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]

1/8*(-4*a*c+3*b^2)*arctanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/c^(5/2)/e-arc
tanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/e/c^(1/2)+1/2*arctan(1/2*(b-(a-c-(a
^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d
)^2)^(1/2))*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-1/2*arctan(1/2*(b-(a-c+(
a^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(a-c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+
d)^2)^(1/2))*(a-c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/e*2^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/2)-3/4*b*(a+b*tan(e*x+d)+c
*tan(e*x+d)^2)^(1/2)/c^2/e+1/2*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)*tan(e*x+d)/c/e

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Rubi [A]  time = 0.82, antiderivative size = 495, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 2.76, 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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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)^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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not checkedWarning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time:
3.29Not invertible Error: Bad Argument Value

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maple [B]  time = 0.82, size = 7491876, normalized size = 15135.10 \[ \text {output too large to display} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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