\(\int \frac {(a+b \tan (e+f x))^2 (A+B \tan (e+f x)+C \tan ^2(e+f x))}{\sqrt {c+d \tan (e+f x)}} \, dx\) [111]

Optimal result
Mathematica [A] (verified)
Rubi [A] (warning: unable to verify)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-1)]
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 47, antiderivative size = 287 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(a-i b)^2 (B+i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {(a+i b)^2 (i A-B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}+\frac {2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^3 f}-\frac {2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f} \] Output:

-(a-I*b)^2*(B+I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I* 
d)^(1/2)/f+(a+I*b)^2*(I*A-B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1 
/2))/(c+I*d)^(1/2)/f+2/15*(12*a^2*C*d^2-10*a*b*d*(-3*B*d+2*C*c)+b^2*(8*c^2 
*C-10*B*c*d+15*(A-C)*d^2))*(c+d*tan(f*x+e))^(1/2)/d^3/f-2/15*b*(-5*B*b*d-4 
*C*a*d+4*C*b*c)*tan(f*x+e)*(c+d*tan(f*x+e))^(1/2)/d^2/f+2/5*C*(a+b*tan(f*x 
+e))^2*(c+d*tan(f*x+e))^(1/2)/d/f
 

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {15 (a-i b)^2 (i A+B-i C) d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {15 i (a+i b)^2 (A+i B-C) d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 \left (12 a^2 C d^2+10 a b d (-2 c C+3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d^2}+\frac {2 b (-4 b c C+5 b B d+4 a C d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{d}+6 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{15 d f} \] Input:

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

Output:

((-15*(a - I*b)^2*(I*A + B - I*C)*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[ 
c - I*d]])/Sqrt[c - I*d] + ((15*I)*(a + I*b)^2*(A + I*B - C)*d*ArcTanh[Sqr 
t[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d] + (2*(12*a^2*C*d^2 + 1 
0*a*b*d*(-2*c*C + 3*B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt 
[c + d*Tan[e + f*x]])/d^2 + (2*b*(-4*b*c*C + 5*b*B*d + 4*a*C*d)*Tan[e + f* 
x]*Sqrt[c + d*Tan[e + f*x]])/d + 6*C*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan 
[e + f*x]])/(15*d*f)
 

Rubi [A] (warning: unable to verify)

Time = 2.93 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.319, Rules used = {3042, 4130, 27, 3042, 4120, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 \int -\frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{5 d}+\frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan (e+f x)^2-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \int -\frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan ^2(e+f x)-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan ^2(e+f x)-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan (e+f x)^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {15 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {15 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15}{2} d^2 (a+i b)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {15}{2} d^2 (a-i b)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15}{2} d^2 (a+i b)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {15}{2} d^2 (a-i b)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15 i d^2 (a-i b)^2 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {15 i d^2 (a+i b)^2 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {15 i d^2 (a-i b)^2 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {15 i d^2 (a+i b)^2 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15 d (a-i b)^2 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {15 d (a+i b)^2 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}-\frac {15 d^2 (a-i b)^2 (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}-\frac {15 d^2 (a+i b)^2 (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{3 d}}{5 d}\)

Input:

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

Output:

(2*C*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) - ((2*b*(4*b 
*c*C - 5*b*B*d - 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + 
 ((-15*(a - I*b)^2*(A - I*B - C)*d^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/( 
Sqrt[c - I*d]*f) - (15*(a + I*b)^2*(A + I*B - C)*d^2*ArcTan[Tan[e + f*x]/S 
qrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*(12*a^2*C*d^2 - 10*a*b*d*(2*c*C - 3* 
B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]] 
)/(d*f))/(3*d))/(5*d)
 

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 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]
 

rule 4130
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*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[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5512\) vs. \(2(254)=508\).

Time = 0.19 (sec) , antiderivative size = 5513, normalized size of antiderivative = 19.21

method result size
parts \(\text {Expression too large to display}\) \(5513\)
derivativedivides \(\text {Expression too large to display}\) \(18289\)
default \(\text {Expression too large to display}\) \(18289\)

Input:

int((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1 
/2),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25627 vs. \(2 (244) = 488\).

Time = 4.71 (sec) , antiderivative size = 25627, normalized size of antiderivative = 89.29 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ 
e))^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate((a+b*tan(f*x+e))**2*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*tan(f* 
x+e))**(1/2),x)
 

Output:

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

Maxima [F(-1)]

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

integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ 
e))^(1/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ 
e))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,19,7]%%%}+%%%{8,[0,17,7]%%%}+%%%{28,[0,15,7]%%%}+ 
%%%{56,[0
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 38.71 (sec) , antiderivative size = 21254, normalized size of antiderivative = 74.06 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

int(((a + b*tan(e + f*x))^2*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + 
d*tan(e + f*x))^(1/2),x)
 

Output:

atan(((((16*(2*C*b^2*d^3*f^2 - 2*C*a^2*d^3*f^2 + 4*C*a*b*c*d^2*f^2))/f^3 - 
 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 
 - 32*C^2*a*b^3*d*f^2 + 32*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - ( 
16*c^2*f^4 + 16*d^2*f^4)*(C^4*a^8 + C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^ 
4 + 4*C^4*a^6*b^2))^(1/2) - 4*C^2*a^4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b 
^3*d*f^2 - 16*C^2*a^3*b*d*f^2 + 24*C^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f 
^4)))^(1/2))*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 
32*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^4) 
*(C^4*a^8 + C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^4 + 4*C^4*a^6*b^2))^(1/2 
) - 4*C^2*a^4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b^3*d*f^2 - 16*C^2*a^3*b* 
d*f^2 + 24*C^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f^4)))^(1/2) - (16*(c + d 
*tan(e + f*x))^(1/2)*(C^2*a^4*d^2 + C^2*b^4*d^2 - 6*C^2*a^2*b^2*d^2))/f^2) 
*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 32*C^2*a^3*b 
*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^4)*(C^4*a^8 + 
C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^4 + 4*C^4*a^6*b^2))^(1/2) - 4*C^2*a^ 
4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b^3*d*f^2 - 16*C^2*a^3*b*d*f^2 + 24*C 
^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f^4)))^(1/2)*1i - (((16*(2*C*b^2*d^3* 
f^2 - 2*C*a^2*d^3*f^2 + 4*C*a*b*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + 
f*x))^(1/2)*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 3 
2*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^...
 

Reduce [F]

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

int((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1 
/2),x)
 

Output:

(2*sqrt(tan(e + f*x)*d + c)*a**3 + int((sqrt(tan(e + f*x)*d + c)*tan(e + f 
*x)**4)/(tan(e + f*x)*d + c),x)*b**2*c*d*f + 2*int((sqrt(tan(e + f*x)*d + 
c)*tan(e + f*x)**3)/(tan(e + f*x)*d + c),x)*a*b*c*d*f + int((sqrt(tan(e + 
f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)*d + c),x)*b**3*d*f - int((sqrt( 
tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*d + c),x)*a**3*d*f + in 
t((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*d + c),x)*a**2* 
c*d*f + 3*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*d + 
 c),x)*a*b**2*d*f + 3*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + 
 f*x)*d + c),x)*a**2*b*d*f)/(d*f)