3.44 \(\int \frac{x^5 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^6}-\frac{2 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^6} \]

[Out]

(d^4*(d + e*x)^2)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (22*d^3*(d + e*x))/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (2*d*(30
*d + 23*e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^6 - (2*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])
/e^6

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Rubi [A]  time = 0.27113, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1635, 1814, 641, 217, 203} \[ \frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^6}-\frac{2 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^6} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^4*(d + e*x)^2)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (22*d^3*(d + e*x))/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (2*d*(30
*d + 23*e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + Sqrt[d^2 - e^2*x^2]/e^6 - (2*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])
/e^6

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 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^5 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d+e x) \left (\frac{2 d^5}{e^5}+\frac{5 d^4 x}{e^4}+\frac{5 d^3 x^2}{e^3}+\frac{5 d^2 x^3}{e^2}+\frac{5 d x^4}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{\frac{16 d^5}{e^5}+\frac{45 d^4 x}{e^4}+\frac{30 d^3 x^2}{e^3}+\frac{15 d^2 x^3}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{\frac{30 d^5}{e^5}+\frac{15 d^4 x}{e^4}}{\sqrt{d^2-e^2 x^2}} \, dx}{15 d^4}\\ &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^6}-\frac{(2 d) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^5}\\ &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^6}-\frac{(2 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^5}\\ &=\frac{d^4 (d+e x)^2}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{22 d^3 (d+e x)}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 d (30 d+23 e x)}{15 e^6 \sqrt{d^2-e^2 x^2}}+\frac{\sqrt{d^2-e^2 x^2}}{e^6}-\frac{2 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.233577, size = 111, normalized size = 0.78 \[ \frac{-32 d^2 e^2 x^2-\frac{30 (d-e x)^3 (d+e x) \sin ^{-1}\left (\frac{e x}{d}\right )}{\sqrt{1-\frac{e^2 x^2}{d^2}}}-82 d^3 e x+56 d^4+76 d e^3 x^3-15 e^4 x^4}{15 e^6 (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(56*d^4 - 82*d^3*e*x - 32*d^2*e^2*x^2 + 76*d*e^3*x^3 - 15*e^4*x^4 - (30*(d - e*x)^3*(d + e*x)*ArcSin[(e*x)/d])
/Sqrt[1 - (e^2*x^2)/d^2])/(15*e^6*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.103, size = 193, normalized size = 1.4 \begin{align*} -{{x}^{6} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+7\,{\frac{{d}^{2}{x}^{4}}{{e}^{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}-{\frac{28\,{d}^{4}{x}^{2}}{3\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{56\,{d}^{6}}{15\,{e}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,d{x}^{5}}{5\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,d{x}^{3}}{3\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{dx}{{e}^{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}-2\,{\frac{d}{{e}^{5}\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-x^6/(-e^2*x^2+d^2)^(5/2)+7/e^2*d^2*x^4/(-e^2*x^2+d^2)^(5/2)-28/3/e^4*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+56/15/e^6*d
^6/(-e^2*x^2+d^2)^(5/2)+2/5*d/e*x^5/(-e^2*x^2+d^2)^(5/2)-2/3*d/e^3*x^3/(-e^2*x^2+d^2)^(3/2)+2*d/e^5*x/(-e^2*x^
2+d^2)^(1/2)-2*d/e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [B]  time = 1.55754, size = 390, normalized size = 2.73 \begin{align*} \frac{2}{15} \, d e 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{x^{6}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} - \frac{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 )}}{3 \, e} + \frac{7 \, d^{2} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{28 \, d^{4} x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{56 \, d^{6}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{6}} + \frac{8 \, d^{3} x}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{5}} - \frac{14 \, d x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} e^{5}} - \frac{2 \, d \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}} e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

2/15*d*e*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)) - x^6/(-e^2*x^2 + d^2)^(5/2) - 2/3*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))/e + 7*d^2*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 28/3*d^4*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) +
56/15*d^6/((-e^2*x^2 + d^2)^(5/2)*e^6) + 8/15*d^3*x/((-e^2*x^2 + d^2)^(3/2)*e^5) - 14/15*d*x/(sqrt(-e^2*x^2 +
d^2)*e^5) - 2*d*arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^5)

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Fricas [A]  time = 2.01491, size = 397, normalized size = 2.78 \begin{align*} \frac{56 \, d e^{4} x^{4} - 112 \, d^{2} e^{3} x^{3} + 112 \, d^{4} e x - 56 \, d^{5} + 60 \,{\left (d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{4} e x - d^{5}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (15 \, e^{4} x^{4} - 76 \, d e^{3} x^{3} + 32 \, d^{2} e^{2} x^{2} + 82 \, d^{3} e x - 56 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (e^{10} x^{4} - 2 \, d e^{9} x^{3} + 2 \, d^{3} e^{7} x - d^{4} e^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(56*d*e^4*x^4 - 112*d^2*e^3*x^3 + 112*d^4*e*x - 56*d^5 + 60*(d*e^4*x^4 - 2*d^2*e^3*x^3 + 2*d^4*e*x - d^5)
*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^4*x^4 - 76*d*e^3*x^3 + 32*d^2*e^2*x^2 + 82*d^3*e*x - 56*d^4
)*sqrt(-e^2*x^2 + d^2))/(e^10*x^4 - 2*d*e^9*x^3 + 2*d^3*e^7*x - d^4*e^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \left (d + e x\right )^{2}}{\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**5*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**5*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.16453, size = 143, normalized size = 1. \begin{align*} -2 \, d \arcsin \left (\frac{x e}{d}\right ) e^{\left (-6\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (56 \, d^{6} e^{\left (-6\right )} +{\left (30 \, d^{5} e^{\left (-5\right )} -{\left (140 \, d^{4} e^{\left (-4\right )} +{\left (70 \, d^{3} e^{\left (-3\right )} -{\left (105 \, d^{2} e^{\left (-2\right )} +{\left (46 \, d e^{\left (-1\right )} - 15 \, x\right )} x\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^5*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-2*d*arcsin(x*e/d)*e^(-6)*sgn(d) - 1/15*(56*d^6*e^(-6) + (30*d^5*e^(-5) - (140*d^4*e^(-4) + (70*d^3*e^(-3) - (
105*d^2*e^(-2) + (46*d*e^(-1) - 15*x)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3