3.1.19 \(\int \frac {\tan ^7(d+e x)}{(a+b \tan (d+e x)+c \tan ^2(d+e x))^{3/2}} \, dx\) [19]

3.1.19.1 Optimal result

Integrand size = 33, antiderivative size = 1190 \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\frac {3 b \text {arctanh}\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 )}{2 c^{5/2} e}-\frac {5 b \left (7 b^2-12 a c\right ) \text {arctanh}\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 )}{16 c^{9/2} e}-\frac {\sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(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} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {\sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(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} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \tan ^2(d+e x) (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}+\frac {2 \tan ^4(d+e x) (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \left (a \left (b^2-2 (a-c) c\right )+b c (a+c) \tan (d+e x)\right )}{\left (b^2+(a-c)^2\right ) \left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}+\frac {\left (7 b^2-16 a c\right ) \tan ^2(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{3 c^2 \left (b^2-4 a c\right ) e}-\frac {2 b \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c \left (b^2-4 a c\right ) e}-\frac {\left (3 b^2-8 a c-2 b c \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{c^2 \left (b^2-4 a c\right ) e}+\frac {\left (105 b^4-460 a b^2 c+256 a^2 c^2-2 b c \left (35 b^2-116 a c\right ) \tan (d+e x)\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{24 c^4 \left (b^2-4 a c\right ) e} \]

output

3/2*b*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-5/16*b*(-12*a*c+7*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^(9/2)/e+1/2*arctanh(1/2*( 
b^2-(a-c)*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))-b*(2*a-2*c+(a^2-2*a*c+b^2+c^2)^( 
1/2))*tan(e*x+d))*2^(1/2)/(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-b 
^2-2*a*c+c^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))*(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)*(a^2-b^2-2*a*c+ 
c^2-(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-2*a*c+b^2+c^2)^(3/2)/e*2^( 
1/2)-1/2*arctanh(1/2*(b^2-(a-c)*(a-c+(a^2-2*a*c+b^2+c^2)^(1/2))-b*(2*a-2*c 
-(a^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(2*a-2*c-(a^2-2*a*c+b^2+c^ 
2)^(1/2))^(1/2)/(a^2-b^2-2*a*c+c^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))*(2*a-2*c-(a^2-2*a*c+b^2+c^2)^(1/2)) 
^(1/2)*(a^2-b^2-2*a*c+c^2+(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-2*a* 
c+b^2+c^2)^(3/2)/e*2^(1/2)+1/3*(-16*a*c+7*b^2)*(a+b*tan(e*x+d)+c*tan(e*x+d 
)^2)^(1/2)*tan(e*x+d)^2/c^2/(-4*a*c+b^2)/e-2*b*(a+b*tan(e*x+d)+c*tan(e*x+d 
)^2)^(1/2)*tan(e*x+d)^3/c/(-4*a*c+b^2)/e+2*(2*a+b*tan(e*x+d))/(-4*a*c+b^2) 
/e/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)-2*tan(e*x+d)^2*(2*a+b*tan(e*x+d)) 
/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)+2*tan(e*x+d)^4*(2*a+ 
b*tan(e*x+d))/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)-(a+b*ta 
n(e*x+d)+c*tan(e*x+d)^2)^(1/2)*(3*b^2-8*a*c-2*b*c*tan(e*x+d))/c^2/(-4*a...
 

3.1.19.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 11.26 (sec) , antiderivative size = 2476, normalized size of antiderivative = 2.08 \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Result too large to show} \]

input

Integrate[Tan[d + e*x]^7/(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^(3/2),x]
 

output

(((-35*a^2*b^3 - 35*b^5 + 60*a^3*b*c + 130*a*b^3*c - 96*a^2*b*c^2 - 11*b^3 
*c^2 + 12*a*b*c^3 + 16*b*c^4)*ArcTanh[(2*Sqrt[c]*Tan[(d + e*x)/2])/(Sqrt[a 
]*(-1 + Tan[(d + e*x)/2]^2) - Sqrt[a*(-1 + Tan[(d + e*x)/2]^2)^2 + 2*Tan[( 
d + e*x)/2]*(b + 2*c*Tan[(d + e*x)/2] - b*Tan[(d + e*x)/2]^2)])]*(1 + Cos[ 
d + e*x])*Sqrt[(1 + Cos[2*(d + e*x)])/(1 + Cos[d + e*x])^2]*Sqrt[(a + c + 
(a - c)*Cos[2*(d + e*x)] + b*Sin[2*(d + e*x)])/(1 + Cos[2*(d + e*x)])]*(-1 
 + Tan[(d + e*x)/2]^2)*(1 + Tan[(d + e*x)/2]^2)*Sqrt[(a*(-1 + Tan[(d + e*x 
)/2]^2)^2 + 2*Tan[(d + e*x)/2]*(b + 2*c*Tan[(d + e*x)/2] - b*Tan[(d + e*x) 
/2]^2))/(1 + Tan[(d + e*x)/2]^2)^2])/(Sqrt[c]*Sqrt[a + c + (a - c)*Cos[2*( 
d + e*x)] + b*Sin[2*(d + e*x)]]*Sqrt[(-1 + Tan[(d + e*x)/2]^2)^2]*Sqrt[a*( 
-1 + Tan[(d + e*x)/2]^2)^2 + 2*Tan[(d + e*x)/2]*(b + 2*c*Tan[(d + e*x)/2] 
- b*Tan[(d + e*x)/2]^2)]) + ((-8*a*c^4 + 8*c^5)*(1 + Cos[d + e*x])*Sqrt[(1 
 + Cos[2*(d + e*x)])/(1 + Cos[d + e*x])^2]*RootSum[a^2 + b^2 + 4*b*Sqrt[c] 
*#1 - 2*a*#1^2 + 4*c*#1^2 + #1^4 & , (-(a*Log[-1 + Tan[(d + e*x)/2]^2]) + 
a*Log[#1 - 2*Sqrt[c]*Tan[(d + e*x)/2] - #1*Tan[(d + e*x)/2]^2 + Sqrt[a + 2 
*b*Tan[(d + e*x)/2] + (-2*a + 4*c)*Tan[(d + e*x)/2]^2 - 2*b*Tan[(d + e*x)/ 
2]^3 + a*Tan[(d + e*x)/2]^4]] + Log[-1 + Tan[(d + e*x)/2]^2]*#1^2 - Log[#1 
 - 2*Sqrt[c]*Tan[(d + e*x)/2] - #1*Tan[(d + e*x)/2]^2 + Sqrt[a + 2*b*Tan[( 
d + e*x)/2] + (-2*a + 4*c)*Tan[(d + e*x)/2]^2 - 2*b*Tan[(d + e*x)/2]^3 + a 
*Tan[(d + e*x)/2]^4]]*#1^2)/(-(b*Sqrt[c]) + a*#1 - 2*c*#1 - #1^3) & ]*S...
 

3.1.19.3 Rubi [A] (verified)

Time = 5.57 (sec) , antiderivative size = 1158, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 4183, 7276, 2009}

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.

\(\displaystyle \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (d+e x)^7}{\left (a+b \tan (d+e x)+c \tan (d+e x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\tan ^7(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^2(d+e x)+b \tan (d+e x)+a\right )^{3/2}}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 (2 a+b \tan (d+e x)) \tan ^4(d+e x)}{\left (b^2-4 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}-\frac {2 b \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a} \tan ^3(d+e x)}{c \left (b^2-4 a c\right )}+\frac {\left (7 b^2-16 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a} \tan ^2(d+e x)}{3 c^2 \left (b^2-4 a c\right )}-\frac {2 (2 a+b \tan (d+e x)) \tan ^2(d+e x)}{\left (b^2-4 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}-\frac {5 b \left (7 b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}\right )}{16 c^{9/2}}+\frac {3 b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}\right )}{2 c^{5/2}}-\frac {\sqrt {2 a-2 c-\sqrt {a^2-2 c a+b^2+c^2}} \sqrt {a^2-2 c a-b^2+c^2+(a-c) \sqrt {a^2-2 c a+b^2+c^2}} \text {arctanh}\left (\frac {b^2-\left (2 a-2 c-\sqrt {a^2-2 c a+b^2+c^2}\right ) \tan (d+e x) b-(a-c) \left (a-c+\sqrt {a^2-2 c a+b^2+c^2}\right )}{\sqrt {2} \sqrt {2 a-2 c-\sqrt {a^2-2 c a+b^2+c^2}} \sqrt {a^2-2 c a-b^2+c^2+(a-c) \sqrt {a^2-2 c a+b^2+c^2}} \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}\right )}{\sqrt {2} \left (a^2-2 c a+b^2+c^2\right )^{3/2}}+\frac {\sqrt {2 a-2 c+\sqrt {a^2-2 c a+b^2+c^2}} \sqrt {a^2-2 c a-b^2+c^2-(a-c) \sqrt {a^2-2 c a+b^2+c^2}} \text {arctanh}\left (\frac {b^2-\left (2 a-2 c+\sqrt {a^2-2 c a+b^2+c^2}\right ) \tan (d+e x) b-(a-c) \left (a-c-\sqrt {a^2-2 c a+b^2+c^2}\right )}{\sqrt {2} \sqrt {2 a-2 c+\sqrt {a^2-2 c a+b^2+c^2}} \sqrt {a^2-2 c a-b^2+c^2-(a-c) \sqrt {a^2-2 c a+b^2+c^2}} \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}\right )}{\sqrt {2} \left (a^2-2 c a+b^2+c^2\right )^{3/2}}-\frac {\left (3 b^2-2 c \tan (d+e x) b-8 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}{c^2 \left (b^2-4 a c\right )}+\frac {\left (105 b^4-460 a c b^2-2 c \left (35 b^2-116 a c\right ) \tan (d+e x) b+256 a^2 c^2\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}{24 c^4 \left (b^2-4 a c\right )}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}-\frac {2 \left (a \left (b^2-2 (a-c) c\right )+b c (a+c) \tan (d+e x)\right )}{\left (b^2+(a-c)^2\right ) \left (b^2-4 a c\right ) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}}{e}\)

input

Int[Tan[d + e*x]^7/(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^(3/2),x]
 

output

((3*b*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + 
c*Tan[d + e*x]^2])])/(2*c^(5/2)) - (5*b*(7*b^2 - 12*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])])/(1 
6*c^(9/2)) - (Sqrt[2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b 
^2 - 2*a*c + c^2 + (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(b^2 - ( 
a - c)*(a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a - 2*c - Sqrt[a^2 + 
 b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2*c - Sqrt[a^2 + b^ 
2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 + (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]*(a^2 
 + b^2 - 2*a*c + c^2)^(3/2)) + (Sqrt[2*a - 2*c + Sqrt[a^2 + b^2 - 2*a*c + 
c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 - (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2] 
]*ArcTanh[(b^2 - (a - c)*(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a 
- 2*c + Sqrt[a^2 + b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2 
*c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 - (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]*(a^2 + b^2 - 2*a*c + c^2)^(3/2)) + (2*(2*a + b*Tan[d + e*x]) 
)/((b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]) - (2*Tan[d + 
 e*x]^2*(2*a + b*Tan[d + e*x]))/((b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e*x] + c 
*Tan[d + e*x]^2]) + (2*Tan[d + e*x]^4*(2*a + b*Tan[d + e*x]))/((b^2 - 4*a* 
c)*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]) - (2*(a*(b^2 - 2*(a - c...
 

3.1.19.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]
 

rule 7276

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

3.1.19.4 Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 2.11 (sec) , antiderivative size = 13068421, normalized size of antiderivative = 10981.87

\[\text {output too large to display}\]

input

int(tan(e*x+d)^7/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x)
 

output

result too large to display
 

3.1.19.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20284 vs. \(2 (1096) = 2192\).

Time = 9.01 (sec) , antiderivative size = 40569, normalized size of antiderivative = 34.09 \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \]

input

integrate(tan(e*x+d)^7/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"fricas")
 

output

Too large to include
 

3.1.19.6 Sympy [F]

\[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {\tan ^{7}{\left (d + e x \right )}}{\left (a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \]

input

integrate(tan(e*x+d)**7/(a+b*tan(e*x+d)+c*tan(e*x+d)**2)**(3/2),x)
 

output

Integral(tan(d + e*x)**7/(a + b*tan(d + e*x) + c*tan(d + e*x)**2)**(3/2), 
x)
 

3.1.19.7 Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \]

input

integrate(tan(e*x+d)^7/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"maxima")
 

output

Timed out
 

3.1.19.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \]

input

integrate(tan(e*x+d)^7/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"giac")
 

output

Timed out
 

3.1.19.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^7(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Hanged} \]

input

int(tan(d + e*x)^7/(a + b*tan(d + e*x) + c*tan(d + e*x)^2)^(3/2),x)
 

output

\text{Hanged}