3.1.5 \(\int \frac {A+B x^2}{(d+e x^2) \sqrt {a+c x^4}} \, dx\) [5]

Optimal. Leaf size=369 \[ -\frac {(B d-A e) \tan ^{-1}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {c d^2+a e^2}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {a+c x^4}}+\frac {a^{3/4} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d e \left (c d^2-a e^2\right ) \sqrt {a+c x^4}} \]

[Out]

-1/2*(-A*e+B*d)*arctan(x*(a*e^2+c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(c*x^4+a)^(1/2))/d^(1/2)/e^(1/2)/(a*e^2+c*d^2)^(1
/2)-1/2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(
1/4)*x/a^(1/4))),1/2*2^(1/2))*(B*a^(1/2)-A*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^
(1/2)/a^(1/4)/c^(1/4)/(-e*a^(1/2)+d*c^(1/2))/(c*x^4+a)^(1/2)+1/4*a^(3/4)*(-A*e+B*d)*(cos(2*arctan(c^(1/4)*x/a^
(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),-1/4*(-e*a^(1/2)
+d*c^(1/2))^2/d/e/a^(1/2)/c^(1/2),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*(e+d*c^(1/2)/a^(1/2))^2*((c*x^4+a)/(a^(1/
2)+x^2*c^(1/2))^2)^(1/2)/c^(1/4)/d/e/(-a*e^2+c*d^2)/(c*x^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1723, 226, 1721} \begin {gather*} \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d e \sqrt {a+c x^4} \left (c d^2-a e^2\right )}-\frac {(B d-A e) \text {ArcTan}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2+c d^2}}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {a} B-A \sqrt {c}\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]),x]

[Out]

-1/2*((B*d - A*e)*ArcTan[(Sqrt[c*d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(Sqrt[d]*Sqrt[e]*Sqrt[c*d
^2 + a*e^2]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*El
lipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[a + c*x^4]) + (a^
(3/4)*((Sqrt[c]*d)/Sqrt[a] + e)^2*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)
^2]*EllipticPi[-1/4*(Sqrt[c]*d - Sqrt[a]*e)^2/(Sqrt[a]*Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*c
^(1/4)*d*e*(c*d^2 - a*e^2)*Sqrt[a + c*x^4])

Rule 226

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

Rule 1721

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, Simp[(-(B*d - A*e))*(ArcTan[Rt[c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + c*x^4])]/(2*d*e*Rt[c*(d/e) + a*(e/d), 2]))
, x] + Simp[(B*d + A*e)*(A + B*x^2)*(Sqrt[A^2*((a + c*x^4)/(a*(A + B*x^2)^2))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 1723

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2
]}, Dist[(A*(c*d + a*e*q) - a*B*(e + d*q))/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] + Dist[a*(B*d - A*e)
*((e + d*q)/(c*d^2 - a*e^2)), Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e, A,
B}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && NeQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx &=-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{\sqrt {c} d-\sqrt {a} e}+\frac {\left (\sqrt {a} (B d-A e)\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{\sqrt {c} d-\sqrt {a} e}\\ &=-\frac {(B d-A e) \tan ^{-1}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {c d^2+a e^2}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) (B d-A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{c} d e \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {a+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 10.22, size = 138, normalized size = 0.37 \begin {gather*} -\frac {i \sqrt {1+\frac {c x^4}{a}} \left (B d F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+(-B d+A e) \Pi \left (-\frac {i \sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} d e \sqrt {a+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]),x]

[Out]

((-I)*Sqrt[1 + (c*x^4)/a]*(B*d*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] + (-(B*d) + A*e)*Elliptic
Pi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1]))/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*d*e*
Sqrt[a + c*x^4])

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 192, normalized size = 0.52

method	result	size
default	\(\frac {B \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\left (A e -B d \right ) \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticPi \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{e d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(192\)
elliptic	\(\frac {B \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticPi \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right ) A}{d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}-\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticPi \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right ) B}{e \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

B/e/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*
EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+(A*e-B*d)/e/d/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/
2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticPi(x*(I/a^(1/2)*c^(1/2))^(1/2),I*a^(1/2)/c^(1/2)*e/
d,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2))

________________________________________________________________________________________

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((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + a)*(x^2*e + d)), x)

________________________________________________________________________________________

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((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x^{2}}{\sqrt {a + c x^{4}} \left (d + e x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(e*x**2+d)/(c*x**4+a)**(1/2),x)

[Out]

Integral((A + B*x**2)/(sqrt(a + c*x**4)*(d + e*x**2)), x)

________________________________________________________________________________________

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((B*x^2+A)/(e*x^2+d)/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + a)*(x^2*e + d)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/((a + c*x^4)^(1/2)*(d + e*x^2)),x)

[Out]

int((A + B*x^2)/((a + c*x^4)^(1/2)*(d + e*x^2)), x)

________________________________________________________________________________________