3.85 \(\int \frac{x^3 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{32 (d+e x)}{15 e^4 \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

(d^2*(d + e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d + e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d +
e*x))/(15*e^4*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^4

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Rubi [A]  time = 0.215959, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1635, 778, 217, 203} \[ \frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{32 (d+e x)}{15 e^4 \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^2*(d + e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d + e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d +
e*x))/(15*e^4*Sqrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^4

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 778

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

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^3 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d+e x)^2 \left (\frac{3 d^3}{e^3}+\frac{5 d^2 x}{e^2}+\frac{5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{\left (\frac{17 d^3}{e^3}+\frac{15 d^2 x}{e^2}\right ) (d+e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{32 (d+e x)}{15 e^4 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^3}\\ &=\frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{32 (d+e x)}{15 e^4 \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^3}\\ &=\frac{d^2 (d+e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{13 d (d+e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{32 (d+e x)}{15 e^4 \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.16662, size = 112, normalized size = 0.95 \[ \frac{(d+e x) \left (d \left (22 d^2-51 d e x+32 e^2 x^2\right ) \sqrt{1-\frac{e^2 x^2}{d^2}}-15 (d-e x)^3 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{15 d e^4 (d-e x)^2 \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(d*(22*d^2 - 51*d*e*x + 32*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^3*ArcSin[(e*x)/d]))/(15*
d*e^4*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [B]  time = 0.093, size = 234, normalized size = 2. \begin{align*}{\frac{e{x}^{5}}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{{x}^{3}}{3\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,x}{5\,{e}^{3}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{1}{{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+3\,{\frac{d{x}^{4}}{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}-{\frac{11\,{d}^{3}{x}^{2}}{3\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{22\,{d}^{5}}{15\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{3\,{d}^{2}{x}^{3}}{2\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{9\,{d}^{4}x}{10\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{3\,{d}^{2}x}{10\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/5*e*x^5/(-e^2*x^2+d^2)^(5/2)-1/3/e*x^3/(-e^2*x^2+d^2)^(3/2)+8/5/e^3*x/(-e^2*x^2+d^2)^(1/2)-1/e^3/(e^2)^(1/2)
*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+3*d*x^4/(-e^2*x^2+d^2)^(5/2)-11/3*d^3/e^2*x^2/(-e^2*x^2+d^2)^(5/2)
+22/15*d^5/e^4/(-e^2*x^2+d^2)^(5/2)+3/2/e*d^2*x^3/(-e^2*x^2+d^2)^(5/2)-9/10/e^3*d^4*x/(-e^2*x^2+d^2)^(5/2)+3/1
0/e^3*d^2*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [B]  time = 1.56242, size = 417, normalized size = 3.53 \begin{align*} \frac{1}{15} \, e^{3} x{\left (\frac{15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{6}}\right )} - \frac{1}{3} \, e x{\left (\frac{3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{2}} - \frac{2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{4}}\right )} + \frac{3 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{3 \, d^{2} x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} - \frac{11 \, d^{3} x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{9 \, d^{4} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{3}} + \frac{22 \, d^{5}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{17 \, d^{2} x}{30 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{3}} + \frac{2 \, x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} e^{3}} - \frac{\arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^3*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 1/3*e*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) + 3*d*x^
4/(-e^2*x^2 + d^2)^(5/2) + 3/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e) - 11/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2)
- 9/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)*e^3) + 22/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^4) + 17/30*d^2*x/((-e^2*x^2 +
d^2)^(3/2)*e^3) + 2/15*x/(sqrt(-e^2*x^2 + d^2)*e^3) - arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^3)

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Fricas [A]  time = 1.69453, size = 336, normalized size = 2.85 \begin{align*} \frac{22 \, e^{3} x^{3} - 66 \, d e^{2} x^{2} + 66 \, d^{2} e x - 22 \, d^{3} + 30 \,{\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (32 \, e^{2} x^{2} - 51 \, d e x + 22 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (e^{7} x^{3} - 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x - d^{3} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(22*e^3*x^3 - 66*d*e^2*x^2 + 66*d^2*e*x - 22*d^3 + 30*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (32*e^2*x^2 - 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 - 3*d*e^6*x
^2 + 3*d^2*e^5*x - d^3*e^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**3*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.18498, size = 128, normalized size = 1.08 \begin{align*} -\arcsin \left (\frac{x e}{d}\right ) e^{\left (-4\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (22 \, d^{5} e^{\left (-4\right )} +{\left (15 \, d^{4} e^{\left (-3\right )} -{\left (55 \, d^{3} e^{\left (-2\right )} +{\left (35 \, d^{2} e^{\left (-1\right )} -{\left (32 \, x e + 45 \, d\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-4)*sgn(d) - 1/15*(22*d^5*e^(-4) + (15*d^4*e^(-3) - (55*d^3*e^(-2) + (35*d^2*e^(-1) - (32*x*
e + 45*d)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3