3.2.95 \(\int \frac {1}{\sqrt [4]{a+b x^4} (c+d x^4)} \, dx\) [195]

Optimal. Leaf size=105 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}} \]

[Out]

1/2*arctan((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(1/4)+1/2*arctanh((-a*d+b*c)^(1/4)*x
/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(1/4)

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Rubi [A]
time = 0.04, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {385, 218, 214, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\tanh ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

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

Rule 214

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 385

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.59, size = 178, normalized size = 1.70 \begin {gather*} \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (\tan ^{-1}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )+\tanh ^{-1}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )\right )}{c^{3/4} \sqrt [4]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (d \,x^{4}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(1/((b*x^4 + a)^(1/4)*(d*x^4 + c)), 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(1/(b*x^4+a)^(1/4)/(d*x^4+c),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 {1}{\sqrt [4]{a + b x^{4}} \left (c + d x^{4}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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