\(\int \frac {\tan ^5(d+e x)}{(a+b \tan ^2(d+e x)+c \tan ^4(d+e x))^{3/2}} \, dx\) [44]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}+\frac {a (2 a-b)+((a-b) b+2 a c) \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \] 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)^(3/2)/e+(a*(2*a-b)+((a-b)*b+2*a*c)*tan( 
e*x+d)^2)/(a-b+c)/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 4.77 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=-\frac {\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}+\frac {2 \sqrt {2} \left (2 a^2-b^2+2 a c+\left (2 a^2+b^2-2 a (b+c)\right ) \cos (2 (d+e x))\right ) \sec ^2(d+e x)}{(a-b+c) \left (-b^2+4 a c\right ) \sqrt {(3 a+b+3 c+4 (a-c) \cos (2 (d+e x))+(a-b+c) \cos (4 (d+e x))) \sec ^4(d+e x)}}}{2 e} \] Input:

Integrate[Tan[d + e*x]^5/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x 
]
 

Output:

-1/2*(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])]/(a - b + c)^(3/2) + (2*Sqrt[2] 
*(2*a^2 - b^2 + 2*a*c + (2*a^2 + b^2 - 2*a*(b + c))*Cos[2*(d + e*x)])*Sec[ 
d + e*x]^2)/((a - b + c)*(-b^2 + 4*a*c)*Sqrt[(3*a + b + 3*c + 4*(a - c)*Co 
s[2*(d + e*x)] + (a - b + c)*Cos[4*(d + e*x)])*Sec[d + e*x]^4]))/e
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4183, 1578, 1264, 27, 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.

\(\displaystyle \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (d+e x)^5}{\left (a+b \tan (d+e x)^2+c \tan (d+e x)^4\right )^{3/2}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\tan ^5(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^4(d+e x)+b \tan ^2(d+e x)+a\right )^{3/2}}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1264

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 \left ((b (a-b)+2 a c) \tan ^2(d+e x)+a (2 a-b)\right )}{(a-b+c) \left (b^2-4 a c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}}{2 e}\)

Input:

Int[Tan[d + e*x]^5/(a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4)^(3/2),x]
 

Output:

(-(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])]/(a - b + c)^(3/2)) + (2*(a*(2*a - 
 b) + ((a - b)*b + 2*a*c)*Tan[d + e*x]^2))/((a - b + c)*(b^2 - 4*a*c)*Sqrt 
[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]))/(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 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 1264

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d + e*x) 
^m*(f + g*x)^n, a + b*x + c*x^2, x], R = Coeff[PolynomialRemainder[(d + e*x 
)^m*(f + g*x)^n, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[ 
(d + e*x)^m*(f + g*x)^n, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + ( 
2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + S 
imp[1/((p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*E 
xpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*R - b*S) 
)/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[n, 1] 
 && LtQ[p, -1] && ILtQ[m, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(508\) vs. \(2(147)=294\).

Time = 0.23 (sec) , antiderivative size = 509, normalized size of antiderivative = 3.20

method	result	size
derivativedivides	\(\frac {-\frac {2 a +b \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}+\frac {2 c \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \sqrt {a -b +c}}-\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}+\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}+\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}-\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}-\frac {b +2 c \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}}{e}\)	\(509\)
default	\(\frac {-\frac {2 a +b \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}+\frac {2 c \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \sqrt {a -b +c}}-\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}+\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}-b +2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}+\frac {2 c \sqrt {c \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2}-\sqrt {-4 a c +b^{2}}\, \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}{\left (\sqrt {-4 a c +b^{2}}+b -2 c \right ) \left (-4 a c +b^{2}\right ) \left (\tan \left (e x +d \right )^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}-\frac {b +2 c \tan \left (e x +d \right )^{2}}{\sqrt {a +b \tan \left (e x +d \right )^{2}+c \tan \left (e x +d \right )^{4}}\, \left (4 a c -b^{2}\right )}}{e}\)	\(509\)

Input:

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (147) = 294\).

Time = 0.47 (sec) , antiderivative size = 1095, normalized size of antiderivative = 6.89 \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorith 
m="fricas")
 

Output:

[-1/4*(((b^2*c - 4*a*c^2)*tan(e*x + d)^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b* 
c)*tan(e*x + d)^2)*sqrt(a - b + c)*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)*s 
qrt(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)) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*a^3 - 3* 
a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)* 
tan(e*x + d)^2 + (2*a^2 - a*b)*c))/((4*a*c^4 + (8*a^2 - 8*a*b - b^2)*c^3 + 
 2*(2*a^3 - 4*a^2*b + a*b^2 + b^3)*c^2 - (a^2*b^2 - 2*a*b^3 + b^4)*c)*e*ta 
n(e*x + d)^4 - (a^2*b^3 - 2*a*b^4 + b^5 - 4*a*b*c^3 - (8*a^2*b - 8*a*b^2 - 
 b^3)*c^2 - 2*(2*a^3*b - 4*a^2*b^2 + a*b^3 + b^4)*c)*e*tan(e*x + d)^2 - (a 
^3*b^2 - 2*a^2*b^3 + a*b^4 - 4*a^2*c^3 - (8*a^3 - 8*a^2*b - a*b^2)*c^2 - 2 
*(2*a^4 - 4*a^3*b + a^2*b^2 + a*b^3)*c)*e), 1/2*(((b^2*c - 4*a*c^2)*tan(e* 
x + d)^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*tan(e*x + d)^2)*sqrt(-a + b - 
 c)*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*t 
an(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)) - 2*sqrt(c*tan(e* 
x + d)^4 + b*tan(e*x + d)^2 + a)*(2*a^3 - 3*a^2*b + a*b^2 + (a^2*b - 2*a*b 
^2 + b^3 + 2*a*c^2 + (2*a^2 - a*b - b^2)*c)*tan(e*x + d)^2 + (2*a^2 - a*b) 
*c))/((4*a*c^4 + (8*a^2 - 8*a*b - b^2)*c^3 + 2*(2*a^3 - 4*a^2*b + a*b^2...
 

Sympy [F]

\[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (d + e x \right )}}{\left (a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(e*x+d)**5/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(3/2),x)
 

Output:

Integral(tan(d + e*x)**5/(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)**(3/2 
), x)
 

Maxima [F]

\[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (e x + d\right )^{5}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorith 
m="maxima")
 

Output:

integrate(tan(e*x + d)^5/(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)^(3/2), 
x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x, algorith 
m="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^5}{{\left (c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int(tan(d + e*x)^5/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^5(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(tan(e*x+d)^5/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(3/2),x)
 

Output:

( - sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**2*b - 2* 
sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*a - 4*int((sqrt(tan(d + e* 
x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e*x)**8*c**2 + 
2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)**4*b**2 + 2*t 
an(d + e*x)**2*a*b + a**2),x)*tan(d + e*x)**4*a*c**2*e + int((sqrt(tan(d + 
 e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e*x)**8*c**2 
 + 2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)**4*b**2 + 
2*tan(d + e*x)**2*a*b + a**2),x)*tan(d + e*x)**4*b**2*c*e - 4*int((sqrt(ta 
n(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e*x)**8 
*c**2 + 2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)**4*b* 
*2 + 2*tan(d + e*x)**2*a*b + a**2),x)*tan(d + e*x)**2*a*b*c*e + int((sqrt( 
tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e*x)* 
*8*c**2 + 2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)**4* 
b**2 + 2*tan(d + e*x)**2*a*b + a**2),x)*tan(d + e*x)**2*b**3*e - 4*int((sq 
rt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e* 
x)**8*c**2 + 2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)* 
*4*b**2 + 2*tan(d + e*x)**2*a*b + a**2),x)*a**2*c*e + int((sqrt(tan(d + e* 
x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x)**3)/(tan(d + e*x)**8*c**2 + 
2*tan(d + e*x)**6*b*c + 2*tan(d + e*x)**4*a*c + tan(d + e*x)**4*b**2 + 2*t 
an(d + e*x)**2*a*b + a**2),x)*a*b**2*e)/(e*(4*tan(d + e*x)**4*a*c**2 - ...