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

Optimal. Leaf size=167 \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} x \sqrt{a d-b c}}-\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} x \sqrt{a d-b c}} \]

[Out]

(a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/
4)], -1])/(Sqrt[d]*Sqrt[-(b*c) + a*d]*x) - (a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c
) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(Sqrt[d]*Sqrt[-(b*c) + a*d]*x)

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Rubi [A]  time = 0.108276, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {399, 490, 1218} \[ \frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} x \sqrt{a d-b c}}-\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} x \sqrt{a d-b c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/
4)], -1])/(Sqrt[d]*Sqrt[-(b*c) + a*d]*x) - (a^(1/4)*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c
) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(Sqrt[d]*Sqrt[-(b*c) + a*d]*x)

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx &=\frac{\left (2 \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{a}} \left (b c-a d+d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{x}\\ &=-\frac{\sqrt{-\frac{b x^2}{a}} \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-b c+a d}-\sqrt{d} x^2\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{\sqrt{d} x}+\frac{\sqrt{-\frac{b x^2}{a}} \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{-b c+a d}+\sqrt{d} x^2\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{\sqrt{d} x}\\ &=\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} \sqrt{-b c+a d} x}-\frac{\sqrt [4]{a} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt{d} \sqrt{-b c+a d} x}\\ \end{align*}

Mathematica [C]  time = 0.0527963, size = 160, normalized size = 0.96 \[ -\frac{6 a c x F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{\sqrt [4]{a+b x^2} \left (c+d x^2\right ) \left (x^2 \left (4 a d F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+b c F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )-6 a c F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*a*c*x*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/((a + b*x^2)^(1/4)*(c + d*x^2)*(-6*a*c*Appel
lF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2, -((b*x^2)/a), -((d*x
^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{2}+c}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt [4]{a + b x^{2}} \left (c + d x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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