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

Optimal. Leaf size=83 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \]

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Rubi [A]  time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {377, 212, 206, 203} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.80 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.33, size = 83, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{2\ 2^{3/4} \sqrt [4]{a} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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