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

Optimal. Leaf size=105 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}{\sqrt {a} x}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}{\sqrt {a} x}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}} \]

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Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 0.96, 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} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.15, size = 163, normalized size = 1.55 \begin {gather*} -\frac {6 b x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {a x^2}{b},\frac {a x^2}{2 b}\right )}{\left (2 b-a x^2\right ) \sqrt [4]{a x^2-b} \left (a x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};\frac {a x^2}{b},\frac {a x^2}{2 b}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};\frac {a x^2}{b},\frac {a x^2}{2 b}\right )\right )+6 b F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {a x^2}{b},\frac {a x^2}{2 b}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 105, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {a} x}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {a} x}\right )}{2 \sqrt {2} \sqrt {a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 116.86, size = 338, normalized size = 3.22 \begin {gather*} -\left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a b^{3} \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {3}{4}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} \sqrt {a x^{2} - b} b \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}}\right )} \sqrt {a b \sqrt {\frac {1}{a^{2} b^{3}}}} - \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}}\right )}}{x}\right ) - \frac {1}{4} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {a x^{2} - b} a^{2} b^{2} x \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {3}{4}} + {\left (a x^{2} - b\right )}^{\frac {1}{4}} a b^{2} \sqrt {\frac {1}{a^{2} b^{3}}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}} + {\left (a x^{2} - b\right )}^{\frac {3}{4}}}{a x^{2} - 2 \, b}\right ) + \frac {1}{4} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {a x^{2} - b} a^{2} b^{2} x \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {3}{4}} - {\left (a x^{2} - b\right )}^{\frac {1}{4}} a b^{2} \sqrt {\frac {1}{a^{2} b^{3}}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x \left (\frac {1}{a^{2} b^{3}}\right )^{\frac {1}{4}} - {\left (a x^{2} - b\right )}^{\frac {3}{4}}}{a x^{2} - 2 \, b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a \,x^{2}-2 b \right ) \left (a \,x^{2}-b \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-int(1/((a*x^2 - b)^(1/4)*(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 {1}{\left (a x^{2} - 2 b\right ) \sqrt [4]{a x^{2} - b}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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