### 3.134 $$\int \frac{x (a+b x^2+c x^4)}{\sqrt{d-e x} \sqrt{d+e x}} \, dx$$

Optimal. Leaf size=109 $-\frac{\sqrt{d-e x} \sqrt{d+e x} \left (a e^4+b d^2 e^2+c d^4\right )}{e^6}+\frac{(d-e x)^{3/2} (d+e x)^{3/2} \left (b e^2+2 c d^2\right )}{3 e^6}-\frac{c (d-e x)^{5/2} (d+e x)^{5/2}}{5 e^6}$

[Out]

-(((c*d^4 + b*d^2*e^2 + a*e^4)*Sqrt[d - e*x]*Sqrt[d + e*x])/e^6) + ((2*c*d^2 + b*e^2)*(d - e*x)^(3/2)*(d + e*x
)^(3/2))/(3*e^6) - (c*(d - e*x)^(5/2)*(d + e*x)^(5/2))/(5*e^6)

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Rubi [A]  time = 0.1221, antiderivative size = 149, normalized size of antiderivative = 1.37, number of steps used = 4, number of rules used = 3, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {520, 1247, 698} $-\frac{\left (d^2-e^2 x^2\right ) \left (a e^4+b d^2 e^2+c d^4\right )}{e^6 \sqrt{d-e x} \sqrt{d+e x}}+\frac{\left (d^2-e^2 x^2\right )^2 \left (b e^2+2 c d^2\right )}{3 e^6 \sqrt{d-e x} \sqrt{d+e x}}-\frac{c \left (d^2-e^2 x^2\right )^3}{5 e^6 \sqrt{d-e x} \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(((c*d^4 + b*d^2*e^2 + a*e^4)*(d^2 - e^2*x^2))/(e^6*Sqrt[d - e*x]*Sqrt[d + e*x])) + ((2*c*d^2 + b*e^2)*(d^2 -
e^2*x^2)^2)/(3*e^6*Sqrt[d - e*x]*Sqrt[d + e*x]) - (c*(d^2 - e^2*x^2)^3)/(5*e^6*Sqrt[d - e*x]*Sqrt[d + e*x])

Rule 520

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1
*a2 + b1*b2*x^n)^FracPart[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [C]  time = 0.703123, size = 194, normalized size = 1.78 $-\frac{\sqrt{d-e x} \sqrt{d+e x} \left (5 \left (3 a e^4+2 b d^2 e^2+b e^4 x^2\right )+c \left (4 d^2 e^2 x^2+8 d^4+3 e^4 x^4\right )\right )+\frac{30 \sqrt{d} \sqrt{d+e x} \sin ^{-1}\left (\frac{\sqrt{d-e x}}{\sqrt{2} \sqrt{d}}\right ) \left (a e^4+b d^2 e^2+c d^4\right )}{\sqrt{\frac{e x}{d}+1}}-30 d \tan ^{-1}\left (\frac{\sqrt{d-e x}}{\sqrt{d+e x}}\right ) \left (a e^4+b d^2 e^2+c d^4\right )}{15 e^6}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[d - e*x]*Sqrt[d + e*x]*(5*(2*b*d^2*e^2 + 3*a*e^4 + b*e^4*x^2) + c*(8*d^4 + 4*d^2*e^2*x^2 + 3*e^4*x^4))
+ (30*Sqrt[d]*(c*d^4 + b*d^2*e^2 + a*e^4)*Sqrt[d + e*x]*ArcSin[Sqrt[d - e*x]/(Sqrt[2]*Sqrt[d])])/Sqrt[1 + (e*x
)/d] - 30*d*(c*d^4 + b*d^2*e^2 + a*e^4)*ArcTan[Sqrt[d - e*x]/Sqrt[d + e*x]])/(15*e^6)

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Maple [A]  time = 0.005, size = 73, normalized size = 0.7 \begin{align*} -{\frac{3\,c{x}^{4}{e}^{4}+5\,b{e}^{4}{x}^{2}+4\,c{d}^{2}{e}^{2}{x}^{2}+15\,a{e}^{4}+10\,b{d}^{2}{e}^{2}+8\,c{d}^{4}}{15\,{e}^{6}}\sqrt{-ex+d}\sqrt{ex+d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-e*x+d)^(1/2)*(e*x+d)^(1/2)*(3*c*e^4*x^4+5*b*e^4*x^2+4*c*d^2*e^2*x^2+15*a*e^4+10*b*d^2*e^2+8*c*d^4)/e^6

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Maxima [A]  time = 1.63594, size = 188, normalized size = 1.72 \begin{align*} -\frac{\sqrt{-e^{2} x^{2} + d^{2}} c x^{4}}{5 \, e^{2}} - \frac{4 \, \sqrt{-e^{2} x^{2} + d^{2}} c d^{2} x^{2}}{15 \, e^{4}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} b x^{2}}{3 \, e^{2}} - \frac{8 \, \sqrt{-e^{2} x^{2} + d^{2}} c d^{4}}{15 \, e^{6}} - \frac{2 \, \sqrt{-e^{2} x^{2} + d^{2}} b d^{2}}{3 \, e^{4}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} a}{e^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(-e^2*x^2 + d^2)*c*x^4/e^2 - 4/15*sqrt(-e^2*x^2 + d^2)*c*d^2*x^2/e^4 - 1/3*sqrt(-e^2*x^2 + d^2)*b*x^2
/e^2 - 8/15*sqrt(-e^2*x^2 + d^2)*c*d^4/e^6 - 2/3*sqrt(-e^2*x^2 + d^2)*b*d^2/e^4 - sqrt(-e^2*x^2 + d^2)*a/e^2

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Fricas [A]  time = 1.81681, size = 162, normalized size = 1.49 \begin{align*} -\frac{{\left (3 \, c e^{4} x^{4} + 8 \, c d^{4} + 10 \, b d^{2} e^{2} + 15 \, a e^{4} +{\left (4 \, c d^{2} e^{2} + 5 \, b e^{4}\right )} x^{2}\right )} \sqrt{e x + d} \sqrt{-e x + d}}{15 \, e^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*c*e^4*x^4 + 8*c*d^4 + 10*b*d^2*e^2 + 15*a*e^4 + (4*c*d^2*e^2 + 5*b*e^4)*x^2)*sqrt(e*x + d)*sqrt(-e*x
+ d)/e^6

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Sympy [C]  time = 77.8217, size = 350, normalized size = 3.21 \begin{align*} - \frac{i a d{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{4} & 0, 0, \frac{1}{2}, 1 \\- \frac{1}{2}, - \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, 0 & \end{matrix} \middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{2}} - \frac{a d{G_{6, 6}^{2, 6}\left (\begin{matrix} -1, - \frac{3}{4}, - \frac{1}{2}, - \frac{1}{4}, 0, 1 & \\- \frac{3}{4}, - \frac{1}{4} & -1, - \frac{1}{2}, - \frac{1}{2}, 0 \end{matrix} \middle |{\frac{d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{2}} - \frac{i b d^{3}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{5}{4}, - \frac{3}{4} & -1, -1, - \frac{1}{2}, 1 \\- \frac{3}{2}, - \frac{5}{4}, -1, - \frac{3}{4}, - \frac{1}{2}, 0 & \end{matrix} \middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{4}} - \frac{b d^{3}{G_{6, 6}^{2, 6}\left (\begin{matrix} -2, - \frac{7}{4}, - \frac{3}{2}, - \frac{5}{4}, -1, 1 & \\- \frac{7}{4}, - \frac{5}{4} & -2, - \frac{3}{2}, - \frac{3}{2}, 0 \end{matrix} \middle |{\frac{d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{4}} - \frac{i c d^{5}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{9}{4}, - \frac{7}{4} & -2, -2, - \frac{3}{2}, 1 \\- \frac{5}{2}, - \frac{9}{4}, -2, - \frac{7}{4}, - \frac{3}{2}, 0 & \end{matrix} \middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{6}} - \frac{c d^{5}{G_{6, 6}^{2, 6}\left (\begin{matrix} -3, - \frac{11}{4}, - \frac{5}{2}, - \frac{9}{4}, -2, 1 & \\- \frac{11}{4}, - \frac{9}{4} & -3, - \frac{5}{2}, - \frac{5}{2}, 0 \end{matrix} \middle |{\frac{d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} e^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*d*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), d**2/(e**2*x**2))/(4*pi**(3/
2)*e**2) - a*d*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), d**2*exp_polar
(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*e**2) - I*b*d**3*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4,
-1, -3/4, -1/2, 0), ()), d**2/(e**2*x**2))/(4*pi**(3/2)*e**4) - b*d**3*meijerg(((-2, -7/4, -3/2, -5/4, -1, 1),
()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), d**2*exp_polar(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*e**4) - I*c*d**5*
meijerg(((-9/4, -7/4), (-2, -2, -3/2, 1)), ((-5/2, -9/4, -2, -7/4, -3/2, 0), ()), d**2/(e**2*x**2))/(4*pi**(3/
2)*e**6) - c*d**5*meijerg(((-3, -11/4, -5/2, -9/4, -2, 1), ()), ((-11/4, -9/4), (-3, -5/2, -5/2, 0)), d**2*exp
_polar(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*e**6)

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Giac [A]  time = 1.15936, size = 153, normalized size = 1.4 \begin{align*} -\frac{1}{276480} \,{\left (15 \, c d^{4} e^{25} + 15 \, b d^{2} e^{27} -{\left (20 \, c d^{3} e^{25} + 10 \, b d e^{27} -{\left (22 \, c d^{2} e^{25} + 3 \,{\left ({\left (x e + d\right )} c e^{25} - 4 \, c d e^{25}\right )}{\left (x e + d\right )} + 5 \, b e^{27}\right )}{\left (x e + d\right )}\right )}{\left (x e + d\right )} + 15 \, a e^{29}\right )} \sqrt{x e + d} \sqrt{-x e + d} e^{\left (-1\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/276480*(15*c*d^4*e^25 + 15*b*d^2*e^27 - (20*c*d^3*e^25 + 10*b*d*e^27 - (22*c*d^2*e^25 + 3*((x*e + d)*c*e^25
- 4*c*d*e^25)*(x*e + d) + 5*b*e^27)*(x*e + d))*(x*e + d) + 15*a*e^29)*sqrt(x*e + d)*sqrt(-x*e + d)*e^(-1)