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

Optimal. Leaf size=136 \[ -\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^2}-\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e} \]

[Out]

-1/6*d*x*(-e^2*x^2+d^2)^(3/2)/e-2/15*(-e^2*x^2+d^2)^(5/2)/e^2-1/3*(-e^2*x^2+d^2)^(7/2)/e^2/(e*x+d)^2-1/4*d^5*a
rctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2-1/4*d^3*x*(-e^2*x^2+d^2)^(1/2)/e

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Rubi [A]  time = 0.06, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {793, 665, 195, 217, 203} \[ -\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*x*Sqrt[d^2 - e^2*x^2])/(4*e) - (d*x*(d^2 - e^2*x^2)^(3/2))/(6*e) - (2*(d^2 - e^2*x^2)^(5/2))/(15*e^2) -
(d^2 - e^2*x^2)^(7/2)/(3*e^2*(d + e*x)^2) - (d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4*e^2)

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 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 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 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 793

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

Rubi steps

\begin {align*} \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {2 \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{3 e}\\ &=-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {(2 d) \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{3 e}\\ &=-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {d^3 \int \sqrt {d^2-e^2 x^2} \, dx}{2 e}\\ &=-\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {d^5 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e}\\ &=-\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {d^5 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e}\\ &=-\frac {d^3 x \sqrt {d^2-e^2 x^2}}{4 e}-\frac {d x \left (d^2-e^2 x^2\right )^{3/2}}{6 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{15 e^2}-\frac {\left (d^2-e^2 x^2\right )^{7/2}}{3 e^2 (d+e x)^2}-\frac {d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 91, normalized size = 0.67 \[ \frac {\sqrt {d^2-e^2 x^2} \left (-28 d^4+15 d^3 e x+16 d^2 e^2 x^2-30 d e^3 x^3+12 e^4 x^4\right )-15 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{60 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-28*d^4 + 15*d^3*e*x + 16*d^2*e^2*x^2 - 30*d*e^3*x^3 + 12*e^4*x^4) - 15*d^5*ArcTan[(e*x)
/Sqrt[d^2 - e^2*x^2]])/(60*e^2)

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fricas [A]  time = 0.81, size = 94, normalized size = 0.69 \[ \frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (12 \, e^{4} x^{4} - 30 \, d e^{3} x^{3} + 16 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x - 28 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{60 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (12*e^4*x^4 - 30*d*e^3*x^3 + 16*d^2*e^2*x^2 + 15*d^3*
e*x - 28*d^4)*sqrt(-e^2*x^2 + d^2))/e^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.01, size = 198, normalized size = 1.46 \[ -\frac {d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{4 \sqrt {e^{2}}\, e}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3} x}{4 e}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d x}{6 e}-\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{15 e^{2}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-1/6/e*d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-1/4/e*d^3*(2*(x+d/
e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-1/4/e*d^5/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)-
1/3/e^4/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)

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maxima [C]  time = 1.00, size = 167, normalized size = 1.23 \[ \frac {i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{4 \, e^{2}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{4 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{4 \, {\left (e^{3} x + d e^{2}\right )}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{2 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{12 \, e^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.57, size = 321, normalized size = 2.36 \[ d^{2} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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