\(\int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\) [26]

Optimal result

Integrand size = 35, antiderivative size = 209 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {\sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}-\frac {\left (b^2+4 b c-4 c (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac {\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e} \] Output:

1/2*(a-b+c)^(1/2)*arctanh(1/2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/( 
a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e-1/16*(b^2+4*b*c-4*c*(a+2*c))*arc 
tanh(1/2*(b+2*c*tan(e*x+d)^2)/c^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1 
/2))/c^(3/2)/e+1/8*(b-4*c+2*c*tan(e*x+d)^2)*(a+b*tan(e*x+d)^2+c*tan(e*x+d) 
^4)^(1/2)/c/e
 

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\frac {8 c^{3/2} \sqrt {a-b+c} \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )-\left (b^2+4 b c-4 c (a+2 c)\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+2 \sqrt {c} \left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^{3/2} e} \] Input:

Integrate[Tan[d + e*x]^3*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

(8*c^(3/2)*Sqrt[a - b + c]*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2 
*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] - (b^2 + 
4*b*c - 4*c*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a 
+ b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] + 2*Sqrt[c]*(b - 4*c + 2*c*Tan[d 
+ e*x]^2)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(16*c^(3/2)*e)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4183, 1578, 1231, 27, 1269, 1092, 219, 1154, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (d+e x)^3 \sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\tan ^3(d+e x) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {\int \frac {\tan ^2(d+e x) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}{\tan ^2(d+e x)+1}d\tan ^2(d+e x)}{2 e}\)

\(\Big \downarrow \) 1231

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\int \frac {b^2-4 c b+\left (b^2+4 c b-4 c (a+2 c)\right ) \tan ^2(d+e x)+4 a c}{2 \left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{4 c}}{2 e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\int \frac {b^2-4 c b+\left (b^2+4 c b-4 c (a+2 c)\right ) \tan ^2(d+e x)+4 a c}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 e}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \int \frac {1}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)+8 c (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 e}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {2 \left (-4 c (a+2 c)+b^2+4 b c\right ) \int \frac {1}{4 c-\tan ^4(d+e x)}d\frac {2 c \tan ^2(d+e x)+b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}+8 c (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{8 c}}{2 e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {8 c (a-b+c) \int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)+\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}}{8 c}}{2 e}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-16 c (a-b+c) \int \frac {1}{4 (a-b+c)-\tan ^4(d+e x)}d\frac {(b-2 c) \tan ^2(d+e x)+2 a-b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}}{8 c}}{2 e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{4 c}-\frac {\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {c}}-8 c \sqrt {a-b+c} \text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{8 c}}{2 e}\)

Input:

Int[Tan[d + e*x]^3*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

(-1/8*(-8*c*Sqrt[a - b + c]*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/( 
2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] + ((b^2 
+ 4*b*c - 4*c*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[ 
a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/Sqrt[c])/c + ((b - 4*c + 2*c*T 
an[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(4*c))/(2*e)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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

rule 1092

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

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1231

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4183

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( 
x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] 
 :> Simp[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[n 
2, 2*n] && NeQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.52

method	result	size
derivativedivides	\(\frac {\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}-\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}-\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\)	\(318\)
default	\(\frac {\frac {\left (b +2 c \tan \left (e x +d \right )^{2}\right ) \sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{8 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{16 c^{\frac {3}{2}}}-\frac {\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{2}-\frac {\left (b -2 c \right ) \ln \left (\frac {\frac {b}{2}-c +\left (1+\tan \left (e x +d \right )^{2}\right ) c}{\sqrt {c}}+\sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2}}{e}\)	\(318\)

Input:

int(tan(e*x+d)^3*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/e*(1/8*(b+2*c*tan(e*x+d)^2)/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)+1/ 
16*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*tan(e*x+d)^2)/c^(1/2)+(a+b*tan(e*x+d)^2 
+c*tan(e*x+d)^4)^(1/2))-1/2*(c*(1+tan(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2) 
+a-b+c)^(1/2)-1/4*(b-2*c)*ln((1/2*b-c+(1+tan(e*x+d)^2)*c)/c^(1/2)+(c*(1+ta 
n(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))/c^(1/2)+1/2*(a-b+c)^( 
1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+c)^(1/2)*(c*(1+tan(e* 
x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))/(1+tan(e*x+d)^2)))
 

Fricas [A] (verification not implemented)

Time = 2.16 (sec) , antiderivative size = 1199, normalized size of antiderivative = 5.74 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(e*x+d)^3*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="fricas")
 

Output:

[1/32*(8*sqrt(a - b + c)*c^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + 
d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + 
 d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a 
- b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 
 + 1)) - (b^2 - 4*(a - b)*c - 8*c^2)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8* 
b*c*tan(e*x + d)^2 + b^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a) 
*(2*c*tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c) + 4*sqrt(c*tan(e*x + d)^4 + b*t 
an(e*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/16*(4* 
sqrt(a - b + c)*c^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2* 
(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b 
*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) 
+ 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) + 
(b^2 - 4*(a - b)*c - 8*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b* 
tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 
+ b*c*tan(e*x + d)^2 + a*c)) + 2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 
+ a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/32*(16*sqrt(-a + b - 
 c)*c^2*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2* 
c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x + 
 d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c)) - (b^2 - 4*(a 
 - b)*c - 8*c^2)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^...
 

Sympy [F]

\[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{3}{\left (d + e x \right )}\, dx \] Input:

integrate(tan(e*x+d)**3*(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
 

Output:

Integral(sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)*tan(d + e*x)**3, 
x)
 

Maxima [F]

\[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{3} \,d x } \] Input:

integrate(tan(e*x+d)^3*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="maxima")
 

Output:

integrate(sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*tan(e*x + d)^3, x)
 

Giac [F]

\[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{3} \,d x } \] Input:

integrate(tan(e*x+d)^3*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="giac")
 

Output:

integrate(sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*tan(e*x + d)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx=\int {\mathrm {tan}\left (d+e\,x\right )}^3\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \] Input:

int(tan(d + e*x)^3*(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)
 

Output:

int(tan(d + e*x)^3*(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)
 

Reduce [F]

\[ \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx =\text {Too large to display} \] Input:

int(tan(e*x+d)^3*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
 

Output:

(sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**2*b + 2*sqr 
t(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**2*c + 2*sqrt(ta 
n(d + e*x)**4*c + tan(d + e*x)**2*b + a)*a - 3*sqrt(tan(d + e*x)**4*c + ta 
n(d + e*x)**2*b + a)*b - 4*int((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b 
 + a)*tan(d + e*x)**5)/(tan(d + e*x)**4*b*c + 2*tan(d + e*x)**4*c**2 + tan 
(d + e*x)**2*b**2 + 2*tan(d + e*x)**2*b*c + a*b + 2*a*c),x)*a*b*c*e - 8*in 
t((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**5)/(tan(d 
 + e*x)**4*b*c + 2*tan(d + e*x)**4*c**2 + tan(d + e*x)**2*b**2 + 2*tan(d + 
 e*x)**2*b*c + a*b + 2*a*c),x)*a*c**2*e + int((sqrt(tan(d + e*x)**4*c + ta 
n(d + e*x)**2*b + a)*tan(d + e*x)**5)/(tan(d + e*x)**4*b*c + 2*tan(d + e*x 
)**4*c**2 + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)**2*b*c + a*b + 2*a*c),x) 
*b**3*e + 6*int((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e 
*x)**5)/(tan(d + e*x)**4*b*c + 2*tan(d + e*x)**4*c**2 + tan(d + e*x)**2*b* 
*2 + 2*tan(d + e*x)**2*b*c + a*b + 2*a*c),x)*b**2*c*e - 16*int((sqrt(tan(d 
 + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**5)/(tan(d + e*x)**4*b* 
c + 2*tan(d + e*x)**4*c**2 + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)**2*b*c 
+ a*b + 2*a*c),x)*c**3*e - 4*int((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2 
*b + a)*tan(d + e*x))/(tan(d + e*x)**4*b*c + 2*tan(d + e*x)**4*c**2 + tan( 
d + e*x)**2*b**2 + 2*tan(d + e*x)**2*b*c + a*b + 2*a*c),x)*a*b**2*e - 12*i 
nt((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x))/(tan(...