3.1.33 \(\int \frac {x^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=173 \[ -\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \begin {gather*} -\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(d + e*x)^2)/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-8*d^3*x^2*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (11*d^2*x^3*Sqrt[d^2 - e^2*x^2])/(24*e^2) - (2*d*x^4*Sqrt[d^2 - e^
2*x^2])/(5*e) - (x^5*Sqrt[d^2 - e^2*x^2])/6 - (d^4*(256*d + 165*e*x)*Sqrt[d^2 - e^2*x^2])/(240*e^5) + (11*d^6*
ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^5)

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^4 (d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^4 \left (-11 d^2 e^2-12 d e^3 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{6 e^2}\\ &=-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x^3 \left (48 d^3 e^3+55 d^2 e^4 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{30 e^4}\\ &=-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {x^2 \left (-165 d^4 e^4-192 d^3 e^5 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{120 e^6}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {x \left (384 d^5 e^5+495 d^4 e^6 x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{360 e^8}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^4}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {\left (11 d^6\right ) \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^4}\\ &=-\frac {8 d^3 x^2 \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {11 d^2 x^3 \sqrt {d^2-e^2 x^2}}{24 e^2}-\frac {2 d x^4 \sqrt {d^2-e^2 x^2}}{5 e}-\frac {1}{6} x^5 \sqrt {d^2-e^2 x^2}-\frac {d^4 (256 d+165 e x) \sqrt {d^2-e^2 x^2}}{240 e^5}+\frac {11 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 103, normalized size = 0.60 \begin {gather*} \frac {165 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (256 d^5+165 d^4 e x+128 d^3 e^2 x^2+110 d^2 e^3 x^3+96 d e^4 x^4+40 e^5 x^5\right )}{240 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(d + e*x)^2)/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(256*d^5 + 165*d^4*e*x + 128*d^3*e^2*x^2 + 110*d^2*e^3*x^3 + 96*d*e^4*x^4 + 40*e^5*x^5)
) + 165*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(240*e^5)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.42, size = 125, normalized size = 0.72 \begin {gather*} \frac {11 d^6 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{16 e^6}+\frac {\sqrt {d^2-e^2 x^2} \left (-256 d^5-165 d^4 e x-128 d^3 e^2 x^2-110 d^2 e^3 x^3-96 d e^4 x^4-40 e^5 x^5\right )}{240 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(d + e*x)^2)/Sqrt[d^2 - e^2*x^2],x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-256*d^5 - 165*d^4*e*x - 128*d^3*e^2*x^2 - 110*d^2*e^3*x^3 - 96*d*e^4*x^4 - 40*e^5*x^5))
/(240*e^5) + (11*d^6*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(16*e^6)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 105, normalized size = 0.61 \begin {gather*} -\frac {330 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 96 \, d e^{4} x^{4} + 110 \, d^{2} e^{3} x^{3} + 128 \, d^{3} e^{2} x^{2} + 165 \, d^{4} e x + 256 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/240*(330*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 + 96*d*e^4*x^4 + 110*d^2*e^3*x^3 + 128
*d^3*e^2*x^2 + 165*d^4*e*x + 256*d^5)*sqrt(-e^2*x^2 + d^2))/e^5

________________________________________________________________________________________

giac [A]  time = 0.27, size = 84, normalized size = 0.49 \begin {gather*} \frac {11}{16} \, d^{6} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) - \frac {1}{240} \, {\left (256 \, d^{5} e^{\left (-5\right )} + {\left (165 \, d^{4} e^{\left (-4\right )} + 2 \, {\left (64 \, d^{3} e^{\left (-3\right )} + {\left (55 \, d^{2} e^{\left (-2\right )} + 4 \, {\left (12 \, d e^{\left (-1\right )} + 5 \, x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

11/16*d^6*arcsin(x*e/d)*e^(-5)*sgn(d) - 1/240*(256*d^5*e^(-5) + (165*d^4*e^(-4) + 2*(64*d^3*e^(-3) + (55*d^2*e
^(-2) + 4*(12*d*e^(-1) + 5*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 174, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x^{5}}{6}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,x^{4}}{5 e}+\frac {11 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}\, e^{4}}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x^{3}}{24 e^{2}}-\frac {8 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x^{2}}{15 e^{3}}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} x}{16 e^{4}}-\frac {16 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}}{15 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/6*x^5*(-e^2*x^2+d^2)^(1/2)-11/24*d^2*x^3*(-e^2*x^2+d^2)^(1/2)/e^2-11/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e^4+11/1
6/e^4*d^6/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-2/5*d*x^4*(-e^2*x^2+d^2)^(1/2)/e-8/15*d^3*x^2
*(-e^2*x^2+d^2)^(1/2)/e^3-16/15*d^5*(-e^2*x^2+d^2)^(1/2)/e^5

________________________________________________________________________________________

maxima [A]  time = 0.98, size = 153, normalized size = 0.88 \begin {gather*} -\frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} x^{5} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d x^{4}}{5 \, e} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x^{3}}{24 \, e^{2}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} x^{2}}{15 \, e^{3}} + \frac {11 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{5}} - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{4}} - \frac {16 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{15 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)^2/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(-e^2*x^2 + d^2)*x^5 - 2/5*sqrt(-e^2*x^2 + d^2)*d*x^4/e - 11/24*sqrt(-e^2*x^2 + d^2)*d^2*x^3/e^2 - 8/
15*sqrt(-e^2*x^2 + d^2)*d^3*x^2/e^3 + 11/16*d^6*arcsin(e*x/d)/e^5 - 11/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e^4 - 16/
15*sqrt(-e^2*x^2 + d^2)*d^5/e^5

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\left (d+e\,x\right )}^2}{\sqrt {d^2-e^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(1/2),x)

[Out]

int((x^4*(d + e*x)^2)/(d^2 - e^2*x^2)^(1/2), x)

________________________________________________________________________________________

sympy [C]  time = 13.48, size = 558, normalized size = 3.23 \begin {gather*} d^{2} \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 ) + 2 d e \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 ) + e^{2} \left (\begin {cases} - \frac {5 i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{7}} + \frac {5 i d^{5} x}{16 e^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d^{3} x^{3}}{48 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{5}}{24 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i 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 {5 d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{7}} - \frac {5 d^{5} x}{16 e^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d^{3} x^{3}}{48 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{5}}{24 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)**2/(-e**2*x**2+d**2)**(1/2),x)

[Out]

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

________________________________________________________________________________________