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

Optimal. Leaf size=121 \[ \frac{d^3 (d+e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{17 d^2 (d+e x)}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (15 d+13 e x)}{15 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

[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 = 0.207675, antiderivative size = 121, 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, 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)^(7/2),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 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 )^{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 = 0.214264, size = 96, normalized size = 0.79 \[ \frac{-15 d (d-e x)^2 \sqrt{1-\frac{e^2 x^2}{d^2}} \sin ^{-1}\left (\frac{e x}{d}\right )-17 d^2 e x+16 d^3-22 d e^2 x^2+26 e^3 x^3}{15 e^5 (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*d^3 - 17*d^2*e*x - 22*d*e^2*x^2 + 26*e^3*x^3 - 15*d*(d - e*x)^2*Sqrt[1 - (e^2*x^2)/d^2]*ArcSin[(e*x)/d])/(
15*e^5*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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Maple [B]  time = 0.083, size = 236, normalized size = 2. \begin{align*}{\frac{{x}^{5}}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{{x}^{3}}{3\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{6\,x}{5\,{e}^{4}}{\frac{1}{\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{x}^{4}}{e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}-{\frac{8\,{d}^{3}{x}^{2}}{3\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{16\,{d}^{5}}{15\,{e}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{{d}^{2}{x}^{3}}{2\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{3\,{d}^{4}x}{10\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{{d}^{2}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)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/5*x^5/(-e^2*x^2+d^2)^(5/2)-1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)+6/5/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*x^4/(-e^2*x^2+d^2)^(5/2)-8/3*d^3/e^3*x^2/(-e^2*x^2+d^2)^(5/2
)+16/15*d^5/e^5/(-e^2*x^2+d^2)^(5/2)+1/2*d^2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/10*d^4/e^4*x/(-e^2*x^2+d^2)^(5/2)+
1/10*d^2/e^4*x/(-e^2*x^2+d^2)^(3/2)

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Maxima [B]  time = 1.54407, size = 420, normalized size = 3.47 \begin{align*} \frac{1}{15} \, 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{1}{3} \, 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{2 \, d x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{d^{2} x^{3}}{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{2}} - \frac{8 \, d^{3} x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{3}} - \frac{3 \, d^{4} x}{10 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{4}} + \frac{16 \, d^{5}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e^{5}} + \frac{11 \, d^{2} x}{30 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} e^{4}} - \frac{4 \, x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} e^{4}} - \frac{\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)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*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)) - 1/3*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)) + 2*d*x^4/
((-e^2*x^2 + d^2)^(5/2)*e) + 1/2*d^2*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 8/3*d^3*x^2/((-e^2*x^2 + d^2)^(5/2)*e^
3) - 3/10*d^4*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 16/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e^5) + 11/30*d^2*x/((-e^2*x^2
+ d^2)^(3/2)*e^4) - 4/15*x/(sqrt(-e^2*x^2 + d^2)*e^4) - arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^4)

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Fricas [A]  time = 1.81285, size = 359, normalized size = 2.97 \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)^(7/2),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 (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**4*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

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

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Giac [A]  time = 1.16284, size = 128, normalized size = 1.06 \begin{align*} -\arcsin \left (\frac{x e}{d}\right ) e^{\left (-5\right )} \mathrm{sgn}\left (d\right ) - \frac{{\left (16 \, d^{5} e^{\left (-5\right )} +{\left (15 \, d^{4} e^{\left (-4\right )} -{\left (40 \, d^{3} e^{\left (-3\right )} +{\left (35 \, d^{2} e^{\left (-2\right )} - 2 \,{\left (15 \, d e^{\left (-1\right )} + 13 \, 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^4*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^(-5)*sgn(d) - 1/15*(16*d^5*e^(-5) + (15*d^4*e^(-4) - (40*d^3*e^(-3) + (35*d^2*e^(-2) - 2*(15*
d*e^(-1) + 13*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^3