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

Optimal. Leaf size=172 \[ \frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}+\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4} \]

[Out]

(3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) + (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/2
))/(7*e^2) - (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d + 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) + (3*d^
8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

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Rubi [A]  time = 0.100889, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {833, 780, 195, 217, 203} \[ \frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}+\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) + (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/2
))/(7*e^2) - (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d + 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) + (3*d^
8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

Rule 833

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

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 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 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])

Rubi steps

\begin{align*} \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx &=-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{\int x^2 \left (-3 d^2 e-8 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{8 e^2}\\ &=-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac{\int x \left (16 d^3 e^2+21 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{56 e^4}\\ &=-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac{d^4 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{16 e^3}\\ &=\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac{\left (3 d^6\right ) \int \sqrt{d^2-e^2 x^2} \, dx}{64 e^3}\\ &=\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}+\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac{\left (3 d^8\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{128 e^3}\\ &=\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}+\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac{\left (3 d^8\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^3}\\ &=\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}+\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}-\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d+35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}+\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4}\\ \end{align*}

Mathematica [A]  time = 0.203894, size = 146, normalized size = 0.85 \[ \frac{\sqrt{d^2-e^2 x^2} \left (105 d^7 \sin ^{-1}\left (\frac{e x}{d}\right )-\sqrt{1-\frac{e^2 x^2}{d^2}} \left (128 d^5 e^2 x^2+70 d^4 e^3 x^3-1024 d^3 e^4 x^4-840 d^2 e^5 x^5+105 d^6 e x+256 d^7+640 d e^6 x^6+560 e^7 x^7\right )\right )}{4480 e^4 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(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]*(256*d^7 + 105*d^6*e*x + 128*d^5*e^2*x^2 + 70*d^4*e^3*x^3 - 10
24*d^3*e^4*x^4 - 840*d^2*e^5*x^5 + 640*d*e^6*x^6 + 560*e^7*x^7)) + 105*d^7*ArcSin[(e*x)/d]))/(4480*e^4*Sqrt[1
- (e^2*x^2)/d^2])

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Maple [A]  time = 0.06, size = 173, normalized size = 1. \begin{align*} -{\frac{{x}^{3}}{8\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{{d}^{2}x}{16\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{d}^{4}x}{64\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{d}^{6}x}{128\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{3\,{d}^{8}}{128\,{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{d{x}^{2}}{7\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{2\,{d}^{3}}{35\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x^3*(-e^2*x^2+d^2)^(5/2)/e-1/16*d^2/e^3*x*(-e^2*x^2+d^2)^(5/2)+1/64*d^4*x*(-e^2*x^2+d^2)^(3/2)/e^3+3/128*
d^6*x*(-e^2*x^2+d^2)^(1/2)/e^3+3/128*d^8/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/7*d*x^2*
(-e^2*x^2+d^2)^(5/2)/e^2-2/35*d^3/e^4*(-e^2*x^2+d^2)^(5/2)

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Maxima [A]  time = 1.5631, size = 223, normalized size = 1.3 \begin{align*} \frac{3 \, d^{8} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{128 \, \sqrt{e^{2}} e^{3}} + \frac{3 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{6} x}{128 \, e^{3}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} x^{3}}{8 \, e} + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d^{4} x}{64 \, e^{3}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} d x^{2}}{7 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} d^{2} x}{16 \, e^{3}} - \frac{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} d^{3}}{35 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*d^8*arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^3) + 3/128*sqrt(-e^2*x^2 + d^2)*d^6*x/e^3 - 1/8*(-e^2*x^2 +
d^2)^(5/2)*x^3/e + 1/64*(-e^2*x^2 + d^2)^(3/2)*d^4*x/e^3 - 1/7*(-e^2*x^2 + d^2)^(5/2)*d*x^2/e^2 - 1/16*(-e^2*
x^2 + d^2)^(5/2)*d^2*x/e^3 - 2/35*(-e^2*x^2 + d^2)^(5/2)*d^3/e^4

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Fricas [A]  time = 1.76217, size = 289, normalized size = 1.68 \begin{align*} -\frac{210 \, d^{8} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (560 \, e^{7} x^{7} + 640 \, d e^{6} x^{6} - 840 \, d^{2} e^{5} x^{5} - 1024 \, d^{3} e^{4} x^{4} + 70 \, d^{4} e^{3} x^{3} + 128 \, d^{5} e^{2} x^{2} + 105 \, d^{6} e x + 256 \, d^{7}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{4480 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4480*(210*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (560*e^7*x^7 + 640*d*e^6*x^6 - 840*d^2*e^5*x^5 -
1024*d^3*e^4*x^4 + 70*d^4*e^3*x^3 + 128*d^5*e^2*x^2 + 105*d^6*e*x + 256*d^7)*sqrt(-e^2*x^2 + d^2))/e^4

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Sympy [A]  time = 19.5484, size = 779, normalized size = 4.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*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*s
qrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) + d**2*e*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)/Abs(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)) - d*e*
*2*Piecewise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt(d**2 - e**2*x**2)/(105*e**4) - d**2
*x**4*sqrt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne(e, 0)), (x**6*sqrt(d**2)/6, True))
- e**3*Piecewise((-5*I*d**8*acosh(e*x/d)/(128*e**7) + 5*I*d**7*x/(128*e**6*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*
*5*x**3/(384*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**5/(192*e**2*sqrt(-1 + e**2*x**2/d**2)) - 7*I*d*x**7/(
48*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**9/(8*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (5
*d**8*asin(e*x/d)/(128*e**7) - 5*d**7*x/(128*e**6*sqrt(1 - e**2*x**2/d**2)) + 5*d**5*x**3/(384*e**4*sqrt(1 - e
**2*x**2/d**2)) + d**3*x**5/(192*e**2*sqrt(1 - e**2*x**2/d**2)) + 7*d*x**7/(48*sqrt(1 - e**2*x**2/d**2)) - e**
2*x**9/(8*d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.31202, size = 143, normalized size = 0.83 \begin{align*} \frac{3}{128} \, d^{8} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-4\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{4480} \,{\left (256 \, d^{7} e^{\left (-4\right )} +{\left (105 \, d^{6} e^{\left (-3\right )} + 2 \,{\left (64 \, d^{5} e^{\left (-2\right )} +{\left (35 \, d^{4} e^{\left (-1\right )} - 4 \,{\left (128 \, d^{3} + 5 \,{\left (21 \, d^{2} e - 2 \,{\left (7 \, x e^{3} + 8 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*d^8*arcsin(x*e/d)*e^(-4)*sgn(d) - 1/4480*(256*d^7*e^(-4) + (105*d^6*e^(-3) + 2*(64*d^5*e^(-2) + (35*d^4*
e^(-1) - 4*(128*d^3 + 5*(21*d^2*e - 2*(7*x*e^3 + 8*d*e^2)*x)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)