3.42 \(\int \frac{\tan ^2(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)

Optimal. Leaf size=436 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}-\frac{\left (\sqrt{a}+\sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

[Out]

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

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Rubi [A]  time = 0.3407, antiderivative size = 436, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3700, 1319, 1103, 1706} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}+\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\left (\sqrt{a}+\sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^2/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

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

Rule 3700

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rule 1319

Int[(x_)^2/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (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 + b*x^2 + 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 + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2
- 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && PosQ[c/a] && NeQ[c*d^2 - a*e^2, 0]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + 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 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{\left (\sqrt{a}-\sqrt{c}\right ) e}-\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}{\left (1+x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{\left (\sqrt{a}-\sqrt{c}\right ) e}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{a-b+c} e}+\frac{\sqrt [4]{a} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}}}{2 \left (\sqrt{a}-\sqrt{c}\right ) \sqrt [4]{c} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\left (\sqrt{a}+\sqrt{c}\right ) \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) \sqrt [4]{c} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\\ \end{align*}

Mathematica [C]  time = 8.99379, size = 311, normalized size = 0.71 \[ -\frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c \tan ^2(d+e x)}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c \tan ^2(d+e x)}{b-\sqrt{b^2-4 a c}}+1} \left (\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \tan (d+e x)\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )-\Pi \left (\frac{b+\sqrt{b^2-4 a c}}{2 c};i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} \tan (d+e x)\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{2} e \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Tan[d + e*x]], (b + Sqrt[b^2 - 4*a*c])/(b -
Sqrt[b^2 - 4*a*c])] - EllipticPi[(b + Sqrt[b^2 - 4*a*c])/(2*c), I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*
c])]*Tan[d + e*x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*Tan[d
+ e*x]^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*Tan[d + e*x]^2)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c/(b
+ Sqrt[b^2 - 4*a*c])]*e*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])

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Maple [A]  time = 0.158, size = 402, normalized size = 0.9 \begin{align*}{\frac{1}{e} \left ({\frac{\sqrt{2}}{4}\sqrt{4-2\,{\frac{ \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ({\frac{\tan \left ( ex+d \right ) \sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}}}}}-{\sqrt{2}\sqrt{1+{\frac{b \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{2\,a}}-{\frac{ \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}}\sqrt{1+{\frac{b \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{2\,a}}+{\frac{ \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}}{\it EllipticPi} \left ({\frac{\tan \left ( ex+d \right ) \sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},-2\,{\frac{a}{-b+\sqrt{-4\,ac+{b}^{2}}}},{\sqrt{2}\sqrt{-{\frac{1}{2\,a} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }}{\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}}} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}+{\frac{1}{a}\sqrt{-4\,ac+{b}^{2}}}}}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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