3.2.62 \(\int \frac {(d^2-e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [162]

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]

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

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {669, 685, 655, 201, 223, 209} \begin {gather*} \frac {5 d^4 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\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} \end {gather*}

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 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 655

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 669

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 685

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
]

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 ) \text {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*}

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Mathematica [A]
time = 0.21, size = 100, normalized size = 0.93 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (16 d^3+9 d^2 e x-16 d e^2 x^2+6 e^3 x^3\right )}{24 e}-\frac {5 d^4 \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)^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))/(24*e) - (5*d^4*Log[-(Sqrt[-e^2]*x) + Sq
rt[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. \(243\) vs. \(2(92)=184\).
time = 0.06, size = 244, normalized size = 2.26

method	result	size
risch	\(\frac {\left (6 e^{3} x^{3}-16 d \,e^{2} x^{2}+9 d^{2} e x +16 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e}+\frac {5 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(83\)
default	\(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}}{e^{2}}\)	\(244\)

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,method=_RETURNVERBOSE)

[Out]

1/e^2*(1/3/d/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(7/2)+5/3*e/d*(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)*arctan((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 = 115, normalized size = 1.06 \begin {gather*} -\frac {5}{8} i \, d^{4} \arcsin \left (\frac {x e}{d} + 2\right ) e^{\left (-1\right )} + \frac {5}{4} \, \sqrt {x^{2} e^{2} + 4 \, d x e + 3 \, d^{2}} d^{3} e^{\left (-1\right )} + \frac {5}{8} \, \sqrt {x^{2} e^{2} + 4 \, d x e + 3 \, d^{2}} d^{2} x + \frac {5}{12} \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{\left (-1\right )} + \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (x e^{2} + d e\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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.34, size = 79, normalized size = 0.73 \begin {gather*} -\frac {1}{24} \, {\left (30 \, d^{4} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (6 \, x^{3} e^{3} - 16 \, d x^{2} e^{2} + 9 \, d^{2} x e + 16 \, d^{3}\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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 4.46, size = 350, normalized size = 3.24 \begin {gather*} 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}\: \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 ) - 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}\: \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 ) \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)**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/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) -
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*Piecewi
se((-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))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (88) = 176\).
time = 1.63, size = 177, normalized size = 1.64 \begin {gather*} -\frac {{\left (240 \, d^{5} \arctan \left (\sqrt {\frac {2 \, d}{x e + d} - 1}\right ) e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + \frac {{\left (15 \, d^{5} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {7}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 73 \, d^{5} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 55 \, d^{5} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 15 \, d^{5} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right )\right )} {\left (x e + d\right )}^{4}}{d^{4}}\right )} e^{\left (-6\right )}}{192 \, d} \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)^2,x, algorithm="giac")

[Out]

-1/192*(240*d^5*arctan(sqrt(2*d/(x*e + d) - 1))*e^5*sgn(1/(x*e + d)) + (15*d^5*(2*d/(x*e + d) - 1)^(7/2)*e^5*s
gn(1/(x*e + d)) - 73*d^5*(2*d/(x*e + d) - 1)^(5/2)*e^5*sgn(1/(x*e + d)) - 55*d^5*(2*d/(x*e + d) - 1)^(3/2)*e^5
*sgn(1/(x*e + d)) - 15*d^5*sqrt(2*d/(x*e + d) - 1)*e^5*sgn(1/(x*e + d)))*(x*e + d)^4/d^4)*e^(-6)/d

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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}}{{\left (d+e\,x\right )}^2} \,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)^2,x)

[Out]

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

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