3.2.7 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{d+e x} \, dx\) [107]

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]

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

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Rubi [A]
time = 0.02, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 = {679, 201, 223, 209} \begin {gather*} \frac {3 d^5 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\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}{8} d^3 x \sqrt {d^2-e^2 x^2} \end {gather*}

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 201

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 679

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.24, size = 111, normalized size = 1.11 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (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\right )}{40 e}-\frac {3 d^5 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{8 \sqrt {-e^2}} \end {gather*}

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))/(40*e) - (3*d^5*Log[-(S
qrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(84)=168\).
time = 0.07, size = 192, normalized size = 1.92

method	result	size
risch	\(\frac {\left (8 e^{4} x^{4}-10 d \,e^{3} x^{3}-16 d^{2} x^{2} e^{2}+25 d^{3} e x +8 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{40 e}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(94\)
default	\(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{e}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 105, normalized size = 1.05 \begin {gather*} -\frac {3}{8} i \, d^{5} \arcsin \left (\frac {x e}{d} + 2\right ) e^{\left (-1\right )} + \frac {3}{4} \, \sqrt {x^{2} e^{2} + 4 \, d x e + 3 \, d^{2}} d^{4} e^{\left (-1\right )} + \frac {3}{8} \, \sqrt {x^{2} e^{2} + 4 \, d x e + 3 \, d^{2}} d^{3} x + \frac {1}{4} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d x + \frac {1}{5} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} e^{\left (-1\right )} \end {gather*}

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

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Fricas [A]
time = 2.91, size = 89, normalized size = 0.89 \begin {gather*} -\frac {1}{40} \, {\left (30 \, d^{5} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (8 \, x^{4} e^{4} - 10 \, d x^{3} e^{3} - 16 \, d^{2} x^{2} e^{2} + 25 \, d^{3} x e + 8 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 4.25, size = 435, normalized size = 4.35 \begin {gather*} 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}\: \left |{\frac {e^{2} x^{2}}{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}\: \left |{\frac {e^{2} x^{2}}{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 {gather*}

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/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) -
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*Piec
ewise((-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/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 [A]
time = 0.97, size = 74, normalized size = 0.74 \begin {gather*} \frac {3}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) + \frac {1}{40} \, {\left (8 \, d^{4} e^{\left (-1\right )} + {\left (25 \, d^{3} - 2 \, {\left (8 \, d^{2} e - {\left (4 \, x e^{3} - 5 \, d e^{2}\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \end {gather*}

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]

3/8*d^5*arcsin(x*e/d)*e^(-1)*sgn(d) + 1/40*(8*d^4*e^(-1) + (25*d^3 - 2*(8*d^2*e - (4*x*e^3 - 5*d*e^2)*x)*x)*x)
*sqrt(-x^2*e^2 + d^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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