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

Optimal. Leaf size=495 \[ \frac{\sqrt{-\sqrt{a^2-2 a c+b^2+c^2}+a-c} \tan ^{-1}\left (\frac{b-\left (-\sqrt{a^2-2 a c+b^2+c^2}+a-c\right ) \tan (d+e x)}{\sqrt{2} \sqrt{-\sqrt{a^2-2 a c+b^2+c^2}+a-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} e \sqrt{a^2-2 a c+b^2+c^2}}-\frac{\sqrt{\sqrt{a^2-2 a c+b^2+c^2}+a-c} \tan ^{-1}\left (\frac{b-\left (\sqrt{a^2-2 a c+b^2+c^2}+a-c\right ) \tan (d+e x)}{\sqrt{2} \sqrt{\sqrt{a^2-2 a c+b^2+c^2}+a-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} e \sqrt{a^2-2 a c+b^2+c^2}}+\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{8 c^{5/2} e}-\frac{3 b \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2 e}+\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}-\frac{\tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{c} e} \]

[Out]

(Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTan[(b - (a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2])*Tan[d + e*x])
/(Sqrt[2]*Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*
Sqrt[a^2 + b^2 - 2*a*c + c^2]*e) - (Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTan[(b - (a - c + Sqrt[a^2
+ b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*
x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*Sqrt[a^2 + b^2 - 2*a*c + c^2]*e) - ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c
]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])]/(Sqrt[c]*e) + ((3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*Tan[d + e*x])/
(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(8*c^(5/2)*e) - (3*b*Sqrt[a + b*Tan[d + e*x] + c*Tan
[d + e*x]^2])/(4*c^2*e) + (Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(2*c*e)

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Rubi [A]  time = 0.820011, antiderivative size = 495, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3700, 6725, 621, 206, 742, 640, 987, 1030, 205} \[ \frac{\sqrt{-\sqrt{a^2-2 a c+b^2+c^2}+a-c} \tan ^{-1}\left (\frac{b-\left (-\sqrt{a^2-2 a c+b^2+c^2}+a-c\right ) \tan (d+e x)}{\sqrt{2} \sqrt{-\sqrt{a^2-2 a c+b^2+c^2}+a-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} e \sqrt{a^2-2 a c+b^2+c^2}}-\frac{\sqrt{\sqrt{a^2-2 a c+b^2+c^2}+a-c} \tan ^{-1}\left (\frac{b-\left (\sqrt{a^2-2 a c+b^2+c^2}+a-c\right ) \tan (d+e x)}{\sqrt{2} \sqrt{\sqrt{a^2-2 a c+b^2+c^2}+a-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} e \sqrt{a^2-2 a c+b^2+c^2}}+\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{8 c^{5/2} e}-\frac{3 b \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2 e}+\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}-\frac{\tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{c} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTan[(b - (a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2])*Tan[d + e*x])
/(Sqrt[2]*Sqrt[a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*
Sqrt[a^2 + b^2 - 2*a*c + c^2]*e) - (Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTan[(b - (a - c + Sqrt[a^2
+ b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*
x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*Sqrt[a^2 + b^2 - 2*a*c + c^2]*e) - ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c
]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])]/(Sqrt[c]*e) + ((3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*Tan[d + e*x])/
(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(8*c^(5/2)*e) - (3*b*Sqrt[a + b*Tan[d + e*x] + c*Tan
[d + e*x]^2])/(4*c^2*e) + (Tan[d + e*x]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/(2*c*e)

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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

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

Rule 987

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

Rule 1030

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
a*g*h, Subst[Int[1/Simp[2*a^2*g*h*c + a*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; F
reeQ[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^4(d+e x)}{\sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt{a+b x+c x^2}}+\frac{x^2}{\sqrt{a+b x+c x^2}}+\frac{1}{\left (1+x^2\right ) \sqrt{a+b x+c x^2}}\right ) \, dx,x,\tan (d+e x)\right )}{e}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}+\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c \tan (d+e x)}{\sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{e}+\frac{\operatorname{Subst}\left (\int \frac{-a-\frac{3 b x}{2}}{\sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 c e}-\frac{\operatorname{Subst}\left (\int \frac{a-c-\sqrt{a^2+b^2-2 a c+c^2}+b x}{\left (1+x^2\right ) \sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 \sqrt{a^2+b^2-2 a c+c^2} e}+\frac{\operatorname{Subst}\left (\int \frac{a-c+\sqrt{a^2+b^2-2 a c+c^2}+b x}{\left (1+x^2\right ) \sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{2 \sqrt{a^2+b^2-2 a c+c^2} e}\\ &=-\frac{\tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{c} e}-\frac{3 b \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2 e}+\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}+\frac{\left (3 b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,\tan (d+e x)\right )}{8 c^2 e}+\frac{\left (b \left (a-c-\sqrt{a^2+b^2-2 a c+c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{2 b \left (a-c-\sqrt{a^2+b^2-2 a c+c^2}\right )+b x^2} \, dx,x,\frac{b-\left (a-c-\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{a^2+b^2-2 a c+c^2} e}-\frac{\left (b \left (a-c+\sqrt{a^2+b^2-2 a c+c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{2 b \left (a-c+\sqrt{a^2+b^2-2 a c+c^2}\right )+b x^2} \, dx,x,\frac{b-\left (a-c+\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{a^2+b^2-2 a c+c^2} e}\\ &=\frac{\sqrt{a-c-\sqrt{a^2+b^2-2 a c+c^2}} \tan ^{-1}\left (\frac{b-\left (a-c-\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{2} \sqrt{a-c-\sqrt{a^2+b^2-2 a c+c^2}} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} \sqrt{a^2+b^2-2 a c+c^2} e}-\frac{\sqrt{a-c+\sqrt{a^2+b^2-2 a c+c^2}} \tan ^{-1}\left (\frac{b-\left (a-c+\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{2} \sqrt{a-c+\sqrt{a^2+b^2-2 a c+c^2}} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} \sqrt{a^2+b^2-2 a c+c^2} e}-\frac{\tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{c} e}-\frac{3 b \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2 e}+\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}+\frac{\left (3 b^2-4 a c\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c \tan (d+e x)}{\sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{4 c^2 e}\\ &=\frac{\sqrt{a-c-\sqrt{a^2+b^2-2 a c+c^2}} \tan ^{-1}\left (\frac{b-\left (a-c-\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{2} \sqrt{a-c-\sqrt{a^2+b^2-2 a c+c^2}} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} \sqrt{a^2+b^2-2 a c+c^2} e}-\frac{\sqrt{a-c+\sqrt{a^2+b^2-2 a c+c^2}} \tan ^{-1}\left (\frac{b-\left (a-c+\sqrt{a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt{2} \sqrt{a-c+\sqrt{a^2+b^2-2 a c+c^2}} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{2} \sqrt{a^2+b^2-2 a c+c^2} e}-\frac{\tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{c} e}+\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{8 c^{5/2} e}-\frac{3 b \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2 e}+\frac{\tan (d+e x) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{2 c e}\\ \end{align*}

Mathematica [C]  time = 2.52434, size = 283, normalized size = 0.57 \[ \frac{\frac{\left (3 b^2-4 c (a+2 c)\right ) \tanh ^{-1}\left (\frac{b+2 c \tan (d+e x)}{2 \sqrt{c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{c^{5/2}}+\frac{2 (2 c \tan (d+e x)-3 b) \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c^2}-\frac{4 i \tanh ^{-1}\left (\frac{2 a+(b-2 i c) \tan (d+e x)-i b}{2 \sqrt{a-i b-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{a-i b-c}}+\frac{4 i \tanh ^{-1}\left (\frac{2 a+(b+2 i c) \tan (d+e x)+i b}{2 \sqrt{a+i b-c} \sqrt{a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt{a+i b-c}}}{8 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)*ArcTanh[(2*a - I*b + (b - (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan
[d + e*x]^2])])/Sqrt[a - I*b - c] + ((4*I)*ArcTanh[(2*a + I*b + (b + (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a + I*b -
c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/Sqrt[a + I*b - c] + ((3*b^2 - 4*c*(a + 2*c))*ArcTanh[(b + 2*
c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/c^(5/2) + (2*(-3*b + 2*c*Tan[d + e*x
])*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])/c^2)/(8*e)

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Maple [B]  time = 0.305, size = 7491919, normalized size = 15135.2 \begin{align*} \text{output 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)+c*tan(e*x+d)^2)^(1/2),x)

[Out]

result too large to display

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

[Out]

Timed out

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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)^4/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(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 ^{4}{\left (d + e x \right )}}{\sqrt{a + b \tan{\left (d + e x \right )} + c \tan ^{2}{\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)+c*tan(e*x+d)**2)**(1/2),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)+c*tan(e*x+d)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError