3.4.25 \(\int \frac {1}{\sqrt [4]{a+b x^2} (c+d x^2)} \, dx\) [325]

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

[Out]

a^(1/4)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/x/d^(1/2)/(a*d
-b*c)^(1/2)-a^(1/4)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/x/d
^(1/2)/(a*d-b*c)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 167, normalized size of antiderivative = 1.00, 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 = {408, 504, 1232} \begin {gather*} \frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\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 .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt {d} x \sqrt {a d-b c}} \end {gather*}

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 408

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 504

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), Int[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 1232

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], 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 ) \text {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}} \text {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}} \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 6.99, size = 160, normalized size = 0.96 \begin {gather*} -\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 (-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 )+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 )\right )} \end {gather*}

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

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 = 0.00, 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^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)] 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^2+a)^(1/4)/(d*x^2+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^{2}} \left (c + d x^{2}\right )}\, dx \end {gather*}

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 = 0.00, 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^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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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