3.171 \(\int \frac {x^4}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\)

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

[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 = 0.24, 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 = {852, 1635, 1814, 12, 217, 203} \[ -\frac {d^3 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {17 d^2 (d-e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (15 d-13 e x)}{15 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*(d - e*x)^2)/(5*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*(15*
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 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 852

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 1635

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 1814

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 {\operatorname {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 = 0.16, size = 106, normalized size = 0.86 \[ \sqrt {d^2-e^2 x^2} \left (-\frac {d^2}{10 e^5 (d+e x)^3}+\frac {31 d}{60 e^5 (d+e x)^2}-\frac {1}{8 e^5 (e x-d)}-\frac {193}{120 e^5 (d+e x)}\right )-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

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]*(-1/8*1/(e^5*(-d + e*x)) - d^2/(10*e^5*(d + e*x)^3) + (31*d)/(60*e^5*(d + e*x)^2) - 193/(1
20*e^5*(d + e*x))) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^5

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fricas [A]  time = 0.95, size = 171, normalized size = 1.39 \[ -\frac {16 \, e^{4} x^{4} + 32 \, d e^{3} x^{3} - 32 \, d^{3} e x - 16 \, d^{4} - 30 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (26 \, e^{3} x^{3} + 22 \, d e^{2} x^{2} - 17 \, d^{2} e x - 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{9} x^{4} + 2 \, d e^{8} x^{3} - 2 \, d^{3} e^{6} x - d^{4} e^{5}\right )}} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.02, size = 198, normalized size = 1.61 \[ \frac {4 x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {34 x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{4}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}}-\frac {d^{3}}{5 \left (x +\frac {d}{e}\right )^{2} \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{7}}+\frac {17 d^{2}}{15 \left (x +\frac {d}{e}\right ) \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{6}}-\frac {2 d}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}} \]

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)

[Out]

4/(-e^2*x^2+d^2)^(1/2)/e^4*x-1/(e^2)^(1/2)/e^4*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-2*d/e^5/(-e^2*x^2+d^
2)^(1/2)+17/15/e^6*d^2/(x+d/e)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-34/15/e^4/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/
2)*x-1/5*d^3/e^7/(x+d/e)^2/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [A]  time = 1.02, size = 170, normalized size = 1.38 \[ -\frac {d^{3}}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{7} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{5}\right )}} + \frac {17 \, d^{2}}{15 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{6} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{5}\right )}} + \frac {26 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{5}} - \frac {2 \, d}{\sqrt {-e^{2} x^{2} + d^{2}} e^{5}} \]

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

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

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

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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