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

Optimal. Leaf size=829 \[ \frac{\sqrt{a-b+c} \tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}\right )}{2 e}+\frac{\sqrt{c} \tan (d+e x) \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{e \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )}-\frac{\sqrt [4]{a} \sqrt [4]{c} E\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{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}-\frac{\sqrt [4]{c} (a-b+c) \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 ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac{\left (b-c+\sqrt{a} \sqrt{c}\right ) \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 ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{2 \sqrt [4]{a} \sqrt [4]{c} e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac{\left (\sqrt{a}+\sqrt{c}\right ) (a-b+c) \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{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) \sqrt [4]{c} e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}} \]

[Out]

(Sqrt[a - b + c]*ArcTan[(Sqrt[a - b + c]*Tan[d + e*x])/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]])/(2*e) +
(Sqrt[c]*Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(e*(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)) -
(a^(1/4)*c^(1/4)*EllipticE[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])/
(e*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((b + Sqrt[a]*Sqrt[c] - c)*EllipticF[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])/(2*a^(1/4)*c^(1/4)*e*Sqrt[a + b*Tan[d + e*x]^2 +
c*Tan[d + e*x]^4]) - (c^(1/4)*(a - b + c)*EllipticF[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])/(2*a^(1/4)*(Sqrt[a] - Sqrt[c])*e*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((Sq
rt[a] + Sqrt[c])*(a - b + c)*EllipticPi[-(Sqrt[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*Ta
n[d + e*x]^2 + c*Tan[d + e*x]^4])

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Rubi [A]  time = 0.551654, antiderivative size = 829, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1208, 1197, 1103, 1195, 1216, 1706} \[ \frac{\sqrt{a-b+c} \tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}\right )}{2 e}+\frac{\sqrt{c} \tan (d+e x) \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{e \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )}-\frac{\sqrt [4]{a} \sqrt [4]{c} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}-\frac{\sqrt [4]{c} (a-b+c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac{\left (b-c+\sqrt{a} \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{2 \sqrt [4]{a} \sqrt [4]{c} e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+\frac{\left (\sqrt{a}+\sqrt{c}\right ) (a-b+c) \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right ) \sqrt{\frac{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}{\left (\sqrt{c} \tan ^2(d+e x)+\sqrt{a}\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) \sqrt [4]{c} e \sqrt{c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - b + c]*ArcTan[(Sqrt[a - b + c]*Tan[d + e*x])/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]])/(2*e) +
(Sqrt[c]*Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(e*(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)) -
(a^(1/4)*c^(1/4)*EllipticE[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])/
(e*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((b + Sqrt[a]*Sqrt[c] - c)*EllipticF[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])/(2*a^(1/4)*c^(1/4)*e*Sqrt[a + b*Tan[d + e*x]^2 +
c*Tan[d + e*x]^4]) - (c^(1/4)*(a - b + c)*EllipticF[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])/(2*a^(1/4)*(Sqrt[a] - Sqrt[c])*e*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((Sq
rt[a] + Sqrt[c])*(a - b + c)*EllipticPi[-(Sqrt[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*Ta
n[d + e*x]^2 + c*Tan[d + e*x]^4])

Rule 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 0]

Rule 1197

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

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 1195

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

Rule 1216

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Di
st[(c*d + a*e*q)/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(a*e*(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] && NeQ[c*d^2 - a*e^2, 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 \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2+c x^4}}{1+x^2} \, dx,x,\tan (d+e x)\right )}{e}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{-b+c-c x^2}{\sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}+\frac{(a-b+c) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\left (b+\sqrt{a} \sqrt{c}-c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}-\frac{\left (\sqrt{a} \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}+\frac{\left (\sqrt{a} (a-b+c)\right ) \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{\left (\sqrt{c} (a-b+c)\right ) \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-b+c} \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}+\frac{\sqrt{c} \tan (d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{e \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )}-\frac{\sqrt [4]{a} \sqrt [4]{c} E\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}}}{e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\left (b+\sqrt{a} \sqrt{c}-c\right ) 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 \sqrt [4]{a} \sqrt [4]{c} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\sqrt [4]{c} (a-b+c) 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 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{c}\right ) e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\left (\sqrt{a}+\sqrt{c}\right ) (a-b+c) \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 = 1.79924, size = 428, normalized size = 0.52 \[ \frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c \tan ^2(d+e x)}{\sqrt{b^2-4 a c}+b}} \sqrt{1-\frac{2 c \tan ^2(d+e x)}{\sqrt{b^2-4 a c}-b}} \left (-\left (\sqrt{b^2-4 a c}+b-2 c\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \tan (d+e x)\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+\left (\sqrt{b^2-4 a c}-b\right ) E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} \tan (d+e x)\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )-2 (a-b+c) \Pi \left (\frac{b+\sqrt{b^2-4 a c}}{2 c};i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} \tan (d+e x)\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )\right )}{2 \sqrt{2} e \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*((-b + Sqrt[b^2 - 4*a*c])*EllipticE[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])] - (b - 2*c + Sqrt[b^2 - 4*a*c])*EllipticF[I*ArcSinh[Sqrt[2]*Sqr
t[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])] - 2*(a - b + c)*E
llipticPi[(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 = 0.153, size = 1497, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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/a+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2)
)^(1/2)*(4+2/a*tan(e*x+d)^2*b+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)*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))*b-1/4*2^(1/2)/(-b/a+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(
1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b+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)*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))*c-1/2*c*a*2^(1/2)/(-b/a+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b-2/a*tan(e*x+d)^2*(-4*a*
c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b+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)/(b+(-4*a*c+b^2)^(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))+1/2*c*a*2^(1/2)/(-b/a+1/a*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d
)^2*b-2/a*tan(e*x+d)^2*(-4*a*c+b^2)^(1/2))^(1/2)*(4+2/a*tan(e*x+d)^2*b+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)/(b+(-4*a*c+b^2)^(1/2))*EllipticE(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))+a*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))-b*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*t
an(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))+c*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)*E
llipticPi(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 \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((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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