3.1.99 \(\int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\) [99]

3.1.99.1 Optimal result

Integrand size = 45, antiderivative size = 273 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(i a+b) (A-i B-C) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (A+i B-C) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {2 (2 b c C-7 b B d-7 a C d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f} \]

output

-(I*a+b)*(A-I*B-C)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1 
/2))/f+(I*a-b)*(A+I*B-C)*(c+I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I 
*d)^(1/2))/f+2*(A*a*d+A*b*c+B*a*c-B*b*d-C*a*d-C*b*c)*(c+d*tan(f*x+e))^(1/2 
)/f+2/3*(A*b+B*a-C*b)*(c+d*tan(f*x+e))^(3/2)/f-2/35*(-7*B*b*d-7*C*a*d+2*C* 
b*c)*(c+d*tan(f*x+e))^(5/2)/d^2/f+2/7*b*C*tan(f*x+e)*(c+d*tan(f*x+e))^(5/2 
)/d/f
 

3.1.99.2 Mathematica [A] (verified)

Time = 4.84 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.95 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\frac {2 (-2 b c C+7 b B d+7 a C d) (c+d \tan (e+f x))^{5/2}}{d}+10 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}+\frac {35}{3} (i a+b) (A-i B-C) d \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )+\frac {35}{3} (-i a+b) (A+i B-C) d \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )}{35 d f} \]

input

Integrate[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f 
*x] + C*Tan[e + f*x]^2),x]
 

output

((2*(-2*b*c*C + 7*b*B*d + 7*a*C*d)*(c + d*Tan[e + f*x])^(5/2))/d + 10*b*C* 
Tan[e + f*x]*(c + d*Tan[e + f*x])^(5/2) + (35*(I*a + b)*(A - I*B - C)*d*(- 
3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c 
 + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/3 + (35*((-I)*a + b) 
*(A + I*B - C)*d*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt 
[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3 
)/(35*d*f)
 

3.1.99.3 Rubi [A] (warning: unable to verify)

Time = 1.67 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4120, 27, 3042, 4113, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {2 \int \frac {1}{2} (c+d \tan (e+f x))^{3/2} \left ((2 b c C-7 a d C-7 b B d) \tan ^2(e+f x)-7 (A b-C b+a B) d \tan (e+f x)+2 b c C-7 a A d\right )dx}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left ((2 b c C-7 a d C-7 b B d) \tan ^2(e+f x)-7 (A b-C b+a B) d \tan (e+f x)+2 b c C-7 a A d\right )dx}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left ((2 b c C-7 a d C-7 b B d) \tan (e+f x)^2-7 (A b-C b+a B) d \tan (e+f x)+2 b c C-7 a A d\right )dx}{7 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} (7 (b B-a (A-C)) d-7 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} (7 (b B-a (A-C)) d-7 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} (7 d (b B c+b (A-C) d-a (A c-C c-B d))-7 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x))dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} (7 d (b B c+b (A-C) d-a (A c-C c-B d))-7 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x))dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \frac {7 d \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )-7 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \frac {7 d \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )-7 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7}{2} d (a+i b) (c+i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {7}{2} d (a-i b) (c-i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7}{2} d (a+i b) (c+i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {7}{2} d (a-i b) (c-i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7 i d (a-i b) (c-i d)^2 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {7 i d (a+i b) (c+i d)^2 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\frac {7 i d (a-i b) (c-i d)^2 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {7 i d (a+i b) (c+i d)^2 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7 (a+i b) (c+i d)^2 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {7 (a-i b) (c-i d)^2 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7 d (a-i b) (c-i d)^{3/2} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {7 d (a+i b) (c+i d)^{3/2} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}-\frac {14 d (a B+A b-b C) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \sqrt {c+d \tan (e+f x)} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-7 a C d-7 b B d+2 b c C) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

input

Int[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x] + 
C*Tan[e + f*x]^2),x]
 

output

(2*b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])^(5/2))/(7*d*f) - ((-7*(a - I*b)*( 
A - I*B - C)*(c - I*d)^(3/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (7* 
(a + I*b)*(A + I*B - C)*(c + I*d)^(3/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d 
]])/f - (14*d*(A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d)*Sqrt[c + d*T 
an[e + f*x]])/f - (14*(A*b + a*B - b*C)*d*(c + d*Tan[e + f*x])^(3/2))/(3*f 
) + (2*(2*b*c*C - 7*b*B*d - 7*a*C*d)*(c + d*Tan[e + f*x])^(5/2))/(5*d*f))/ 
(7*d)
 

3.1.99.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 3042

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

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4120

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 

3.1.99.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5106\) vs. \(2(239)=478\).

Time = 0.19 (sec) , antiderivative size = 5107, normalized size of antiderivative = 18.71

method	result	size
parts	\(\text {Expression too large to display}\)	\(5107\)
derivativedivides	\(\text {Expression too large to display}\)	\(5149\)
default	\(\text {Expression too large to display}\)	\(5149\)

input

int((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2 
),x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.1.99.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31081 vs. \(2 (232) = 464\).

Time = 16.16 (sec) , antiderivative size = 31081, normalized size of antiderivative = 113.85 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.1.99.6 Sympy [F]

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

input

integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e)+C*tan(f 
*x+e)**2),x)
 

output

Integral((a + b*tan(e + f*x))*(c + d*tan(e + f*x))**(3/2)*(A + B*tan(e + f 
*x) + C*tan(e + f*x)**2), x)
 

3.1.99.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.1.99.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.1.99.9 Mupad [F(-1)]

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

input

int((a + b*tan(e + f*x))*(c + d*tan(e + f*x))^(3/2)*(A + B*tan(e + f*x) + 
C*tan(e + f*x)^2),x)
 

output

\text{Hanged}