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

Optimal. Leaf size=120 \[ -\frac {\tan ^{-1}\left (\frac {a^{3/4} \left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}+1\right )}{\sqrt {b} x \sqrt [4]{a+b x^2}}\right )}{2 a^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {a^{3/4} \left (1-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {b} x \sqrt [4]{a+b x^2}}\right )}{2 a^{3/4} \sqrt {b}} \]

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Rubi [A]  time = 0.02, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {397} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {a^{3/4} \left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}+1\right )}{\sqrt {b} x \sqrt [4]{a+b x^2}}\right )}{2 a^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {a^{3/4} \left (1-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {b} x \sqrt [4]{a+b x^2}}\right )}{2 a^{3/4} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 397

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

Rubi steps

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

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Mathematica [C]  time = 0.15, size = 165, normalized size = 1.38 \begin {gather*} \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 )}{\sqrt [4]{a+b x^2} \left (2 a+b x^2\right ) \left (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 )-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 )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 136, normalized size = 1.13 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {2 \sqrt [4]{a} \sqrt {b} x \sqrt [4]{a+b x^2}}{2 \sqrt {a} \sqrt {a+b x^2}+b x^2}\right )}{4 a^{3/4} \sqrt {b}}-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [4]{a} \sqrt {a+b x^2}}{\sqrt {b}}-\frac {\sqrt {b} x^2}{2 \sqrt [4]{a}}}{x \sqrt [4]{a+b x^2}}\right )}{4 a^{3/4} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 137.83, size = 337, normalized size = 2.81 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(1/4)/(b*x^2+2*a),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)*s
qrt(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 \begin {gather*} \int \frac {1}{{\left (b x^{2} + 2 \, a\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (b \,x^{2}+2 a \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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