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

Optimal. Leaf size=113 \[ \frac {1}{8} d^2 (8 d-3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[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)

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Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {850, 815, 844, 217, 203, 266, 63, 208} \[ \frac {1}{8} d^2 (8 d-3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[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 63

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 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])

Rule 208

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

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 266

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 815

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 844

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 850

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*} \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x (d+e x)} \, dx &=\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 \operatorname {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 ) \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.09, size = 108, normalized size = 0.96 \[ d^4 \log (x)-d^4 \log \left (\sqrt {d^2-e^2 x^2}+d\right )-\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{24} \sqrt {d^2-e^2 x^2} \left (32 d^3-15 d^2 e x-8 d e^2 x^2+6 e^3 x^3\right ) \]

Antiderivative was successfully verified.

[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 - e^2*
x^2]])/8 + d^4*Log[x] - d^4*Log[d + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.86, size = 107, normalized size = 0.95 \[ \frac {3}{4} \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \frac {1}{24} \, {\left (6 \, e^{3} x^{3} - 8 \, d e^{2} x^{2} - 15 \, d^{2} e x + 32 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -3/8*d^4*sign(d)*asin(x*exp(2)/d/exp(1))
+1/2*(-12*d^4*exp(1)^4*exp(2)^2+8*d^4*exp(2)^4+4*d^4*exp(1)^6*exp(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^5/exp(1)-d^4*exp(2)*ln(1/2*
abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/exp(1)^2+2*(((24*exp(1)^7*1/192/exp(1)^4*x-32*ex
p(1)^6*d*1/192/exp(1)^4)*x-60*exp(1)^5*d^2*1/192/exp(1)^4)*x+128*exp(1)^4*d^3*1/192/exp(1)^4)*sqrt(d^2-x^2*exp
(2))

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maple [B]  time = 0.01, size = 245, normalized size = 2.17 \[ -\frac {d^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}-\frac {3 d^{4} e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2} e x}{8}+\sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e x}{4}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}{3}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 d}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 124, normalized size = 1.10 \[ -\frac {3}{8} \, d^{4} \arcsin \left (\frac {e x}{d}\right ) - d^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {3}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e x + \sqrt {-e^{2} x^{2} + d^{2}} d^{3} - \frac {1}{4} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e x + \frac {1}{3} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*d^4*arcsin(e*x/d) - 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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [C]  time = 25.65, size = 469, normalized size = 4.15 \[ d^{3} \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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((-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*e**2*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e
**2), True)) + e**3*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/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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