3.2.41 \(\int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) [141]

Optimal. Leaf size=102 \[ -\frac {a \sqrt {d-e x} \sqrt {d+e x}}{d^2 x}+\frac {c x (-d+e x) \sqrt {d+e x}}{2 e^2 \sqrt {d-e x}}-\frac {\left (c d^2+2 b e^2\right ) \tan ^{-1}\left (\frac {\sqrt {d-e x}}{\sqrt {d+e x}}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 155, normalized size of antiderivative = 1.52, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {534, 1279, 396, 223, 209} \begin {gather*} -\frac {a \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}+\frac {\sqrt {d^2-e^2 x^2} \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (2 b e^2+c d^2\right )}{2 e^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x \left (d^2-e^2 x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 534

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 1279

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {d^2-e^2 x^2} \int \frac {a+b x^2+c x^4}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {\sqrt {d^2-e^2 x^2} \int \frac {-b d^2-c d^2 x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x \left (d^2-e^2 x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (\left (2 b+\frac {c d^2}{e^2}\right ) \sqrt {d^2-e^2 x^2}\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x \left (d^2-e^2 x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (\left (2 b+\frac {c d^2}{e^2}\right ) \sqrt {d^2-e^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {c x \left (d^2-e^2 x^2\right )}{2 e^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (c d^2+2 b e^2\right ) \sqrt {d^2-e^2 x^2} \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 86, normalized size = 0.84 \begin {gather*} \frac {-\frac {e \sqrt {d-e x} \sqrt {d+e x} \left (2 a e^2+c d^2 x^2\right )}{d^2 x}+2 \left (c d^2+2 b e^2\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e x}}\right )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((e*Sqrt[d - e*x]*Sqrt[d + e*x]*(2*a*e^2 + c*d^2*x^2))/(d^2*x)) + 2*(c*d^2 + 2*b*e^2)*ArcTan[Sqrt[d + e*x]/S
qrt[d - e*x]])/(2*e^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.16, size = 148, normalized size = 1.45

method	result	size
risch	\(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (c \,d^{2} x^{2}+2 a \,e^{2}\right )}{2 e^{2} d^{2} x}+\frac {\left (\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) b}{\sqrt {e^{2}}}+\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) c \,d^{2}}{2 e^{2} \sqrt {e^{2}}}\right ) \sqrt {\left (e x +d \right ) \left (-e x +d \right )}}{\sqrt {e x +d}\, \sqrt {-e x +d}}\)	\(139\)
default	\(-\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \left (\mathrm {csgn}\left (e \right ) c \,d^{2} e \,x^{2} \sqrt {-e^{2} x^{2}+d^{2}}-2 \arctan \left (\frac {\mathrm {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) b \,d^{2} e^{2} x -\arctan \left (\frac {\mathrm {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) c \,d^{4} x +2 \sqrt {-e^{2} x^{2}+d^{2}}\, \mathrm {csgn}\left (e \right ) e^{3} a \right ) \mathrm {csgn}\left (e \right )}{2 d^{2} e^{3} \sqrt {-e^{2} x^{2}+d^{2}}\, x}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 70, normalized size = 0.69 \begin {gather*} \frac {1}{2} \, c d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} c x e^{\left (-2\right )} + b \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} a}{d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*d^2*arcsin(x*e/d)*e^(-3) - 1/2*sqrt(-x^2*e^2 + d^2)*c*x*e^(-2) + b*arcsin(x*e/d)*e^(-1) - sqrt(-x^2*e^2
+ d^2)*a/(d^2*x)

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Fricas [A]
time = 0.36, size = 90, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (c d^{4} + 2 \, b d^{2} e^{2}\right )} x \arctan \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{e x}\right ) + {\left (c d^{2} e x^{2} + 2 \, a e^{3}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{2 \, d^{2} e^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(c*d^4 + 2*b*d^2*e^2)*x*arctan((sqrt(e*x + d)*sqrt(-e*x + d) - d)/(e*x)) + (c*d^2*e*x^2 + 2*a*e^3)*sqr
t(e*x + d)*sqrt(-e*x + d))/(d^2*e^3*x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (92) = 184\).
time = 5.66, size = 251, normalized size = 2.46 \begin {gather*} \frac {1}{2} \, {\left ({\left (\pi + 2 \, \arctan \left (\frac {\sqrt {x e + d} {\left (\frac {{\left (\sqrt {2} \sqrt {d} - \sqrt {-x e + d}\right )}^{2}}{x e + d} - 1\right )}}{2 \, {\left (\sqrt {2} \sqrt {d} - \sqrt {-x e + d}\right )}}\right )\right )} {\left (c d^{2} + 2 \, b e^{2}\right )} - {\left ({\left (x e + d\right )} c - c d\right )} \sqrt {x e + d} \sqrt {-x e + d} - \frac {8 \, a {\left (\frac {\sqrt {2} \sqrt {d} - \sqrt {-x e + d}}{\sqrt {x e + d}} - \frac {\sqrt {x e + d}}{\sqrt {2} \sqrt {d} - \sqrt {-x e + d}}\right )} e^{4}}{{\left ({\left (\frac {\sqrt {2} \sqrt {d} - \sqrt {-x e + d}}{\sqrt {x e + d}} - \frac {\sqrt {x e + d}}{\sqrt {2} \sqrt {d} - \sqrt {-x e + d}}\right )}^{2} - 4\right )} d^{2}}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((pi + 2*arctan(1/2*sqrt(x*e + d)*((sqrt(2)*sqrt(d) - sqrt(-x*e + d))^2/(x*e + d) - 1)/(sqrt(2)*sqrt(d) -
sqrt(-x*e + d))))*(c*d^2 + 2*b*e^2) - ((x*e + d)*c - c*d)*sqrt(x*e + d)*sqrt(-x*e + d) - 8*a*((sqrt(2)*sqrt(d)
- sqrt(-x*e + d))/sqrt(x*e + d) - sqrt(x*e + d)/(sqrt(2)*sqrt(d) - sqrt(-x*e + d)))*e^4/((((sqrt(2)*sqrt(d) -
sqrt(-x*e + d))/sqrt(x*e + d) - sqrt(x*e + d)/(sqrt(2)*sqrt(d) - sqrt(-x*e + d)))^2 - 4)*d^2))*e^(-3)

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Mupad [B]
time = 7.00, size = 306, normalized size = 3.00 \begin {gather*} \frac {\frac {14\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^3}-\frac {14\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^5}+\frac {2\,c\,d^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^7}-\frac {2\,c\,d^2\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {d-e\,x}-\sqrt {d}}}{e^3\,{\left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}+1\right )}^4}-\frac {4\,b\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {d-e\,x}-\sqrt {d}\right )}{\sqrt {e^2}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {e^2}}+\frac {2\,c\,d^2\,\mathrm {atan}\left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{e^3}-\frac {\left (\frac {a}{d}+\frac {a\,e\,x}{d^2}\right )\,\sqrt {d-e\,x}}{x\,\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((14*c*d^2*((d + e*x)^(1/2) - d^(1/2))^3)/((d - e*x)^(1/2) - d^(1/2))^3 - (14*c*d^2*((d + e*x)^(1/2) - d^(1/2)
)^5)/((d - e*x)^(1/2) - d^(1/2))^5 + (2*c*d^2*((d + e*x)^(1/2) - d^(1/2))^7)/((d - e*x)^(1/2) - d^(1/2))^7 - (
2*c*d^2*((d + e*x)^(1/2) - d^(1/2)))/((d - e*x)^(1/2) - d^(1/2)))/(e^3*(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*
x)^(1/2) - d^(1/2))^2 + 1)^4) - (4*b*atan((e*((d - e*x)^(1/2) - d^(1/2)))/((e^2)^(1/2)*((d + e*x)^(1/2) - d^(1
/2)))))/(e^2)^(1/2) + (2*c*d^2*atan(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)^(1/2) - d^(1/2))))/e^3 - ((a/d + (a
*e*x)/d^2)*(d - e*x)^(1/2))/(x*(d + e*x)^(1/2))

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