3.2.71 \(\int \frac {x^4}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [171]

Optimal. Leaf size=123 \[ -\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

-1/5*d^3*(-e*x+d)^2/e^5/(-e^2*x^2+d^2)^(5/2)+17/15*d^2*(-e*x+d)/e^5/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+
d^2)^(1/2))/e^5-2/15*(-13*e*x+15*d)/e^5/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {866, 1649, 1828, 12, 223, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {d^3 (d-e x)^2}{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)^2*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

-1/5*(d^3*(d - e*x)^2)/(e^5*(d^2 - e^2*x^2)^(5/2)) + (17*d^2*(d - e*x))/(15*e^5*(d^2 - e^2*x^2)^(3/2)) - (2*(1
5*d - 13*e*x))/(15*e^5*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^4}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac {x^4 (d-e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x) \left (\frac {2 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)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {11 d^4}{e^4}-\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {15 d^4}{e^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^4}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\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)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\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 = 0.41, size = 113, normalized size = 0.92 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (16 d^3+17 d^2 e x-22 d e^2 x^2-26 e^3 x^3\right )}{15 e^5 (-d+e x) (d+e x)^3}+\frac {\log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^4 \sqrt {-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(16*d^3 + 17*d^2*e*x - 22*d*e^2*x^2 - 26*e^3*x^3))/(15*e^5*(-d + e*x)*(d + e*x)^3) + Log[
-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]/(e^4*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(109)=218\).
time = 0.08, size = 363, normalized size = 2.95

method	result	size
default	\(\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}-\frac {2 d}{e^{5} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 x}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {4 d^{3} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{5}}+\frac {d^{4} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{6}}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))-2*d/e^5/(-e^2*
x^2+d^2)^(1/2)+3/e^4*x/(-e^2*x^2+d^2)^(1/2)-4/e^5*d^3*(-1/3/d/e/(x+d/e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1
/3/e/d^3*(-2*e^2*(x+d/e)+2*d*e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))+1/e^6*d^4*(-1/5/d/e/(x+d/e)^2/(-(x+d/e)^
2*e^2+2*d*e*(x+d/e))^(1/2)+3/5*e/d*(-1/3/d/e/(x+d/e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/e/d^3*(-2*e^2*(x
+d/e)+2*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 = 156, normalized size = 1.27 \begin {gather*} -\frac {d^{3}}{5 \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{7} + 2 \, \sqrt {-x^{2} e^{2} + d^{2}} d x e^{6} + \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{5}\right )}} - \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} + \frac {26 \, x e^{\left (-4\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}}} - \frac {2 \, d e^{\left (-5\right )}}{\sqrt {-x^{2} e^{2} + d^{2}}} + \frac {17 \, d^{2}}{15 \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x e^{6} + \sqrt {-x^{2} e^{2} + d^{2}} d e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*d^3/(sqrt(-x^2*e^2 + d^2)*x^2*e^7 + 2*sqrt(-x^2*e^2 + d^2)*d*x*e^6 + sqrt(-x^2*e^2 + d^2)*d^2*e^5) - arcs
in(x*e/d)*e^(-5) + 26/15*x*e^(-4)/sqrt(-x^2*e^2 + d^2) - 2*d*e^(-5)/sqrt(-x^2*e^2 + d^2) + 17/15*d^2/(sqrt(-x^
2*e^2 + d^2)*x*e^6 + sqrt(-x^2*e^2 + d^2)*d*e^5)

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Fricas [A]
time = 3.63, size = 161, normalized size = 1.31 \begin {gather*} -\frac {16 \, x^{4} e^{4} + 32 \, d x^{3} e^{3} - 32 \, d^{3} x e - 16 \, d^{4} - 30 \, {\left (x^{4} e^{4} + 2 \, d x^{3} e^{3} - 2 \, 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 (26 \, x^{3} e^{3} + 22 \, d x^{2} e^{2} - 17 \, d^{2} x e - 16 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{4} e^{9} + 2 \, d x^{3} e^{8} - 2 \, d^{3} x e^{6} - d^{4} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(16*x^4*e^4 + 32*d*x^3*e^3 - 32*d^3*x*e - 16*d^4 - 30*(x^4*e^4 + 2*d*x^3*e^3 - 2*d^3*x*e - d^4)*arctan(-
(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (26*x^3*e^3 + 22*d*x^2*e^2 - 17*d^2*x*e - 16*d^3)*sqrt(-x^2*e^2 + d^2))
/(x^4*e^9 + 2*d*x^3*e^8 - 2*d^3*x*e^6 - d^4*e^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/exp(1)*(1/32768*(4096/5*sqrt(2*sageVAR
d*(sageVARx*exp(1)+sageVARd)^-1/exp(1)*exp(1)-1)*(2*sageVARd*(sageVARx*exp(1)+sageVARd)^-1/exp(1)*exp(1)-1)^2*
exp(1)^16*(-sign((sag

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\left (d^2-e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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