3.10.13 \(\int \frac {(-b+a x^2) \sqrt {b^2+a^2 x^4}}{x^2 (b+a x^2)} \, dx\)

Optimal. Leaf size=69 \[ \frac {\sqrt {a^2 x^4+b^2}}{x}+\sqrt {2} \sqrt {a} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {a^2 x^4+b^2}}\right ) \]

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Rubi [A]  time = 0.97, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {6725, 277, 305, 220, 1196, 1209, 1211, 1699, 205} \begin {gather*} \frac {\sqrt {a^2 x^4+b^2}}{x}+\sqrt {2} \sqrt {a} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {a^2 x^4+b^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[b^2 + a^2*x^4]/x + Sqrt[2]*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[b]*x)/Sqrt[b^2 + a^2*x^4]]

Rule 205

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1209

Int[((a_) + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d - c*e*x^2)*(a +
c*x^4)^(p - 1), x], x] + Dist[(c*d^2 + a*e^2)/e^2, Int[(a + c*x^4)^(p - 1)/(d + e*x^2), x], x] /; FreeQ[{a, c,
d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p + 1/2, 0]

Rule 1211

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C]  time = 0.29, size = 147, normalized size = 2.13 \begin {gather*} \frac {\sqrt {\frac {i a}{b}} \left (a^2 x^4+b^2\right )+2 i a b x \sqrt {\frac {a^2 x^4}{b^2}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i a}{b}} x\right )\right |-1\right )-4 i a b x \sqrt {\frac {a^2 x^4}{b^2}+1} \Pi \left (-i;\left .i \sinh ^{-1}\left (\sqrt {\frac {i a}{b}} x\right )\right |-1\right )}{x \sqrt {\frac {i a}{b}} \sqrt {a^2 x^4+b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(I*a)/b]*(b^2 + a^2*x^4) + (2*I)*a*b*x*Sqrt[1 + (a^2*x^4)/b^2]*EllipticF[I*ArcSinh[Sqrt[(I*a)/b]*x], -1]
- (4*I)*a*b*x*Sqrt[1 + (a^2*x^4)/b^2]*EllipticPi[-I, I*ArcSinh[Sqrt[(I*a)/b]*x], -1])/(Sqrt[(I*a)/b]*x*Sqrt[b
^2 + a^2*x^4])

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IntegrateAlgebraic [A]  time = 0.45, size = 69, normalized size = 1.00 \begin {gather*} \frac {\sqrt {b^2+a^2 x^4}}{x}+\sqrt {2} \sqrt {a} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {b^2+a^2 x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-b + a*x^2)*Sqrt[b^2 + a^2*x^4])/(x^2*(b + a*x^2)),x]

[Out]

Sqrt[b^2 + a^2*x^4]/x + Sqrt[2]*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[b]*x)/Sqrt[b^2 + a^2*x^4]]

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fricas [A]  time = 1.34, size = 162, normalized size = 2.35 \begin {gather*} \left [\frac {\sqrt {2} \sqrt {-a b} x \log \left (\frac {a^{2} x^{4} - 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} \sqrt {-a b} x + b^{2}}{a^{2} x^{4} + 2 \, a b x^{2} + b^{2}}\right ) + 2 \, \sqrt {a^{2} x^{4} + b^{2}}}{2 \, x}, -\frac {\sqrt {2} \sqrt {a b} x \arctan \left (\frac {\sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} \sqrt {a b}}{2 \, a b x}\right ) - \sqrt {a^{2} x^{4} + b^{2}}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*sqrt(-a*b)*x*log((a^2*x^4 - 2*a*b*x^2 - 2*sqrt(2)*sqrt(a^2*x^4 + b^2)*sqrt(-a*b)*x + b^2)/(a^2*x
^4 + 2*a*b*x^2 + b^2)) + 2*sqrt(a^2*x^4 + b^2))/x, -(sqrt(2)*sqrt(a*b)*x*arctan(1/2*sqrt(2)*sqrt(a^2*x^4 + b^2
)*sqrt(a*b)/(a*b*x)) - sqrt(a^2*x^4 + b^2))/x]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a^{2} x^{4} + b^{2}} {\left (a x^{2} - b\right )}}{{\left (a x^{2} + b\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a^2*x^4 + b^2)*(a*x^2 - b)/((a*x^2 + b)*x^2), x)

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maple [A]  time = 0.18, size = 63, normalized size = 0.91

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {a^{2} x^{4}+b^{2}}\, \sqrt {2}}{x}-\frac {2 a b \arctan \left (\frac {\sqrt {a^{2} x^{4}+b^{2}}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\right ) \sqrt {2}}{2}\)	\(63\)
risch	\(\frac {\sqrt {a^{2} x^{4}+b^{2}}}{x}-2 b a \left (\frac {\sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticF \left (x \sqrt {\frac {i a}{b}}, i\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}-\frac {2 \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticPi \left (x \sqrt {\frac {i a}{b}}, i, \frac {\sqrt {-\frac {i a}{b}}}{\sqrt {\frac {i a}{b}}}\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}\right )\)	\(174\)
default	\(2 a \left (-\frac {b \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticF \left (x \sqrt {\frac {i a}{b}}, i\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}+\frac {i b \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticF \left (x \sqrt {\frac {i a}{b}}, i\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}-\frac {i b \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticE \left (x \sqrt {\frac {i a}{b}}, i\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}+\frac {2 b \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticPi \left (x \sqrt {\frac {i a}{b}}, i, \frac {\sqrt {-\frac {i a}{b}}}{\sqrt {\frac {i a}{b}}}\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}\right )+\frac {\sqrt {a^{2} x^{4}+b^{2}}}{x}-\frac {2 i a b \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \left (\EllipticF \left (x \sqrt {\frac {i a}{b}}, i\right )-\EllipticE \left (x \sqrt {\frac {i a}{b}}, i\right )\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*x^4+b^2)^(1/2)*2^(1/2)/x-2*a*b/(a*b)^(1/2)*arctan(1/2*(a^2*x^4+b^2)^(1/2)*2^(1/2)/x/(a*b)^(1/2)))*2^
(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a^{2} x^{4} + b^{2}} {\left (a x^{2} - b\right )}}{{\left (a x^{2} + b\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a^2*x^4 + b^2)*(a*x^2 - b)/((a*x^2 + b)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x^{2} - b\right ) \sqrt {a^{2} x^{4} + b^{2}}}{x^{2} \left (a x^{2} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**2 - b)*sqrt(a**2*x**4 + b**2)/(x**2*(a*x**2 + b)), x)

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