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

Optimal. Leaf size=102 \[ \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{a x^3+b x}}{\sqrt {a x^3+b x}+x^2}\right )-\sqrt {2} \tan ^{-1}\left (\frac {\frac {\sqrt {a x^3+b x}}{\sqrt {2}}-\frac {x^2}{\sqrt {2}}}{x \sqrt [4]{a x^3+b x}}\right ) \]

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Rubi [F]  time = 2.00, 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+a x^2}{\left (b+a x^2+x^3\right ) \sqrt [4]{b x+a x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(12*b*x^(1/4)*(b + a*x^2)^(1/4)*Defer[Subst][Defer[Int][x^2/((b + a*x^8)^(1/4)*(b + a*x^8 + x^12)), x], x, x^(
1/4)])/(b*x + a*x^3)^(1/4) + (4*a*x^(1/4)*(b + a*x^2)^(1/4)*Defer[Subst][Defer[Int][x^10/((b + a*x^8)^(1/4)*(b
+ a*x^8 + x^12)), x], x, x^(1/4)])/(b*x + a*x^3)^(1/4)

Rubi steps

\begin {align*} \int \frac {3 b+a x^2}{\left (b+a x^2+x^3\right ) \sqrt [4]{b x+a x^3}} \, dx &=\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^2}\right ) \int \frac {3 b+a x^2}{\sqrt [4]{x} \sqrt [4]{b+a x^2} \left (b+a x^2+x^3\right )} \, dx}{\sqrt [4]{b x+a x^3}}\\ &=\frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (3 b+a x^8\right )}{\sqrt [4]{b+a x^8} \left (b+a x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^3}}\\ &=\frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {3 b x^2}{\sqrt [4]{b+a x^8} \left (b+a x^8+x^{12}\right )}+\frac {a x^{10}}{\sqrt [4]{b+a x^8} \left (b+a x^8+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^3}}\\ &=\frac {\left (4 a \sqrt [4]{x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{b+a x^8} \left (b+a x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^3}}+\frac {\left (12 b \sqrt [4]{x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{b+a x^8} \left (b+a x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^3}}\\ \end {align*}

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Mathematica [F]  time = 0.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 b+a x^2}{\left (b+a x^2+x^3\right ) \sqrt [4]{b x+a x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 102, normalized size = 1.00 \begin {gather*} -\sqrt {2} \tan ^{-1}\left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {b x+a x^3}}{\sqrt {2}}}{x \sqrt [4]{b x+a x^3}}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{b x+a x^3}}{x^2+\sqrt {b x+a x^3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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((a*x^2+3*b)/(a*x^2+x^3+b)/(a*x^3+b*x)^(1/4),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 {a x^{2} + 3 \, b}{{\left (a x^{3} + b x\right )}^{\frac {1}{4}} {\left (a x^{2} + x^{3} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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