\(\int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\) [34]

Optimal result

Integrand size = 35, antiderivative size = 182 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 c^{3/2} e}+\frac {\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 c e} \] Output:

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

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\frac {2 \text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}+\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{c^{3/2}}-\frac {2 \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{c}}{4 e} \] Input:

Integrate[Tan[d + e*x]^5/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

-1/4*((2*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*S 
qrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/Sqrt[a - b + c] + ((b + 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])])/c^(3/2) - (2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e 
*x]^4])/c)/e
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.99, 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, 1267, 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.

\(\displaystyle \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (d+e x)^5}{\sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\tan ^5(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1267

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[Tan[d + e*x]^5/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

(-1/2*((2*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])])/Sqrt[a - b + c] + ((b + 
 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 + Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d 
+ e*x]^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 1267

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

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 = 0.22 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{4 c^{\frac {3}{2}}}-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}}{e}\)	\(232\)
default	\(\frac {\frac {\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}}{2 c}-\frac {b \ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{4 c^{\frac {3}{2}}}-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 \sqrt {a -b +c}}-\frac {\ln \left (\frac {\frac {b}{2}+c \tan \left (e x +d \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\right )}{2 \sqrt {c}}}{e}\)	\(232\)

Input:

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/e*(1/2/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)-1/4*b/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/(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+t 
an(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))-1/ 
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 
))/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 1226, normalized size of antiderivative = 6.74 \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Too large to display} \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="fricas")
 

Output:

[1/8*(2*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)) + (a*b - b^2 + (2*a - b)*c + 2*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*tan(e*x + d)^2 + a)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e), 1/4*(s 
qrt(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)) + ( 
a*b - b^2 + (2*a - b)*c + 2*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)*((a - b)*c + c^2))/(((a - b)*c^2 + c^3)*e), -1/8*(4*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)) - (a*b - b^2 
 + (2*a - b)*c + 2*c^2)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*...
 

Sympy [F]

\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\tan ^{5}{\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \] Input:

integrate(tan(e*x+d)**5/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
 

Output:

Integral(tan(d + e*x)**5/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), 
x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int { \frac {\tan \left (e x + d\right )^{5}}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}} \,d x } \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorith 
m="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \] Input:

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)
 

Output:

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^5(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\sqrt {\tan \left (e x +d \right )^{4} c +\tan \left (e x +d \right )^{2} b +a}\, \tan \left (e x +d \right )^{5}}{\tan \left (e x +d \right )^{4} c +\tan \left (e x +d \right )^{2} b +a}d x \] Input:

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
 

Output:

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*c + tan(d + e*x)**2*b + a),x)