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

Optimal. Leaf size=278 \[ \frac {\sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{2 c d \left (a+b x^2\right )^{3/4}}+\frac {x \sqrt [4]{a+b x^2}}{2 c \left (c+d x^2\right )}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-2 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d x (b c-a d)}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-2 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d x (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 278, 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 = {412, 530, 233, 231, 401, 108, 409, 1218} \[ \frac {x \sqrt [4]{a+b x^2}}{2 c \left (c+d x^2\right )}+\frac {\sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c d \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-2 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d x (b c-a d)}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-2 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d x (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^2)^(1/4))/(2*c*(c + d*x^2)) + (Sqrt[a]*Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/
Sqrt[a]]/2, 2])/(2*c*d*(a + b*x^2)^(3/4)) - (a^(1/4)*(b*c - 2*a*d)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sq
rt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(4*c*d*(b*c - a*d)*x) - (a^(1/4)*(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*d*(b*c - a*d)*x)

Rule 108

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - (d*e)/f + (d*x^4)/f]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-(f/(d*e - c*f)), 0]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 401

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[Sqrt[-((b*x^2)/a)]/(2*x), Subst[I
nt[1/(Sqrt[-((b*x)/a)]*(a + b*x)^(3/4)*(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0]

Rule 409

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

Rule 412

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

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

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Mathematica [C]  time = 0.19, size = 232, normalized size = 0.83 \[ \frac {x \left (\frac {6 \left (\frac {a+b x^2}{c}-\frac {6 a^2 F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{x^2 \left (4 a d F_1\left (\frac {3}{2};\frac {3}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c F_1\left (\frac {3}{2};\frac {7}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )-6 a c F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}\right )}{c+d x^2}+\frac {b x^2 \left (\frac {b x^2}{a}+1\right )^{3/4} F_1\left (\frac {3}{2};\frac {3}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c^2}\right )}{12 \left (a+b x^2\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*((b*x^2*(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/c^2 + (6*((a + b*x^2)
/c - (6*a^2*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/(-6*a*c*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^
2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*AppellF1[3/2,
7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(c + d*x^2)))/(12*(a + b*x^2)^(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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 \[ \int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (d\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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