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

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]

-(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

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Rubi [A]  time = 0.237506, antiderivative size = 123, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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])

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*}

Mathematica [A]  time = 0.172455, 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*e^5*(-d + e*x)) - d^2/(10*e^5*(d + e*x)^3) + (31*d)/(60*e^5*(d + e*x)^2) - 193/(120
*e^5*(d + e*x))) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^5

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Maple [A]  time = 0.069, size = 198, normalized size = 1.6 \begin{align*} 4\,{\frac{x}{{e}^{4}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}-{\frac{1}{{e}^{4}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-2\,{\frac{d}{{e}^{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}+{\frac{17\,{d}^{2}}{15\,{e}^{6}} \left ({\frac{d}{e}}+x \right ) ^{-1}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{34\,x}{15\,{e}^{4}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{{d}^{3}}{5\,{e}^{7}} \left ({\frac{d}{e}}+x \right ) ^{-2}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.59017, size = 360, normalized size = 2.93 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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