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

Optimal. Leaf size=336 \[ \frac {b x}{2 c \sqrt [4]{a+b x^2} (b c-a d)}-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c \left (c+d x^2\right ) (b c-a d)}-\frac {\sqrt {a} \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c \sqrt [4]{a+b x^2} (b c-a d)}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (3 b c-2 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c \sqrt {d} x (a d-b c)^{3/2}}+\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (3 b c-2 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c \sqrt {d} x (a d-b c)^{3/2}} \]

[Out]

1/2*b*x/c/(-a*d+b*c)/(b*x^2+a)^(1/4)-1/2*d*x*(b*x^2+a)^(3/4)/c/(-a*d+b*c)/(d*x^2+c)-1/2*(1+b*x^2/a)^(1/4)*(cos
(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)
/a^(1/2))),2^(1/2))*a^(1/2)*b^(1/2)/c/(-a*d+b*c)/(b*x^2+a)^(1/4)-1/4*a^(1/4)*(-2*a*d+3*b*c)*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)/c/(a*d-b*c)^(3/2)/x/d^(1/2)+1/4*a^(1/4)*
(-2*a*d+3*b*c)*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)/c/(a*d-b
*c)^(3/2)/x/d^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {414, 530, 229, 227, 196, 399, 490, 1218} \[ \frac {b x}{2 c \sqrt [4]{a+b x^2} (b c-a d)}-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c \left (c+d x^2\right ) (b c-a d)}-\frac {\sqrt {a} \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c \sqrt [4]{a+b x^2} (b c-a d)}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (3 b c-2 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c \sqrt {d} x (a d-b c)^{3/2}}+\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (3 b c-2 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c \sqrt {d} x (a d-b c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(2*c*(b*c - a*d)*(a + b*x^2)^(1/4)) - (d*x*(a + b*x^2)^(3/4))/(2*c*(b*c - a*d)*(c + d*x^2)) - (Sqrt[a]*S
qrt[b]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*c*(b*c - a*d)*(a + b*x^2)^(1/4))
- (a^(1/4)*(3*b*c - 2*a*d)*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])/(4*c*Sqrt[d]*(-(b*c) + a*d)^(3/2)*x) + (a^(1/4)*(3*b*c - 2*a*d)*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])/(4*c*Sqrt[d]*(-(b*c)
+ a*d)^(3/2)*x)

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

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 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

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 530

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

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 )^2} \, dx &=-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}+\frac {\int \frac {2 b c-a d+\frac {1}{2} b d x^2}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}+\frac {b \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{4 c (b c-a d)}+\frac {(3 b c-2 a d) \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{4 c (b c-a d)}\\ &=-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}+\frac {\left ((3 b c-2 a d) \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 )}{2 c (b c-a d) x}+\frac {\left (b \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{4 c (b c-a d) \sqrt [4]{a+b x^2}}\\ &=\frac {b x}{2 c (b c-a d) \sqrt [4]{a+b x^2}}-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}-\frac {\left ((3 b c-2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \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 )}{4 c \sqrt {d} (b c-a d) x}+\frac {\left ((3 b c-2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \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 )}{4 c \sqrt {d} (b c-a d) x}-\frac {\left (b \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{4 c (b c-a d) \sqrt [4]{a+b x^2}}\\ &=\frac {b x}{2 c (b c-a d) \sqrt [4]{a+b x^2}}-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c (b c-a d) \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{a} (3 b c-2 a d) \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 )}{4 c \sqrt {d} (-b c+a d)^{3/2} x}+\frac {\sqrt [4]{a} (3 b c-2 a d) \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 )}{4 c \sqrt {d} (-b c+a d)^{3/2} x}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 392, normalized size = 1.17 \[ \frac {-6 a c x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right ) \left (b d x^2 \sqrt [4]{\frac {b x^2}{a}+1} \left (c+d x^2\right ) F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-6 c \left (2 a d-2 b c+b d x^2\right )\right )-d x^3 \left (6 c \left (a+b x^2\right )-b x^2 \sqrt [4]{\frac {b x^2}{a}+1} \left (c+d x^2\right ) F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right ) \left (4 a d F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{12 c^2 \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (b c-a d) \left (x^2 \left (4 a d F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )-6 a c F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*a*c*x*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)]*(-6*c*(-2*b*c + 2*a*d + b*d*x^2) + b*d*x^2*(1
+ (b*x^2)/a)^(1/4)*(c + d*x^2)*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]) - d*x^3*(6*c*(a + b*x^
2) - b*x^2*(1 + (b*x^2)/a)^(1/4)*(c + d*x^2)*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])*(4*a*d*Ap
pellF1[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)]))/(12*c^2*(b*c - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)*(-6*a*c*AppellF1[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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (d \,x^{2}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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