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

Optimal. Leaf size=108 \[ \frac{5}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{5 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

(5*d^2*x*Sqrt[d^2 - e^2*x^2])/8 + (5*d*(d^2 - e^2*x^2)^(3/2))/(12*e) + ((d - e*x)*(d^2 - e^2*x^2)^(3/2))/(4*e)
+ (5*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

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Rubi [A]  time = 0.0418491, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {655, 671, 641, 195, 217, 203} \[ \frac{5}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{5 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*d^2*x*Sqrt[d^2 - e^2*x^2])/8 + (5*d*(d^2 - e^2*x^2)^(3/2))/(12*e) + ((d - e*x)*(d^2 - e^2*x^2)^(3/2))/(4*e)
+ (5*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

Rule 655

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

Rule 671

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

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 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{\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int (d-e x)^2 \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{1}{4} (5 d) \int (d-e x) \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{1}{4} \left (5 d^2\right ) \int \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{5}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{1}{8} \left (5 d^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{5}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{1}{8} \left (5 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )\\ &=\frac{5}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{5 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e}\\ \end{align*}

Mathematica [A]  time = 0.0578734, size = 80, normalized size = 0.74 \[ \frac{\sqrt{d^2-e^2 x^2} \left (9 d^2 e x+16 d^3-16 d e^2 x^2+6 e^3 x^3\right )+15 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{24 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(16*d^3 + 9*d^2*e*x - 16*d*e^2*x^2 + 6*e^3*x^3) + 15*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]
])/(24*e)

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Maple [B]  time = 0.054, size = 194, normalized size = 1.8 \begin{align*}{\frac{1}{3\,{e}^{3}d} \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}}+{\frac{1}{3\,de} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{5\,x}{12} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{d}^{2}x}{8}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{5\,{d}^{4}}{8}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/e^3/d/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+1/3/e/d*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+5/12*(-(
d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+5/8*d^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x+5/8*d^4/(e^2)^(1/2)*arctan(
(e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

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Maxima [C]  time = 1.7967, size = 161, normalized size = 1.49 \begin{align*} -\frac{5 i \, d^{4} \arcsin \left (\frac{e x}{d} + 2\right )}{8 \, e} + \frac{5}{8} \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x + \frac{5 \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e} + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}{4 \,{\left (e^{2} x + d e\right )}} + \frac{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d}{12 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*I*d^4*arcsin(e*x/d + 2)/e + 5/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^2*x + 5/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^
2)*d^3/e + 1/4*(-e^2*x^2 + d^2)^(5/2)/(e^2*x + d*e) + 5/12*(-e^2*x^2 + d^2)^(3/2)*d/e

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Fricas [A]  time = 1.55863, size = 177, normalized size = 1.64 \begin{align*} -\frac{30 \, d^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (6 \, e^{3} x^{3} - 16 \, d e^{2} x^{2} + 9 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(30*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*e^3*x^3 - 16*d*e^2*x^2 + 9*d^2*e*x + 16*d^3)*sqrt
(-e^2*x^2 + d^2))/e

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Sympy [C]  time = 8.56547, size = 354, normalized size = 3.28 \begin{align*} d^{2} \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e} - \frac{i d x}{2 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{3}}{2 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^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e} + \frac{d x \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}}{2} & \text{otherwise} \end{cases}\right ) - 2 d e \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 ) + e^{2} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 +
e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, Tru
e)) - 2*d*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + e**2*Pi
ecewise((-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*sqrt(1 - e**2*x**2/d**2)), 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((-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

sage0*x