3.84 \(\int \frac{x^4 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac{d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{24 d (d+e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

(d^3*(d + e*x)^3)/(5*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

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Rubi [A]  time = 0.324808, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1635, 641, 217, 203} \[ \frac{d^3 (d+e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{6 d^2 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{24 d (d+e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^3*(d + e*x)^3)/(5*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 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 641

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 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)^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) \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)^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*}

Mathematica [A]  time = 0.211784, size = 119, normalized size = 0.84 \[ \frac{(d+e x) \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (-57 d^2 e x+24 d^3+39 d e^2 x^2-5 e^3 x^3\right )-15 (d-e x)^3 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{5 e^5 (d-e x)^2 \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(Sqrt[1 - (e^2*x^2)/d^2]*(24*d^3 - 57*d^2*e*x + 39*d*e^2*x^2 - 5*e^3*x^3) - 15*(d - e*x)^3*ArcSin[(
e*x)/d]))/(5*e^5*(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 = 0.114, size = 262, normalized size = 1.9 \begin{align*} -{e{x}^{6} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+9\,{\frac{{d}^{2}{x}^{4}}{e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}-12\,{\frac{{d}^{4}{x}^{2}}{{e}^{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}+{\frac{24\,{d}^{6}}{5\,{e}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{3\,d{x}^{5}}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{d{x}^{3}}{{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,dx}{5\,{e}^{4}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-3\,{\frac{d}{{e}^{4}\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}} \right ) }+{\frac{{d}^{3}{x}^{3}}{2\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{3\,{d}^{5}x}{10\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{{d}^{3}x}{10\,{e}^{4}} \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^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-e*x^6/(-e^2*x^2+d^2)^(5/2)+9/e*d^2*x^4/(-e^2*x^2+d^2)^(5/2)-12/e^3*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+24/5/e^5*d^6/
(-e^2*x^2+d^2)^(5/2)+3/5*d*x^5/(-e^2*x^2+d^2)^(5/2)-d/e^2*x^3/(-e^2*x^2+d^2)^(3/2)+16/5*d/e^4*x/(-e^2*x^2+d^2)
^(1/2)-3*d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+1/2*d^3*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/1
0*d^5/e^4*x/(-e^2*x^2+d^2)^(5/2)+1/10*d^3/e^4*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [B]  time = 1.52637, size = 455, normalized size = 3.2 \begin{align*} \frac{1}{5} \, d e^{2} 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{e x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - d 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{9 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{d^{3} x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{12 \, d^{4} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{3}} - \frac{3 \, d^{5} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{24 \, d^{6}}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{5}} + \frac{9 \, d^{3} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{4}} - \frac{6 \, d x}{5 \, \sqrt{-e^{2} x^{2} + d^{2}} e^{4}} - \frac{3 \, d \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}} e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/5*d*e^2*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)) - e*x^6/(-e^2*x^2 + d^2)^(5/2) - d*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)) + 9*d^2*x^4/((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^3*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 12*d^4
*x^2/((-e^2*x^2 + d^2)^(5/2)*e^3) - 3/10*d^5*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 24/5*d^6/((-e^2*x^2 + d^2)^(5/2)
*e^5) + 9/10*d^3*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 6/5*d*x/(sqrt(-e^2*x^2 + d^2)*e^4) - 3*d*arcsin(e^2*x/sqrt(d
^2*e^2))/(sqrt(e^2)*e^4)

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Fricas [A]  time = 1.66556, size = 367, normalized size = 2.58 \begin{align*} \frac{24 \, d e^{3} x^{3} - 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x - 24 \, d^{4} + 30 \,{\left (d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x - d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (5 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 57 \, d^{2} e x - 24 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (e^{8} x^{3} - 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x - d^{3} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/5*(24*d*e^3*x^3 - 72*d^2*e^2*x^2 + 72*d^3*e*x - 24*d^4 + 30*(d*e^3*x^3 - 3*d^2*e^2*x^2 + 3*d^3*e*x - d^4)*ar
ctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^3*x^3 - 39*d*e^2*x^2 + 57*d^2*e*x - 24*d^3)*sqrt(-e^2*x^2 + d^2
))/(e^8*x^3 - 3*d*e^7*x^2 + 3*d^2*e^6*x - d^3*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \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**4*(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**4*(d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.16533, size = 144, normalized size = 1.01 \begin{align*} -3 \, d \arcsin \left (\frac{x e}{d}\right ) e^{\left (-5\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (24 \, d^{6} e^{\left (-5\right )} +{\left (15 \, d^{5} e^{\left (-4\right )} -{\left (60 \, d^{4} e^{\left (-3\right )} +{\left (35 \, d^{3} e^{\left (-2\right )} -{\left (45 \, d^{2} e^{\left (-1\right )} -{\left (5 \, x e - 24 \, d\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}}}{5 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-3*d*arcsin(x*e/d)*e^(-5)*sgn(d) - 1/5*(24*d^6*e^(-5) + (15*d^5*e^(-4) - (60*d^4*e^(-3) + (35*d^3*e^(-2) - (45
*d^2*e^(-1) - (5*x*e - 24*d)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3