3.1079 \(\int \frac {x^2}{(-2 a+b x^2) (-a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=96 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {442} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b x^2-a}}\right )}{\sqrt {2} \sqrt [4]{a} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 442

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

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 68, normalized size = 0.71 \[ -\frac {x^3 \left (1-\frac {b x^2}{a}\right )^{3/4} F_1\left (\frac {3}{2};\frac {3}{4},1;\frac {5}{2};\frac {b x^2}{a},\frac {b x^2}{2 a}\right )}{6 a \left (b x^2-a\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.92, size = 207, normalized size = 2.16 \[ 2 \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{6}}\right )^{\frac {1}{4}} \arctan \left (\frac {4 \, {\left (\sqrt {\frac {1}{2}} \left (\frac {1}{4}\right )^{\frac {3}{4}} a b^{4} x \sqrt {\frac {b^{4} x^{2} \sqrt {\frac {1}{a b^{6}}} + 2 \, \sqrt {b x^{2} - a}}{x^{2}}} \left (\frac {1}{a b^{6}}\right )^{\frac {3}{4}} - \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (b x^{2} - a\right )}^{\frac {1}{4}} a b^{4} \left (\frac {1}{a b^{6}}\right )^{\frac {3}{4}}\right )}}{x}\right ) - \frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{6}}\right )^{\frac {1}{4}} \log \left (\frac {\left (\frac {1}{4}\right )^{\frac {1}{4}} b^{2} x \left (\frac {1}{a b^{6}}\right )^{\frac {1}{4}} + {\left (b x^{2} - a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a b^{6}}\right )^{\frac {1}{4}} \log \left (-\frac {\left (\frac {1}{4}\right )^{\frac {1}{4}} b^{2} x \left (\frac {1}{a b^{6}}\right )^{\frac {1}{4}} - {\left (b x^{2} - a\right )}^{\frac {1}{4}}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/4)^(1/4)*(1/(a*b^6))^(1/4)*arctan(4*(sqrt(1/2)*(1/4)^(3/4)*a*b^4*x*sqrt((b^4*x^2*sqrt(1/(a*b^6)) + 2*sqrt
(b*x^2 - a))/x^2)*(1/(a*b^6))^(3/4) - (1/4)^(3/4)*(b*x^2 - a)^(1/4)*a*b^4*(1/(a*b^6))^(3/4))/x) - 1/2*(1/4)^(1
/4)*(1/(a*b^6))^(1/4)*log(((1/4)^(1/4)*b^2*x*(1/(a*b^6))^(1/4) + (b*x^2 - a)^(1/4))/x) + 1/2*(1/4)^(1/4)*(1/(a
*b^6))^(1/4)*log(-((1/4)^(1/4)*b^2*x*(1/(a*b^6))^(1/4) - (b*x^2 - a)^(1/4))/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b x^{2} - a\right )}^{\frac {3}{4}} {\left (b x^{2} - 2 \, a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (b \,x^{2}-2 a \right ) \left (b \,x^{2}-a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b x^{2} - a\right )}^{\frac {3}{4}} {\left (b x^{2} - 2 \, a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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