3.34 \(\int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^4}-\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4} \]

[Out]

3/4*d^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4-3/5*d^2*x^2*(-e^2*x^2+d^2)^(1/2)/e^2-1/2*d*x^3*(-e^2*x^2+d^2)^(1/
2)/e-1/5*x^4*(-e^2*x^2+d^2)^(1/2)-3/20*d^3*(5*e*x+8*d)*(-e^2*x^2+d^2)^(1/2)/e^4

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Rubi [A]  time = 0.19, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1809, 833, 780, 217, 203} \[ -\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4}-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(5*e^2) - (d*x^3*Sqrt[d^2 - e^2*x^2])/(2*e) - (x^4*Sqrt[d^2 - e^2*x^2])/5 - (
3*d^3*(8*d + 5*e*x)*Sqrt[d^2 - e^2*x^2])/(20*e^4) + (3*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4*e^4)

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]

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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {x^3 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^3 \left (-9 d^2 e^2-10 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x^2 \left (30 d^3 e^3+36 d^2 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{20 e^4}\\ &=-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x \left (-72 d^4 e^4-90 d^3 e^5 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{60 e^6}\\ &=-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4}+\frac {\left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^3}\\ &=-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4}+\frac {\left (3 d^5\right ) \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^3}\\ &=-\frac {3 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {3 d^3 (8 d+5 e x) \sqrt {d^2-e^2 x^2}}{20 e^4}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 92, normalized size = 0.64 \[ \frac {15 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (24 d^4+15 d^3 e x+12 d^2 e^2 x^2+10 d e^3 x^3+4 e^4 x^4\right )}{20 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(24*d^4 + 15*d^3*e*x + 12*d^2*e^2*x^2 + 10*d*e^3*x^3 + 4*e^4*x^4)) + 15*d^5*ArcTan[(e*x
)/Sqrt[d^2 - e^2*x^2]])/(20*e^4)

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fricas [A]  time = 0.91, size = 94, normalized size = 0.65 \[ -\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (4 \, e^{4} x^{4} + 10 \, d e^{3} x^{3} + 12 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x + 24 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{20 \, e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/20*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (4*e^4*x^4 + 10*d*e^3*x^3 + 12*d^2*e^2*x^2 + 15*d^3*
e*x + 24*d^4)*sqrt(-e^2*x^2 + d^2))/e^4

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giac [A]  time = 0.26, size = 73, normalized size = 0.51 \[ \frac {3}{4} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} \mathrm {sgn}\relax (d) - \frac {1}{20} \, {\left (24 \, d^{4} e^{\left (-4\right )} + {\left (15 \, d^{3} e^{\left (-3\right )} + 2 \, {\left (6 \, d^{2} e^{\left (-2\right )} + {\left (5 \, d e^{\left (-1\right )} + 2 \, x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

3/4*d^5*arcsin(x*e/d)*e^(-4)*sgn(d) - 1/20*(24*d^4*e^(-4) + (15*d^3*e^(-3) + 2*(6*d^2*e^(-2) + (5*d*e^(-1) + 2
*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.01, size = 149, normalized size = 1.03 \[ -\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{4}}{5}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{4 \sqrt {e^{2}}\, e^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d \,x^{3}}{2 e}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x^{2}}{5 e^{2}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x}{4 e^{3}}-\frac {6 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4}}{5 e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.97, size = 128, normalized size = 0.89 \[ -\frac {1}{5} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{4} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} d x^{3}}{2 \, e} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x^{2}}{5 \, e^{2}} + \frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{4 \, e^{4}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x}{4 \, e^{3}} - \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{5 \, e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(1/2),x)

[Out]

int((x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(1/2), x)

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sympy [A]  time = 7.87, size = 357, normalized size = 2.48 \[ d^{2} \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) + 2 d e \left (\begin {cases} - \frac {3 i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{5}} + \frac {3 i d^{3} x}{8 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{3}}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{5}} - \frac {3 d^{3} x}{8 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{3}}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {8 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{6}} - \frac {4 d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)**2/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**2*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x*
*4/(4*sqrt(d**2)), True)) + 2*d*e*Piecewise((-3*I*d**4*acosh(e*x/d)/(8*e**5) + 3*I*d**3*x/(8*e**4*sqrt(-1 + e*
*2*x**2/d**2)) - I*d*x**3/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - I*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**
2*x**2/d**2) > 1), (3*d**4*asin(e*x/d)/(8*e**5) - 3*d**3*x/(8*e**4*sqrt(1 - e**2*x**2/d**2)) + d*x**3/(8*e**2*
sqrt(1 - e**2*x**2/d**2)) + x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + e**2*Piecewise((-8*d**4*sqrt(d**2 -
e**2*x**2)/(15*e**6) - 4*d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**4) - x**4*sqrt(d**2 - e**2*x**2)/(5*e**2), Ne
(e, 0)), (x**6/(6*sqrt(d**2)), True))

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