\(\int \frac {(d^2-e^2 x^2)^{5/2}}{x (d+e x)} \, dx\) [108]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=\frac {1}{8} d^2 (8 d-3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {3}{8} d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {864, 829, 858, 223, 209, 272, 65, 214} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=-\frac {3}{8} d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {1}{8} d^2 (8 d-3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 214

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 829

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

Rule 858

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

Rule 864

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

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.26 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=\frac {1}{24} \sqrt {d^2-e^2 x^2} \left (32 d^3-15 d^2 e x-8 d e^2 x^2+6 e^3 x^3\right )+\frac {3}{4} d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )-d^3 \sqrt {d^2} \log (x)+d^3 \sqrt {d^2} \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right ) \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(32*d^3 - 15*d^2*e*x - 8*d*e^2*x^2 + 6*e^3*x^3))/24 + (3*d^4*ArcTan[(e*x)/(Sqrt[d^2] - Sq
rt[d^2 - e^2*x^2])])/4 - d^3*Sqrt[d^2]*Log[x] + d^3*Sqrt[d^2]*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(296\) vs. \(2(99)=198\).

Time = 0.44 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.63

method	result	size
default	\(\frac {\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5}+d^{2} \left (\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3}+d^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )\right )}{d}-\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{d}\)	\(297\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=\frac {3}{4} \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \frac {1}{24} \, {\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} - 15 \, d^{2} e x + 32 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} \]

[In]

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

[Out]

3/4*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + d^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + 1/24*(6*e^3*x^3 -
8*d*e^2*x^2 - 15*d^2*e*x + 32*d^3)*sqrt(-e^2*x^2 + d^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.77 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.54 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=d^{3} \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} \frac {d^{2} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {d^{2} - e^{2} x^{2}}}{2} & \text {for}\: e^{2} \neq 0 \\x \sqrt {d^{2}} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} - \frac {d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} + \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3} & \text {for}\: e^{2} \neq 0 \\\frac {x^{2} \sqrt {d^{2}}}{2} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} - \frac {d^{2} x \sqrt {d^{2} - e^{2} x^{2}}}{8 e^{2}} + \frac {x^{3} \sqrt {d^{2} - e^{2} x^{2}}}{4} & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**3*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**2/(e**2*x**2) - 1), Abs
(d**2/(e**2*x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asin(d/(e*x)) + I*e*x/sqrt(-d**2/(e*
*2*x**2) + 1), True)) - d**2*e*Piecewise((d**2*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2)
)/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/2 + x*sqrt(d**2 - e**2*x**2)/2, Ne(e**2, 0)),
(x*sqrt(d**2), True)) - d*e**2*Piecewise((-d**2*sqrt(d**2 - e**2*x**2)/(3*e**2) + x**2*sqrt(d**2 - e**2*x**2)/
3, Ne(e**2, 0)), (x**2*sqrt(d**2)/2, True)) + e**3*Piecewise((d**4*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sq
rt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(8*e**2) - d**2*x*sqrt(d**
2 - e**2*x**2)/(8*e**2) + x**3*sqrt(d**2 - e**2*x**2)/4, Ne(e**2, 0)), (x**3*sqrt(d**2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=-\frac {3 \, d^{4} e \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}}} - d^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {3}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e x + \sqrt {-e^{2} x^{2} + d^{2}} d^{3} - \frac {1}{4} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e x + \frac {1}{3} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=-\frac {3 \, d^{4} e \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, {\left | e \right |}} - \frac {d^{4} e \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{{\left | e \right |}} + \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (32 \, d^{3} - {\left (15 \, d^{2} e - 2 \, {\left (3 \, e^{3} x - 4 \, d e^{2}\right )} x\right )} x\right )} \]

[In]

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

[Out]

-3/8*d^4*e*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - d^4*e*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2
*abs(x)))/abs(e) + 1/24*sqrt(-e^2*x^2 + d^2)*(32*d^3 - (15*d^2*e - 2*(3*e^3*x - 4*d*e^2)*x)*x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x\,\left (d+e\,x\right )} \,d x \]

[In]

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

[Out]

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