3.7.19 \(\int \frac {-3 b+2 a x}{\sqrt [4]{-b x+a x^2} (b-a x+x^3)} \, dx\)

Optimal. Leaf size=49 \[ 2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^2-b x}}{x}\right )-2 \tanh ^{-1}\left (\frac {\left (a x^2-b x\right )^{3/4}}{a x-b}\right ) \]

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Rubi [F]  time = 2.11, 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 {-3 b+2 a x}{\sqrt [4]{-b x+a x^2} \left (b-a x+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(-8*a*x^(1/4)*(-b + a*x)^(1/4)*Defer[Subst][Defer[Int][x^6/((-b + a*x^4)^(1/4)*(-b + a*x^4 - x^12)), x], x, x^
(1/4)])/(-(b*x) + a*x^2)^(1/4) - (12*b*x^(1/4)*(-b + a*x)^(1/4)*Defer[Subst][Defer[Int][x^2/((-b + a*x^4)^(1/4
)*(b - a*x^4 + x^12)), x], x, x^(1/4)])/(-(b*x) + a*x^2)^(1/4)

Rubi steps

\begin {align*} \int \frac {-3 b+2 a x}{\sqrt [4]{-b x+a x^2} \left (b-a x+x^3\right )} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{-b+a x}\right ) \int \frac {-3 b+2 a x}{\sqrt [4]{x} \sqrt [4]{-b+a x} \left (b-a x+x^3\right )} \, dx}{\sqrt [4]{-b x+a x^2}}\\ &=\frac {\left (4 \sqrt [4]{x} \sqrt [4]{-b+a x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-3 b+2 a x^4\right )}{\sqrt [4]{-b+a x^4} \left (b-a x^4+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^2}}\\ &=\frac {\left (4 \sqrt [4]{x} \sqrt [4]{-b+a x}\right ) \operatorname {Subst}\left (\int \left (-\frac {2 a x^6}{\sqrt [4]{-b+a x^4} \left (-b+a x^4-x^{12}\right )}-\frac {3 b x^2}{\sqrt [4]{-b+a x^4} \left (b-a x^4+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^2}}\\ &=-\frac {\left (8 a \sqrt [4]{x} \sqrt [4]{-b+a x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{-b+a x^4} \left (-b+a x^4-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^2}}-\frac {\left (12 b \sqrt [4]{x} \sqrt [4]{-b+a x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{-b+a x^4} \left (b-a x^4+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x+a x^2}}\\ \end {align*}

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Mathematica [F]  time = 1.48, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3 b+2 a x}{\sqrt [4]{-b x+a x^2} \left (b-a x+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 49, normalized size = 1.00 \begin {gather*} 2 \tan ^{-1}\left (\frac {\sqrt [4]{-b x+a x^2}}{x}\right )-2 \tanh ^{-1}\left (\frac {\left (-b x+a x^2\right )^{3/4}}{-b+a x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  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((2*a*x-3*b)/(a*x^2-b*x)^(1/4)/(x^3-a*x+b),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {2 a x -3 b}{\left (a \,x^{2}-b x \right )^{\frac {1}{4}} \left (x^{3}-a x +b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {3\,b-2\,a\,x}{{\left (a\,x^2-b\,x\right )}^{1/4}\,\left (x^3-a\,x+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  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((2*a*x-3*b)/(a*x**2-b*x)**(1/4)/(x**3-a*x+b),x)

[Out]

Timed out

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