3.318 \(\int \frac {1}{(-2 a+b x^2) \sqrt [4]{-a+b x^2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}} \]

[Out]

-1/4*arctan(1/2*x*b^(1/2)/a^(1/4)/(b*x^2-a)^(1/4)*2^(1/2))/a^(3/4)*2^(1/2)/b^(1/2)-1/4*arctanh(1/2*x*b^(1/2)/a
^(1/4)/(b*x^2-a)^(1/4)*2^(1/2))/a^(3/4)*2^(1/2)/b^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {398} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*(-a + b*x^2)^(1/4))]/(2*Sqrt[2]*a^(3/4)*Sqrt[b]) - ArcTanh[(Sqrt[b]*x)/(S
qrt[2]*a^(1/4)*(-a + b*x^2)^(1/4))]/(2*Sqrt[2]*a^(3/4)*Sqrt[b])

Rule 398

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b^2/a), 4]}, Simp[(b*Ar
cTan[(q*x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(2*Sqrt[2]*a*d*q), x] + Simp[(b*ArcTanh[(q*x)/(Sqrt[2]*(a + b*x^2)^(1
/4))])/(2*Sqrt[2]*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (-2 a+b x^2\right ) \sqrt [4]{-a+b x^2}} \, dx &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{-a+b x^2}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{-a+b x^2}}\right )}{2 \sqrt {2} a^{3/4} \sqrt {b}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 163, normalized size = 1.61 \[ -\frac {6 a x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {b x^2}{a},\frac {b x^2}{2 a}\right )}{\left (2 a-b x^2\right ) \sqrt [4]{b x^2-a} \left (b x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};\frac {b x^2}{a},\frac {b x^2}{2 a}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};\frac {b x^2}{a},\frac {b x^2}{2 a}\right )\right )+6 a F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {b x^2}{a},\frac {b x^2}{2 a}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*a*x*AppellF1[1/2, 1/4, 1, 3/2, (b*x^2)/a, (b*x^2)/(2*a)])/((2*a - b*x^2)*(-a + b*x^2)^(1/4)*(6*a*AppellF1[
1/2, 1/4, 1, 3/2, (b*x^2)/a, (b*x^2)/(2*a)] + b*x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, (b*x^2)/a, (b*x^2)/(2*a)] +
AppellF1[3/2, 5/4, 1, 5/2, (b*x^2)/a, (b*x^2)/(2*a)])))

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fricas [B]  time = 50.67, size = 338, normalized size = 3.35 \[ -\left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{3} b \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {3}{4}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} \sqrt {b x^{2} - a} a \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}}\right )} \sqrt {a b \sqrt {\frac {1}{a^{3} b^{2}}}} - \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (b x^{2} - a\right )}^{\frac {1}{4}} a \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}}\right )}}{x}\right ) - \frac {1}{4} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {b x^{2} - a} a^{2} b^{2} x \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {3}{4}} + {\left (b x^{2} - a\right )}^{\frac {1}{4}} a^{2} b \sqrt {\frac {1}{a^{3} b^{2}}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}} + {\left (b x^{2} - a\right )}^{\frac {3}{4}}}{b x^{2} - 2 \, a}\right ) + \frac {1}{4} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {b x^{2} - a} a^{2} b^{2} x \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {3}{4}} - {\left (b x^{2} - a\right )}^{\frac {1}{4}} a^{2} b \sqrt {\frac {1}{a^{3} b^{2}}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x \left (\frac {1}{a^{3} b^{2}}\right )^{\frac {1}{4}} - {\left (b x^{2} - a\right )}^{\frac {3}{4}}}{b x^{2} - 2 \, a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/4)^(1/4)*(1/(a^3*b^2))^(1/4)*arctan(2*(sqrt(1/2)*(2*(1/4)^(3/4)*a^3*b*(1/(a^3*b^2))^(3/4) + (1/4)^(1/4)*sq
rt(b*x^2 - a)*a*(1/(a^3*b^2))^(1/4))*sqrt(a*b*sqrt(1/(a^3*b^2))) - (1/4)^(1/4)*(b*x^2 - a)^(1/4)*a*(1/(a^3*b^2
))^(1/4))/x) - 1/4*(1/4)^(1/4)*(1/(a^3*b^2))^(1/4)*log((2*(1/4)^(3/4)*sqrt(b*x^2 - a)*a^2*b^2*x*(1/(a^3*b^2))^
(3/4) + (b*x^2 - a)^(1/4)*a^2*b*sqrt(1/(a^3*b^2)) + (1/4)^(1/4)*a*b*x*(1/(a^3*b^2))^(1/4) + (b*x^2 - a)^(3/4))
/(b*x^2 - 2*a)) + 1/4*(1/4)^(1/4)*(1/(a^3*b^2))^(1/4)*log(-(2*(1/4)^(3/4)*sqrt(b*x^2 - a)*a^2*b^2*x*(1/(a^3*b^
2))^(3/4) - (b*x^2 - a)^(1/4)*a^2*b*sqrt(1/(a^3*b^2)) + (1/4)^(1/4)*a*b*x*(1/(a^3*b^2))^(1/4) - (b*x^2 - a)^(3
/4))/(b*x^2 - 2*a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} - a\right )}^{\frac {1}{4}} {\left (b x^{2} - 2 \, a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}-2 a \right ) \left (b \,x^{2}-a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} - a\right )}^{\frac {1}{4}} {\left (b x^{2} - 2 \, a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- 2 a + b x^{2}\right ) \sqrt [4]{- a + b x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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