3.26.10 \(\int \frac {-b+a x^4}{(b+a x^4) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx\) [2510]

Optimal. Leaf size=209 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}{\sqrt {2 a b-c} x^2-\sqrt {b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}}-\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [4]{2 a b-c} x^2}{\sqrt {2}}+\frac {\sqrt {b^2+c x^4+a^2 x^8}}{\sqrt {2} \sqrt [4]{2 a b-c}}}{x \sqrt [4]{b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}} \]

[Out]

1/4*arctan(2^(1/2)*(2*a*b-c)^(1/4)*x*(a^2*x^8+c*x^4+b^2)^(1/4)/((2*a*b-c)^(1/2)*x^2-(a^2*x^8+c*x^4+b^2)^(1/2))
)*2^(1/2)/(2*a*b-c)^(1/4)-1/4*arctanh((1/2*(2*a*b-c)^(1/4)*x^2*2^(1/2)+1/2*(a^2*x^8+c*x^4+b^2)^(1/2)*2^(1/2)/(
2*a*b-c)^(1/4))/x/(a^2*x^8+c*x^4+b^2)^(1/4))*2^(1/2)/(2*a*b-c)^(1/4)

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Rubi [F]
time = 0.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-b + a*x^4)/((b + a*x^4)*(b^2 + c*x^4 + a^2*x^8)^(1/4)),x]

[Out]

(x*(1 + (2*a^2*x^4)/(c - Sqrt[-4*a^2*b^2 + c^2]))^(1/4)*(1 + (2*a^2*x^4)/(c + Sqrt[-4*a^2*b^2 + c^2]))^(1/4)*A
ppellF1[1/4, 1/4, 1/4, 5/4, (-2*a^2*x^4)/(c - Sqrt[-4*a^2*b^2 + c^2]), (-2*a^2*x^4)/(c + Sqrt[-4*a^2*b^2 + c^2
])])/(b^2 + c*x^4 + a^2*x^8)^(1/4) - 2*b*Defer[Int][1/((b + a*x^4)*(b^2 + c*x^4 + a^2*x^8)^(1/4)), x]

Rubi steps

\begin {align*} \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx &=\int \left (\frac {1}{\sqrt [4]{b^2+c x^4+a^2 x^8}}-\frac {2 b}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}}\right ) \, dx\\ &=-\left ((2 b) \int \frac {1}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx\right )+\int \frac {1}{\sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx\\ &=-\left ((2 b) \int \frac {1}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx\right )+\frac {\left (\sqrt [4]{1+\frac {2 a^2 x^4}{c-\sqrt {-4 a^2 b^2+c^2}}} \sqrt [4]{1+\frac {2 a^2 x^4}{c+\sqrt {-4 a^2 b^2+c^2}}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {2 a^2 x^4}{c-\sqrt {-4 a^2 b^2+c^2}}} \sqrt [4]{1+\frac {2 a^2 x^4}{c+\sqrt {-4 a^2 b^2+c^2}}}} \, dx}{\sqrt [4]{b^2+c x^4+a^2 x^8}}\\ &=\frac {x \sqrt [4]{1+\frac {2 a^2 x^4}{c-\sqrt {-4 a^2 b^2+c^2}}} \sqrt [4]{1+\frac {2 a^2 x^4}{c+\sqrt {-4 a^2 b^2+c^2}}} F_1\left (\frac {1}{4};\frac {1}{4},\frac {1}{4};\frac {5}{4};-\frac {2 a^2 x^4}{c-\sqrt {-4 a^2 b^2+c^2}},-\frac {2 a^2 x^4}{c+\sqrt {-4 a^2 b^2+c^2}}\right )}{\sqrt [4]{b^2+c x^4+a^2 x^8}}-(2 b) \int \frac {1}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx\\ \end {align*}

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Mathematica [A]
time = 1.98, size = 184, normalized size = 0.88 \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}{\sqrt {2 a b-c} x^2-\sqrt {b^2+c x^4+a^2 x^8}}\right )-\tanh ^{-1}\left (\frac {\sqrt {2 a b-c} x^2+\sqrt {b^2+c x^4+a^2 x^8}}{\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-b + a*x^4)/((b + a*x^4)*(b^2 + c*x^4 + a^2*x^8)^(1/4)),x]

[Out]

(ArcTan[(Sqrt[2]*(2*a*b - c)^(1/4)*x*(b^2 + c*x^4 + a^2*x^8)^(1/4))/(Sqrt[2*a*b - c]*x^2 - Sqrt[b^2 + c*x^4 +
a^2*x^8])] - ArcTanh[(Sqrt[2*a*b - c]*x^2 + Sqrt[b^2 + c*x^4 + a^2*x^8])/(Sqrt[2]*(2*a*b - c)^(1/4)*x*(b^2 + c
*x^4 + a^2*x^8)^(1/4))])/(2*Sqrt[2]*(2*a*b - c)^(1/4))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}-b}{\left (a \,x^{4}+b \right ) \left (a^{2} x^{8}+c \,x^{4}+b^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)/(a*x^4+b)/(a^2*x^8+c*x^4+b^2)^(1/4),x)

[Out]

int((a*x^4-b)/(a*x^4+b)/(a^2*x^8+c*x^4+b^2)^(1/4),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((a*x^4-b)/(a*x^4+b)/(a^2*x^8+c*x^4+b^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^4 - b)/((a^2*x^8 + c*x^4 + b^2)^(1/4)*(a*x^4 + b)), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)/(a*x^4+b)/(a^2*x^8+c*x^4+b^2)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - b}{\left (a x^{4} + b\right ) \sqrt [4]{a^{2} x^{8} + b^{2} + c x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)/(a*x**4+b)/(a**2*x**8+c*x**4+b**2)**(1/4),x)

[Out]

Integral((a*x**4 - b)/((a*x**4 + b)*(a**2*x**8 + b**2 + c*x**4)**(1/4)), 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((a*x^4-b)/(a*x^4+b)/(a^2*x^8+c*x^4+b^2)^(1/4),x, algorithm="giac")

[Out]

integrate((a*x^4 - b)/((a^2*x^8 + c*x^4 + b^2)^(1/4)*(a*x^4 + b)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {b-a\,x^4}{\left (a\,x^4+b\right )\,{\left (a^2\,x^8+b^2+c\,x^4\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b - a*x^4)/((b + a*x^4)*(c*x^4 + b^2 + a^2*x^8)^(1/4)),x)

[Out]

int(-(b - a*x^4)/((b + a*x^4)*(c*x^4 + b^2 + a^2*x^8)^(1/4)), x)

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