3.11.75 \(\int \frac {x^2}{(-2+b x^2) (-1+b x^2)^{3/4}} \, dx\) [1075]

Optimal. Leaf size=72 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{-1+b x^2}}\right )}{\sqrt {2} b^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{-1+b x^2}}\right )}{\sqrt {2} b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {453} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{b x^2-1}}\right )}{\sqrt {2} b^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{b x^2-1}}\right )}{\sqrt {2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((-2 + b*x^2)*(-1 + b*x^2)^(3/4)),x]

[Out]

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

Rule 453

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-b/(Sqrt[2]*a*d*Rt[-b^2/a,
4]^3))*ArcTan[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] + Simp[(b/(Sqrt[2]*a*d*Rt[-b^2/a, 4]^3))*ArcT
anh[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && Neg
Q[b^2/a]

Rubi steps

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

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Mathematica [A]
time = 2.02, size = 62, normalized size = 0.86 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{-1+b x^2}}\right )-\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{-1+b x^2}}\right )}{\sqrt {2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/((-2 + b*x^2)*(-1 + b*x^2)^(3/4)),x]

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (b \,x^{2}-2\right ) \left (b \,x^{2}-1\right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2-2)/(b*x^2-1)^(3/4),x)

[Out]

int(x^2/(b*x^2-2)/(b*x^2-1)^(3/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2-2)/(b*x^2-1)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^2/((b*x^2 - 1)^(3/4)*(b*x^2 - 2)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (55) = 110\).
time = 0.58, size = 275, normalized size = 3.82 \begin {gather*} \left [-\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {1}{4}}}{\sqrt {b} x}\right ) - \sqrt {2} \sqrt {b} \log \left (-\frac {b^{2} x^{4} - 2 \, \sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {1}{4}} b^{\frac {3}{2}} x^{3} + 4 \, \sqrt {b x^{2} - 1} b x^{2} + 4 \, b x^{2} - 4 \, \sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {3}{4}} \sqrt {b} x - 4}{b^{2} x^{4} - 4 \, b x^{2} + 4}\right )}{4 \, b^{2}}, \frac {2 \, \sqrt {2} \sqrt {-b} \arctan \left (\frac {\sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {1}{4}} \sqrt {-b}}{b x}\right ) - \sqrt {2} \sqrt {-b} \log \left (-\frac {b^{2} x^{4} - 2 \, \sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {1}{4}} \sqrt {-b} b x^{3} - 4 \, \sqrt {b x^{2} - 1} b x^{2} + 4 \, b x^{2} + 4 \, \sqrt {2} {\left (b x^{2} - 1\right )}^{\frac {3}{4}} \sqrt {-b} x - 4}{b^{2} x^{4} - 4 \, b x^{2} + 4}\right )}{4 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2-2)/(b*x^2-1)^(3/4),x, algorithm="fricas")

[Out]

[-1/4*(2*sqrt(2)*sqrt(b)*arctan(sqrt(2)*(b*x^2 - 1)^(1/4)/(sqrt(b)*x)) - sqrt(2)*sqrt(b)*log(-(b^2*x^4 - 2*sqr
t(2)*(b*x^2 - 1)^(1/4)*b^(3/2)*x^3 + 4*sqrt(b*x^2 - 1)*b*x^2 + 4*b*x^2 - 4*sqrt(2)*(b*x^2 - 1)^(3/4)*sqrt(b)*x
- 4)/(b^2*x^4 - 4*b*x^2 + 4)))/b^2, 1/4*(2*sqrt(2)*sqrt(-b)*arctan(sqrt(2)*(b*x^2 - 1)^(1/4)*sqrt(-b)/(b*x))
- sqrt(2)*sqrt(-b)*log(-(b^2*x^4 - 2*sqrt(2)*(b*x^2 - 1)^(1/4)*sqrt(-b)*b*x^3 - 4*sqrt(b*x^2 - 1)*b*x^2 + 4*b*
x^2 + 4*sqrt(2)*(b*x^2 - 1)^(3/4)*sqrt(-b)*x - 4)/(b^2*x^4 - 4*b*x^2 + 4)))/b^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (b x^{2} - 2\right ) \left (b x^{2} - 1\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**2-2)/(b*x**2-1)**(3/4),x)

[Out]

Integral(x**2/((b*x**2 - 2)*(b*x**2 - 1)**(3/4)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2-2)/(b*x^2-1)^(3/4),x, algorithm="giac")

[Out]

integrate(x^2/((b*x^2 - 1)^(3/4)*(b*x^2 - 2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\left (b\,x^2-1\right )}^{3/4}\,\left (b\,x^2-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((b*x^2 - 1)^(3/4)*(b*x^2 - 2)),x)

[Out]

int(x^2/((b*x^2 - 1)^(3/4)*(b*x^2 - 2)), x)

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