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

Optimal. Leaf size=136 \[ -\frac{d^3 x \sqrt{d^2-e^2 x^2}}{4 e}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{4 e^2} \]

[Out]

-(d^3*x*Sqrt[d^2 - e^2*x^2])/(4*e) - (d*x*(d^2 - e^2*x^2)^(3/2))/(6*e) - (2*(d^2 - e^2*x^2)^(5/2))/(15*e^2) -
(d^2 - e^2*x^2)^(7/2)/(3*e^2*(d + e*x)^2) - (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4*e^2)

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Rubi [A]  time = 0.0566443, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {793, 665, 195, 217, 203} \[ -\frac{d^3 x \sqrt{d^2-e^2 x^2}}{4 e}-\frac{d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac{d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*x*Sqrt[d^2 - e^2*x^2])/(4*e) - (d*x*(d^2 - e^2*x^2)^(3/2))/(6*e) - (2*(d^2 - e^2*x^2)^(5/2))/(15*e^2) -
(d^2 - e^2*x^2)^(7/2)/(3*e^2*(d + e*x)^2) - (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4*e^2)

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.0808211, size = 91, normalized size = 0.67 \[ \frac{\sqrt{d^2-e^2 x^2} \left (16 d^2 e^2 x^2+15 d^3 e x-28 d^4-30 d e^3 x^3+12 e^4 x^4\right )-15 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{60 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-28*d^4 + 15*d^3*e*x + 16*d^2*e^2*x^2 - 30*d*e^3*x^3 + 12*e^4*x^4) - 15*d^5*ArcTan[(e*x)
/Sqrt[d^2 - e^2*x^2]])/(60*e^2)

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Maple [A]  time = 0.059, size = 198, normalized size = 1.5 \begin{align*} -{\frac{2}{15\,{e}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{dx}{6\,e} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{3}x}{4\,e}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{{d}^{5}}{4\,e}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{1}{3\,{e}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.4719, size = 225, normalized size = 1.65 \begin{align*} \frac{i \, d^{5} \arcsin \left (\frac{e x}{d} + 2\right )}{4 \, e^{2}} - \frac{\sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{4 \, e} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} d}{4 \,{\left (e^{3} x + d e^{2}\right )}} - \frac{\sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{2 \, e^{2}} + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d x}{4 \, e} - \frac{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d^{2}}{12 \, e^{2}} + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}{5 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*d^5*arcsin(e*x/d + 2)/e^2 - 1/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^3*x/e - 1/4*(-e^2*x^2 + d^2)^(5/2)*d/(
e^3*x + d*e^2) - 1/2*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4/e^2 + 1/4*(-e^2*x^2 + d^2)^(3/2)*d*x/e - 5/12*(-e^2*x
^2 + d^2)^(3/2)*d^2/e^2 + 1/5*(-e^2*x^2 + d^2)^(5/2)/e^2

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Fricas [A]  time = 1.60258, size = 204, normalized size = 1.5 \begin{align*} \frac{30 \, d^{5} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (12 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 16 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x - 28 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{60 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (12*e^4*x^4 - 30*d*e^3*x^3 + 16*d^2*e^2*x^2 + 15*d^3*
e*x - 28*d^4)*sqrt(-e^2*x^2 + d^2))/e^2

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Sympy [A]  time = 11.1704, size = 323, normalized size = 2.38 \begin{align*} d^{2} \left (\begin{cases} \frac{x^{2} \sqrt{d^{2}}}{2} & \text{for}\: e^{2} = 0 \\- \frac{\left (d^{2} - e^{2} x^{2}\right )^{\frac{3}{2}}}{3 e^{2}} & \text{otherwise} \end{cases}\right ) - 2 d e \left (\begin{cases} - \frac{i d^{4} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{8 e^{3}} + \frac{i d^{3} x}{8 e^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{3 i d x^{3}}{8 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{5}}{4 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{4} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{8 e^{3}} - \frac{d^{3} x}{8 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{3 d x^{3}}{8 \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} - \frac{e^{2} x^{5}}{4 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\right ) + e^{2} \left (\begin{cases} - \frac{2 d^{4} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac{d^{2} x^{2} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac{x^{4} \sqrt{d^{2} - e^{2} x^{2}}}{5} & \text{for}\: e \neq 0 \\\frac{x^{4} \sqrt{d^{2}}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) - 2*d*e*Piecewis
e((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*
x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*asin(e*x/d)/(8
*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqr
t(1 - e**2*x**2/d**2)), True)) + e**2*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**
2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True))

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

[Out]

sage0*x