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

Optimal. Leaf size=662 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}-2 \sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 c^{3/4} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{c^{3/4} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}+\frac{\tan (d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{\sqrt{c} e \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )}+\frac{\left (\sqrt{a}+\sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

[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) + (
Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(Sqrt[c]*e*(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)) - (
a^(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]*Ta
n[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])/(c^(3/4)*e
*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + (a^(1/4)*(Sqrt[a] - 2*Sqrt[c])*EllipticF[2*ArcTan[(c^(1/4)*T
an[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*(Sqrt[a] - Sqrt[c])*c^(3/4)*e*Sqrt[a + b*Tan
[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((Sqrt[a] + Sqrt[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)*Sqr
t[(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])

________________________________________________________________________________________

Rubi [A]  time = 0.438926, antiderivative size = 662, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3700, 1325, 1103, 1195, 1706} \[ \frac{\sqrt [4]{a} \left (\sqrt{a}-2 \sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 c^{3/4} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{c^{3/4} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b+c} \tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}+\frac{\tan (d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{\sqrt{c} e \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )}+\frac{\left (\sqrt{a}+\sqrt{c}\right ) \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right ) \sqrt{\frac{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}{\left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{c}\right )^2}{4 \sqrt{a} \sqrt{c}};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \tan (d+e x)}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt{a}-\sqrt{c}\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^4/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) + (
Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(Sqrt[c]*e*(Sqrt[a] + Sqrt[c]*Tan[d + e*x]^2)) - (
a^(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]*Ta
n[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])/(c^(3/4)*e
*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]) + (a^(1/4)*(Sqrt[a] - 2*Sqrt[c])*EllipticF[2*ArcTan[(c^(1/4)*T
an[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*(Sqrt[a] - Sqrt[c])*c^(3/4)*e*Sqrt[a + b*Tan
[d + e*x]^2 + c*Tan[d + e*x]^4]) + ((Sqrt[a] + Sqrt[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)*Sqr
t[(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 1325

Int[(x_)^4/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
}, -Dist[(2*c*d - a*e*q)/(c*e*(e - d*q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + (-Dist[1/(e*q), Int[(1 - q*x
^2)/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[d^2/(e*(e - d*q)), 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] && 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 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 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 ^4(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^4}{\left (1+x^2\right ) \sqrt{a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\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 (1-\frac{\sqrt{c}}{\sqrt{a}}\right ) e}-\frac{\sqrt{a} \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 )}{\sqrt{c} e}+\frac{\left (\sqrt{a} \left (\sqrt{a}-2 \sqrt{c}\right )\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 ) \sqrt{c} 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{\tan (d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{\sqrt{c} e \left (\sqrt{a}+\sqrt{c} \tan ^2(d+e x)\right )}-\frac{\sqrt [4]{a} 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}}}{c^{3/4} e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\sqrt [4]{a} \left (\sqrt{a}-2 \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{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 ) c^{3/4} 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 = 22.5769, size = 533, normalized size = 0.81 \[ \frac{\frac{\sin (2 (d+e x)) \sqrt{\sec ^4(d+e x) ((a-b+c) \cos (4 (d+e x))+4 (a-c) \cos (2 (d+e x))+3 a+b+3 c)}}{\sqrt{2}}+\frac{-4 \sin (d+e x) \cos (d+e x) \left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )+\frac{i \sqrt{2} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c \tan ^2(d+e x)}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c \tan ^2(d+e x)}{b-\sqrt{b^2-4 a c}}+1} \left (\left (-\sqrt{b^2-4 a c}+b+2 c\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \tan (d+e x)\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+\left (\sqrt{b^2-4 a c}-b\right ) E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} \tan (d+e x)\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )-2 c \Pi \left (\frac{b+\sqrt{b^2-4 a c}}{2 c};i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} \tan (d+e x)\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}}}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}}{4 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[(3*a + b + 3*c + 4*(a - c)*Cos[2*(d + e*x)] + (a - b + c)*Cos[4*(d + e*x)])*Sec[d + e*x]^4]*Sin[2*(d +
e*x)])/Sqrt[2] + ((I*Sqrt[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]*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])] - 2*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[c/(b + Sqrt[b^2 - 4*a*c])]
- 4*Cos[d + e*x]*Sin[d + e*x]*(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4))/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d
+ e*x]^4])/(4*c*e)

________________________________________________________________________________________

Maple [A]  time = 0.184, size = 646, normalized size = 1. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(-1/2*a*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)/(b+(-4*a*c+b^2)^(1/2))*(E
llipticF(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
))-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)))-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)))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\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)**4/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{4}}{\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)^4/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")

[Out]

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