3.2.36 \(\int \frac {a+b x^2+c x^4}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) [136]

Optimal. Leaf size=99 \[ -\frac {c \sqrt {d-e x} \sqrt {d+e x}}{e^2}-\frac {a \sqrt {d-e x} \sqrt {d+e x}}{2 d^2 x^2}-\frac {\left (2 b d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d-e x} \sqrt {d+e x}}{d}\right )}{2 d^3} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 155, normalized size of antiderivative = 1.57, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {534, 1265, 911, 1171, 396, 214} \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2} \left (a e^2+2 b d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {c \left (d^2-e^2 x^2\right )}{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^3*Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

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

Rule 214

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

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 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1171

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

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^3 \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^3 \sqrt {d^2-e^2 x^2}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {\sqrt {d^2-e^2 x^2} \text {Subst}\left (\int \frac {a+b x+c x^2}{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} \text {Subst}\left (\int \frac {\frac {c d^4+b d^2 e^2+a e^4}{e^4}-\frac {\left (2 c d^2+b e^2\right ) x^2}{e^4}+\frac {c x^4}{e^4}}{\left (\frac {d^2}{e^2}-\frac {x^2}{e^2}\right )^2} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\sqrt {d^2-e^2 x^2} \text {Subst}\left (\int \frac {-a-\frac {2 \left (c d^4+b d^2 e^2\right )}{e^4}+\frac {2 c d^2 x^2}{e^4}}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {c \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (e^2 \left (\frac {2 c d^4}{e^6}+\frac {-a-\frac {2 \left (c d^4+b d^2 e^2\right )}{e^4}}{e^2}\right ) \sqrt {d^2-e^2 x^2}\right ) \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {c \left (d^2-e^2 x^2\right )}{e^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {a \left (d^2-e^2 x^2\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (2 b d^2+a e^2\right ) \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 85, normalized size = 0.86 \begin {gather*} -\frac {\frac {\sqrt {d-e x} \sqrt {d+e x} \left (a d e^2+2 c d^3 x^2\right )}{e^2 x^2}+2 \left (2 b d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d-e x}}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Sqrt[d - e*x]*Sqrt[d + e*x]*(a*d*e^2 + 2*c*d^3*x^2))/(e^2*x^2) + 2*(2*b*d^2 + a*e^2)*ArcTanh[Sqrt[d + e
*x]/Sqrt[d - e*x]])/d^3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.16, size = 163, normalized size = 1.65

method	result	size
default	\(-\frac {\sqrt {-e x +d}\, \sqrt {e x +d}\, \left (2 \,\mathrm {csgn}\left (d \right ) c \,d^{3} x^{2} \sqrt {-e^{2} x^{2}+d^{2}}+\ln \left (\frac {2 d \left (\sqrt {-e^{2} x^{2}+d^{2}}\, \mathrm {csgn}\left (d \right )+d \right )}{x}\right ) a \,e^{4} x^{2}+2 \ln \left (\frac {2 d \left (\sqrt {-e^{2} x^{2}+d^{2}}\, \mathrm {csgn}\left (d \right )+d \right )}{x}\right ) b \,d^{2} e^{2} x^{2}+\mathrm {csgn}\left (d \right ) a d \,e^{2} \sqrt {-e^{2} x^{2}+d^{2}}\right ) \mathrm {csgn}\left (d \right )}{2 d^{3} \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x^{2}}\)	\(163\)
risch	\(-\frac {a \sqrt {-e x +d}\, \sqrt {e x +d}}{2 d^{2} x^{2}}+\frac {\left (-\frac {c \sqrt {-\left (e x -d \right ) \left (e x +d \right )}}{e^{2}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right ) a \,e^{2}}{2 d^{2} \sqrt {d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right ) b}{\sqrt {d^{2}}}\right ) \sqrt {\left (e x +d \right ) \left (-e x +d \right )}}{\sqrt {e x +d}\, \sqrt {-e x +d}}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.49, size = 117, normalized size = 1.18 \begin {gather*} -\sqrt {-x^{2} e^{2} + d^{2}} c e^{\left (-2\right )} - \frac {b \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d} - \frac {a e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{3}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} a}{2 \, d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 98, normalized size = 0.99 \begin {gather*} -\frac {2 \, c d^{4} x^{2} - {\left (2 \, b d^{2} e^{2} + a e^{4}\right )} x^{2} \log \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{x}\right ) + {\left (2 \, c d^{3} x^{2} + a d e^{2}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{2 \, d^{3} e^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: schur row 1 7.37652e-11Francis algorithm
not precise enough for[1.0,-772.794735208,124419.104743,-7478366.70813,154801136.25]schur row 1 1.06106e-10Fr
ancis algorithm not p

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Mupad [B]
time = 5.15, size = 422, normalized size = 4.26 \begin {gather*} \frac {b\,\left (\ln \left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-1\right )-\ln \left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )\right )}{d}-\frac {\left (\frac {c\,d}{e^2}+\frac {c\,x}{e}\right )\,\sqrt {d-e\,x}}{\sqrt {d+e\,x}}-\frac {\frac {a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-\frac {a\,e^2}{2}+\frac {15\,a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{2\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^4}}{\frac {16\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-\frac {32\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^4}+\frac {16\,d^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^6}}-\frac {a\,e^2\,\ln \left (\frac {\sqrt {d+e\,x}-\sqrt {d}}{\sqrt {d-e\,x}-\sqrt {d}}\right )}{2\,d^3}+\frac {a\,e^2\,\ln \left (\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2}-1\right )}{2\,d^3}+\frac {a\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{32\,d^3\,{\left (\sqrt {d-e\,x}-\sqrt {d}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*(log(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 - 1) - log(((d + e*x)^(1/2) - d^(1/2))/((d
- e*x)^(1/2) - d^(1/2)))))/d - (((c*d)/e^2 + (c*x)/e)*(d - e*x)^(1/2))/(d + e*x)^(1/2) - ((a*e^2*((d + e*x)^(
1/2) - d^(1/2))^2)/((d - e*x)^(1/2) - d^(1/2))^2 - (a*e^2)/2 + (15*a*e^2*((d + e*x)^(1/2) - d^(1/2))^4)/(2*((d
- e*x)^(1/2) - d^(1/2))^4))/((16*d^3*((d + e*x)^(1/2) - d^(1/2))^2)/((d - e*x)^(1/2) - d^(1/2))^2 - (32*d^3*(
(d + e*x)^(1/2) - d^(1/2))^4)/((d - e*x)^(1/2) - d^(1/2))^4 + (16*d^3*((d + e*x)^(1/2) - d^(1/2))^6)/((d - e*x
)^(1/2) - d^(1/2))^6) - (a*e^2*log(((d + e*x)^(1/2) - d^(1/2))/((d - e*x)^(1/2) - d^(1/2))))/(2*d^3) + (a*e^2*
log(((d + e*x)^(1/2) - d^(1/2))^2/((d - e*x)^(1/2) - d^(1/2))^2 - 1))/(2*d^3) + (a*e^2*((d + e*x)^(1/2) - d^(1
/2))^2)/(32*d^3*((d - e*x)^(1/2) - d^(1/2))^2)

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