3.2.80 \(\int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [180]

Optimal. Leaf size=146 \[ -\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

-1/5*d^3*(-e*x+d)^3/e^5/(-e^2*x^2+d^2)^(5/2)+6/5*d^2*(-e*x+d)^2/e^5/(-e^2*x^2+d^2)^(3/2)-3*d*arctan(e*x/(-e^2*
x^2+d^2)^(1/2))/e^5-24/5*d*(-e*x+d)/e^5/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/e^5

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Rubi [A]
time = 0.22, antiderivative size = 146, 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 = {866, 1649, 655, 223, 209} \begin {gather*} -\frac {3 d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-1/5*(d^3*(d - e*x)^3)/(e^5*(d^2 - e^2*x^2)^(5/2)) + (6*d^2*(d - e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) - (24*d
*(d - e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 866

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

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^4 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^2 \left (\frac {3 d^4}{e^4}-\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}-\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d-e x) \left (\frac {27 d^4}{e^4}-\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {45 d^4}{e^4}-\frac {15 d^3 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 106, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-24 d^3-57 d^2 e x-39 d e^2 x^2-5 e^3 x^3\right )}{5 e^5 (d+e x)^3}+\frac {3 d \left (-e^2\right )^{3/2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-24*d^3 - 57*d^2*e*x - 39*d*e^2*x^2 - 5*e^3*x^3))/(5*e^5*(d + e*x)^3) + (3*d*(-e^2)^(3/2
)*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^8

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(339\) vs. \(2(130)=260\).
time = 0.09, size = 340, normalized size = 2.33

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {24 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{6} \left (x +\frac {d}{e}\right )}+\frac {6 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{7} \left (x +\frac {d}{e}\right )^{2}}-\frac {d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{8} \left (x +\frac {d}{e}\right )^{3}}\)	\(187\)
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}}-\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{4} \sqrt {e^{2}}}-\frac {6 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{6} \left (x +\frac {d}{e}\right )}+\frac {d^{4} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{7}}-\frac {4 d^{3} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{6}}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 146, normalized size = 1.00 \begin {gather*} -3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} d^{3}}{5 \, {\left (x^{3} e^{8} + 3 \, d x^{2} e^{7} + 3 \, d^{2} x e^{6} + d^{3} e^{5}\right )}} + \frac {6 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2}}{5 \, {\left (x^{2} e^{7} + 2 \, d x e^{6} + d^{2} e^{5}\right )}} - \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-5\right )} - \frac {24 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{5 \, {\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)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-3*d*arcsin(x*e/d)*e^(-5) - 1/5*sqrt(-x^2*e^2 + d^2)*d^3/(x^3*e^8 + 3*d*x^2*e^7 + 3*d^2*x*e^6 + d^3*e^5) + 6/5
*sqrt(-x^2*e^2 + d^2)*d^2/(x^2*e^7 + 2*d*x*e^6 + d^2*e^5) - sqrt(-x^2*e^2 + d^2)*e^(-5) - 24/5*sqrt(-x^2*e^2 +
d^2)*d/(x*e^6 + d*e^5)

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Fricas [A]
time = 2.74, size = 164, normalized size = 1.12 \begin {gather*} -\frac {24 \, d x^{3} e^{3} + 72 \, d^{2} x^{2} e^{2} + 72 \, d^{3} x e + 24 \, d^{4} - 30 \, {\left (d x^{3} e^{3} + 3 \, d^{2} x^{2} e^{2} + 3 \, 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 (5 \, x^{3} e^{3} + 39 \, d x^{2} e^{2} + 57 \, d^{2} x e + 24 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (x^{3} e^{8} + 3 \, d x^{2} e^{7} + 3 \, d^{2} x e^{6} + d^{3} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(24*d*x^3*e^3 + 72*d^2*x^2*e^2 + 72*d^3*x*e + 24*d^4 - 30*(d*x^3*e^3 + 3*d^2*x^2*e^2 + 3*d^3*x*e + d^4)*a
rctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (5*x^3*e^3 + 39*d*x^2*e^2 + 57*d^2*x*e + 24*d^3)*sqrt(-x^2*e^2 +
d^2))/(x^3*e^8 + 3*d*x^2*e^7 + 3*d^2*x*e^6 + d^3*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 )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

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Giac [A]
time = 2.12, size = 194, normalized size = 1.33 \begin {gather*} -3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\left (d\right ) - \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-5\right )} + \frac {2 \, {\left (\frac {80 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{\left (-2\right )}}{x} + \frac {120 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{\left (-4\right )}}{x^{2}} + \frac {70 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{\left (-6\right )}}{x^{3}} + \frac {15 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d e^{\left (-8\right )}}{x^{4}} + 19 \, d\right )} e^{\left (-5\right )}}{5 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

-3*d*arcsin(x*e/d)*e^(-5)*sgn(d) - sqrt(-x^2*e^2 + d^2)*e^(-5) + 2/5*(80*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^(-
2)/x + 120*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^(-4)/x^2 + 70*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^(-6)/x^3 +
15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d*e^(-8)/x^4 + 19*d)*e^(-5)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^
5

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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 )}^3} \,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)^3),x)

[Out]

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

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