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]

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

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Rubi [A]  time = 0.12, 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 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]))

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]

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*}

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Mathematica [C]  time = 0.69, size = 194, normalized size = 1.78 \[ -\frac {\sqrt {d-e x} \sqrt {d+e x} \left (5 \left (3 a e^4+2 b d^2 e^2+b e^4 x^2\right )+c \left (8 d^4+4 d^2 e^2 x^2+3 e^4 x^4\right )\right )+\frac {30 \sqrt {d} \sqrt {d+e x} \sin ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {2} \sqrt {d}}\right ) \left (a e^4+b d^2 e^2+c d^4\right )}{\sqrt {\frac {e x}{d}+1}}-30 d \tan ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right ) \left (a e^4+b d^2 e^2+c d^4\right )}{15 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(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]])/e^6

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fricas [A]  time = 1.03, size = 71, normalized size = 0.65 \[ -\frac {{\left (3 \, c e^{4} x^{4} + 8 \, c d^{4} + 10 \, b d^{2} e^{2} + 15 \, a e^{4} + {\left (4 \, c d^{2} e^{2} + 5 \, b e^{4}\right )} x^{2}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{15 \, e^{6}} \]

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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giac [A]  time = 0.63, size = 121, normalized size = 1.11 \[ -\frac {1}{15} \, {\left ({\left ({\left (3 \, {\left ({\left (x e + d\right )} c e^{\left (-5\right )} - 4 \, c d e^{\left (-5\right )}\right )} {\left (x e + d\right )} + {\left (22 \, c d^{2} e^{25} + 5 \, b e^{27}\right )} e^{\left (-30\right )}\right )} {\left (x e + d\right )} - 10 \, {\left (2 \, c d^{3} e^{25} + b d e^{27}\right )} e^{\left (-30\right )}\right )} {\left (x e + d\right )} + 15 \, {\left (c d^{4} e^{25} + b d^{2} e^{27} + a e^{29}\right )} e^{\left (-30\right )}\right )} \sqrt {x e + d} \sqrt {-x e + d} e^{\left (-1\right )} \]

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]

-1/15*(((3*((x*e + d)*c*e^(-5) - 4*c*d*e^(-5))*(x*e + d) + (22*c*d^2*e^25 + 5*b*e^27)*e^(-30))*(x*e + d) - 10*
(2*c*d^3*e^25 + b*d*e^27)*e^(-30))*(x*e + d) + 15*(c*d^4*e^25 + b*d^2*e^27 + a*e^29)*e^(-30))*sqrt(x*e + d)*sq
rt(-x*e + d)*e^(-1)

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maple [A]  time = 0.01, size = 73, normalized size = 0.67 \[ -\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (3 c \,x^{4} e^{4}+5 b \,e^{4} x^{2}+4 c \,d^{2} e^{2} x^{2}+15 a \,e^{4}+10 b \,d^{2} e^{2}+8 c \,d^{4}\right )}{15 e^{6}} \]

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 = 1.03, size = 139, normalized size = 1.28 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} c x^{4}}{5 \, e^{2}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{2} x^{2}}{15 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b x^{2}}{3 \, e^{2}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{4}}{15 \, e^{6}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} b d^{2}}{3 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{e^{2}} \]

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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mupad [B]  time = 1.38, size = 143, normalized size = 1.31 \[ -\frac {\sqrt {d-e\,x}\,\left (\frac {8\,c\,d^5+10\,b\,d^3\,e^2+15\,a\,d\,e^4}{15\,e^6}+\frac {x^3\,\left (4\,c\,d^2\,e^3+5\,b\,e^5\right )}{15\,e^6}+\frac {c\,x^5}{5\,e}+\frac {x^2\,\left (4\,c\,d^3\,e^2+5\,b\,d\,e^4\right )}{15\,e^6}+\frac {x\,\left (8\,c\,d^4\,e+10\,b\,d^2\,e^3+15\,a\,e^5\right )}{15\,e^6}+\frac {c\,d\,x^4}{5\,e^2}\right )}{\sqrt {d+e\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 90.67, size = 350, normalized size = 3.21 \[ - \frac {i a d {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{2}} - \frac {a d {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{2}} - \frac {i b d^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{4}} - \frac {b d^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{4}} - \frac {i c d^{5} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{4}, - \frac {7}{4} & -2, -2, - \frac {3}{2}, 1 \\- \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, - \frac {3}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{6}} - \frac {c d^{5} {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {11}{4}, - \frac {5}{2}, - \frac {9}{4}, -2, 1 & \\- \frac {11}{4}, - \frac {9}{4} & -3, - \frac {5}{2}, - \frac {5}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{6}} \]

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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