6 \, {\left (x e^{6} + d e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(16*d^3*x*e + 16*d^4 - 18*(d^3*x*e + d^4)*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (2*x^3*e^3 - d*x
^2*e^2 + 7*d^2*x*e + 16*d^3)*sqrt(-x^2*e^2 + d^2))/(x*e^6 + d*e^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.50, size = 92, normalized size = 0.78 \begin {gather*} -\frac {3}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\left (d\right ) + \frac {2 \, d^{3} e^{\left (-5\right )}}{\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1} - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (10 \, d^{2} e^{\left (-5\right )} + {\left (2 \, x e^{\left (-3\right )} - 3 \, d e^{\left (-4\right )}\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*d^3*arcsin(x*e/d)*e^(-5)*sgn(d) + 2*d^3*e^(-5)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1) - 1/6*sqrt(-
x^2*e^2 + d^2)*(10*d^2*e^(-5) + (2*x*e^(-3) - 3*d*e^(-4))*x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{\sqrt {d^2-e^2\,x^2}\,\left (d+e\,x\right )} \,d x \end {gather*}

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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