3.399 \(\int \frac{\tan ^2(x)}{\sqrt{a+b \tan ^4(x)}} \, dx\)

Optimal. Leaf size=291 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right ) \sqrt{\frac{a+b \tan ^4(x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (x)}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right ) \sqrt{\frac{a+b \tan ^4(x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(x)}} \]

[Out]

-ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a + b*Tan[x]^4]]/(2*Sqrt[a + b]) + (a^(1/4)*EllipticF[2*ArcTan[(b^(1/4)*Tan[
x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[x]^2)*Sqrt[(a + b*Tan[x]^4)/(Sqrt[a] + Sqrt[b]*Tan[x]^2)^2])/(2*(Sqr
t[a] - Sqrt[b])*b^(1/4)*Sqrt[a + b*Tan[x]^4]) - ((Sqrt[a] + Sqrt[b])*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt
[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[x]^2)*Sqrt[(a + b*Tan[x]^4)/(Sqr
t[a] + Sqrt[b]*Tan[x]^2)^2])/(4*a^(1/4)*(Sqrt[a] - Sqrt[b])*b^(1/4)*Sqrt[a + b*Tan[x]^4])

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Rubi [A]  time = 0.240282, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {3670, 1320, 220, 1707} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}+\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right ) \sqrt{\frac{a+b \tan ^4(x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(x)}}-\frac{\left (\sqrt{a}+\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right ) \sqrt{\frac{a+b \tan ^4(x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2/Sqrt[a + b*Tan[x]^4],x]

[Out]

-ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a + b*Tan[x]^4]]/(2*Sqrt[a + b]) + (a^(1/4)*EllipticF[2*ArcTan[(b^(1/4)*Tan[
x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[x]^2)*Sqrt[(a + b*Tan[x]^4)/(Sqrt[a] + Sqrt[b]*Tan[x]^2)^2])/(2*(Sqr
t[a] - Sqrt[b])*b^(1/4)*Sqrt[a + b*Tan[x]^4]) - ((Sqrt[a] + Sqrt[b])*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt
[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[x]^2)*Sqrt[(a + b*Tan[x]^4)/(Sqr
t[a] + Sqrt[b]*Tan[x]^2)^2])/(4*a^(1/4)*(Sqrt[a] - Sqrt[b])*b^(1/4)*Sqrt[a + b*Tan[x]^4])

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 1320

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1707

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

Rubi steps

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

Mathematica [C]  time = 2.0611, size = 122, normalized size = 0.42 \[ -\frac{i \sqrt{\frac{b \tan ^4(x)}{a}+1} \left (\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (x)\right ),-1\right )-\Pi \left (-\frac{i \sqrt{a}}{\sqrt{b}};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (x)\right )\right |-1\right )\right )}{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b \tan ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^2/Sqrt[a + b*Tan[x]^4],x]

[Out]

((-I)*(EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[x]], -1] - EllipticPi[((-I)*Sqrt[a])/Sqrt[b], I*ArcSi
nh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[x]], -1])*Sqrt[1 + (b*Tan[x]^4)/a])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*Sqrt[a + b*Tan
[x]^4])

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Maple [C]  time = 0.059, size = 179, normalized size = 0.6 \begin{align*}{\sqrt{1-{i \left ( \tan \left ( x \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i \left ( \tan \left ( x \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( \tan \left ( x \right ) \sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}}}-{\sqrt{1-{i \left ( \tan \left ( x \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i \left ( \tan \left ( x \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( \tan \left ( x \right ) \sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},{i\sqrt{a}{\frac{1}{\sqrt{b}}}},{\sqrt{{-i\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(x)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(x)^2)^(1/2)/(a+b*tan
(x)^4)^(1/2)*EllipticF(tan(x)*(I/a^(1/2)*b^(1/2))^(1/2),I)-1/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*ta
n(x)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(x)^2)^(1/2)/(a+b*tan(x)^4)^(1/2)*EllipticPi(tan(x)*(I/a^(1/2)*b^(1/2))^
(1/2),I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)^2/sqrt(b*tan(x)^4 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\tan \left (x\right )^{2}}{\sqrt{b \tan \left (x\right )^{4} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(tan(x)^2/sqrt(b*tan(x)^4 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**2/sqrt(a + b*tan(x)**4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)^2/sqrt(b*tan(x)^4 + a), x)