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 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {a-b+c} \tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}}+\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} e \left (\sqrt {a}-\sqrt {c}\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\left (\sqrt {a}+\sqrt {c}\right ) \left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right ) \sqrt {\frac {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt {a}+\sqrt {c} \tan ^2(d+e x)\right )^2}} \Pi \left (-\frac {\left (\sqrt {a}-\sqrt {c}\right )^2}{4 \sqrt {a} \sqrt {c}};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt {a}-\sqrt {c}\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

[Out]

-1/2*arctan((a-b+c)^(1/2)*tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/(a-b+c)^(1/2)+1/2*a^(1/4)*(cos
(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4)))*EllipticF(sin(2*arct
an(c^(1/4)*tan(e*x+d)/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)/(a^(1/2)+c
^(1/2)*tan(e*x+d)^2)^2)^(1/2)*(a^(1/2)+c^(1/2)*tan(e*x+d)^2)/c^(1/4)/e/(a^(1/2)-c^(1/2))/(a+b*tan(e*x+d)^2+c*t
an(e*x+d)^4)^(1/2)-1/4*(cos(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*tan(e*x+d)/a^(
1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*tan(e*x+d)/a^(1/4))),-1/4*(a^(1/2)-c^(1/2))^2/a^(1/2)/c^(1/2),1/2*(2-b/
a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+c^(1/2))*((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)/(a^(1/2)+c^(1/2)*tan(e*x+d)^2)^2)
^(1/2)*(a^(1/2)+c^(1/2)*tan(e*x+d)^2)/a^(1/4)/c^(1/4)/e/(a^(1/2)-c^(1/2))/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1
/2)

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Rubi [A]  time = 0.34, antiderivative size = 436, normalized size of antiderivative = 1.00, 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 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 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 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]

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]

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*}

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Mathematica [C]  time = 11.19, size = 311, normalized size = 0.71 \[ -\frac {i \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c \tan ^2(d+e x)}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {2 c \tan ^2(d+e x)}{b-\sqrt {b^2-4 a c}}+1} \left (F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \tan (d+e x)\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\Pi \left (\frac {b+\sqrt {b^2-4 a c}}{2 c};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \tan (d+e x)\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {2} e \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

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

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)

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maple [A]  time = 0.42, size = 402, normalized size = 0.92 \[ \frac {\frac {\sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 c a +b^{2}}\right ) \left (\tan ^{2}\left (e x +d \right )\right )}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 c a +b^{2}}\right ) \left (\tan ^{2}\left (e x +d \right )\right )}{a}}\, \EllipticF \left (\frac {\tan \left (e x +d \right ) \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 c a +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 c a +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 c a +b^{2}}}{a}}\, \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}}-\frac {\sqrt {2}\, \sqrt {1+\frac {\left (\tan ^{2}\left (e x +d \right )\right ) b}{2 a}-\frac {\left (\tan ^{2}\left (e x +d \right )\right ) \sqrt {-4 c a +b^{2}}}{2 a}}\, \sqrt {1+\frac {\left (\tan ^{2}\left (e x +d \right )\right ) b}{2 a}+\frac {\left (\tan ^{2}\left (e x +d \right )\right ) \sqrt {-4 c a +b^{2}}}{2 a}}\, \EllipticPi \left (\frac {\tan \left (e x +d \right ) \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 c a +b^{2}}}{a}}}{2}, -\frac {2 a}{-b +\sqrt {-4 c a +b^{2}}}, \frac {\sqrt {-\frac {b +\sqrt {-4 c a +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {-4 c a +b^{2}}}{a}}}\right )}{\sqrt {-\frac {b}{a}+\frac {\sqrt {-4 c a +b^{2}}}{a}}\, \sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}}}{e} \]

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)/(-1/a*b+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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )^{2}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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