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

Optimal. Leaf size=100 \[ \frac{3}{8} d^3 x \sqrt{d^2-e^2 x^2}+\frac{1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac{3 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

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

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Rubi [A]  time = 0.0305591, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {665, 195, 217, 203} \[ \frac{3}{8} d^3 x \sqrt{d^2-e^2 x^2}+\frac{1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac{\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac{3 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0611791, size = 91, normalized size = 0.91 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-16 d^2 e^2 x^2+25 d^3 e x+8 d^4-10 d e^3 x^3+8 e^4 x^4\right )+15 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{40 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(8*d^4 + 25*d^3*e*x - 16*d^2*e^2*x^2 - 10*d*e^3*x^3 + 8*e^4*x^4) + 15*d^5*ArcTan[(e*x)/Sq
rt[d^2 - e^2*x^2]])/(40*e)

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Maple [A]  time = 0.051, size = 147, normalized size = 1.5 \begin{align*}{\frac{1}{5\,e} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{dx}{4} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{d}^{3}x}{8}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{3\,{d}^{5}}{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),x)

[Out]

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

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Maxima [C]  time = 1.49222, size = 147, normalized size = 1.47 \begin{align*} -\frac{3 i \, d^{5} \arcsin \left (\frac{e x}{d} + 2\right )}{8 \, e} + \frac{3}{8} \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x + \frac{3 \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{4 \, e} + \frac{1}{4} \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d x + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}}{5 \, 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60922, size = 200, normalized size = 2. \begin{align*} -\frac{30 \, d^{5} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (8 \, e^{4} x^{4} - 10 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 25 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{40 \, 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),x, algorithm="fricas")

[Out]

-1/40*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (8*e^4*x^4 - 10*d*e^3*x^3 - 16*d^2*e^2*x^2 + 25*d^3*
e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/e

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Sympy [C]  time = 10.6007, size = 439, normalized size = 4.39 \begin{align*} d^{3} \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 ) - d^{2} 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 ) - d 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 ) + e^{3} \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((-e**2*x**2+d**2)**(5/2)/(e*x+d),x)

[Out]

d**3*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)) - d**2*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) - d*e**2
*Piecewise((-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)) + e**3*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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError