3.5 \(\int x (d+e x) (d^2-e^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=116 \[ \frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}+\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e} \]

[Out]

1/24*d^2*x*(-e^2*x^2+d^2)^(3/2)/e-1/30*(5*e*x+6*d)*(-e^2*x^2+d^2)^(5/2)/e^2+1/16*d^6*arctan(e*x/(-e^2*x^2+d^2)
^(1/2))/e^2+1/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e

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Rubi [A]  time = 0.03, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {780, 195, 217, 203} \[ \frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x)*(d^2 - e^2*x^2)^(3/2),x]

[Out]

(d^4*x*Sqrt[d^2 - e^2*x^2])/(16*e) + (d^2*x*(d^2 - e^2*x^2)^(3/2))/(24*e) - ((6*d + 5*e*x)*(d^2 - e^2*x^2)^(5/
2))/(30*e^2) + (d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx &=-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^2 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e}\\ &=\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^4 \int \sqrt {d^2-e^2 x^2} \, dx}{8 e}\\ &=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e}\\ &=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e}\\ &=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 124, normalized size = 1.07 \[ \frac {\sqrt {d^2-e^2 x^2} \left (15 d^5 \sin ^{-1}\left (\frac {e x}{d}\right )-\sqrt {1-\frac {e^2 x^2}{d^2}} \left (48 d^5+15 d^4 e x-96 d^3 e^2 x^2-70 d^2 e^3 x^3+48 d e^4 x^4+40 e^5 x^5\right )\right )}{240 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x)*(d^2 - e^2*x^2)^(3/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-(Sqrt[1 - (e^2*x^2)/d^2]*(48*d^5 + 15*d^4*e*x - 96*d^3*e^2*x^2 - 70*d^2*e^3*x^3 + 48*d*
e^4*x^4 + 40*e^5*x^5)) + 15*d^5*ArcSin[(e*x)/d]))/(240*e^2*Sqrt[1 - (e^2*x^2)/d^2])

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fricas [A]  time = 1.02, size = 105, normalized size = 0.91 \[ -\frac {30 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 48 \, d e^{4} x^{4} - 70 \, d^{2} e^{3} x^{3} - 96 \, d^{3} e^{2} x^{2} + 15 \, d^{4} e x + 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")

[Out]

-1/240*(30*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 + 48*d*e^4*x^4 - 70*d^2*e^3*x^3 - 96*d^
3*e^2*x^2 + 15*d^4*e*x + 48*d^5)*sqrt(-e^2*x^2 + d^2))/e^2

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giac [A]  time = 0.21, size = 84, normalized size = 0.72 \[ \frac {1}{16} \, d^{6} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\relax (d) - \frac {1}{240} \, {\left (48 \, d^{5} e^{\left (-2\right )} + {\left (15 \, d^{4} e^{\left (-1\right )} - 2 \, {\left (48 \, d^{3} + {\left (35 \, d^{2} e - 4 \, {\left (5 \, x e^{3} + 6 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")

[Out]

1/16*d^6*arcsin(x*e/d)*e^(-2)*sgn(d) - 1/240*(48*d^5*e^(-2) + (15*d^4*e^(-1) - 2*(48*d^3 + (35*d^2*e - 4*(5*x*
e^3 + 6*d*e^2)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.02, size = 123, normalized size = 1.06 \[ \frac {d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x}{16 e}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2} x}{24 e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x}{6 e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d}{5 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),x)

[Out]

-1/6*x*(-e^2*x^2+d^2)^(5/2)/e+1/24*d^2*x*(-e^2*x^2+d^2)^(3/2)/e+1/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e+1/16*d^6/e/(
e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/5*d/e^2*(-e^2*x^2+d^2)^(5/2)

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maxima [A]  time = 0.98, size = 102, normalized size = 0.88 \[ \frac {d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x+d)*(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")

[Out]

1/16*d^6*arcsin(e*x/d)/e^2 + 1/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e + 1/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x/e - 1/6*(-e
^2*x^2 + d^2)^(5/2)*x/e - 1/5*(-e^2*x^2 + d^2)^(5/2)*d/e^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(d^2 - e^2*x^2)^(3/2)*(d + e*x),x)

[Out]

int(x*(d^2 - e^2*x^2)^(3/2)*(d + e*x), x)

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sympy [A]  time = 12.06, size = 580, normalized size = 5.00 \[ d^{3} \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^{2} 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 ) - d 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 ) - e^{3} \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 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(x*(e*x+d)*(-e**2*x**2+d**2)**(3/2),x)

[Out]

d**3*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + d**2*e*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)) - d*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)) - e**3*Piecewise
((-I*d**6*acosh(e*x/d)/(16*e**5) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1
+ e**2*x**2/d**2)) - 5*I*d*x**5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)),
Abs(e**2*x**2/d**2) > 1), (d**6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3
/(48*e**2*sqrt(1 - e**2*x**2/d**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x*
*2/d**2)), True))

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