\(\int \frac {\sqrt {c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{a+b \tan (e+f x)} \, dx\) [94]

Optimal result

Integrand size = 47, antiderivative size = 234 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=-\frac {(i A+B-i C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b) f}+\frac {(i A-B-i C) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b) f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{b f} \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\frac {b^{3/2} (-i a+b) (A-i B-C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+b^{3/2} (i a+b) (A+i B-C) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 \left (A b^2+a (-b B+a C)\right ) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )+2 \sqrt {b} \left (a^2+b^2\right ) C \sqrt {c+d \tan (e+f x)}}{b^{3/2} \left (a^2+b^2\right ) f} \] Input:

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

Output:

(b^(3/2)*((-I)*a + b)*(A - I*B - C)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e 
 + f*x]]/Sqrt[c - I*d]] + b^(3/2)*(I*a + b)*(A + I*B - C)*Sqrt[c + I*d]*Ar 
cTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] - 2*(A*b^2 + a*(-(b*B) + a*C 
))*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a 
*d]] + 2*Sqrt[b]*(a^2 + b^2)*C*Sqrt[c + d*Tan[e + f*x]])/(b^(3/2)*(a^2 + b 
^2)*f)
 

Rubi [A] (warning: unable to verify)

Time = 3.00 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.96, 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, 4136, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 C \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {(b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {b (a+i b) \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 C \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {(b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {b (a+i b) \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 C \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {2 (b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}+\frac {\frac {b (a+i b) \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 C \sqrt {c+d \tan (e+f x)}}{b f}+\frac {-\frac {2 \sqrt {b c-a d} \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} f \left (a^2+b^2\right )}+\frac {\frac {b (a+i b) \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

Input:

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

Output:

((((a + I*b)*b*(A - I*B - C)*Sqrt[c - I*d]*ArcTan[Tan[e + f*x]/Sqrt[c - I* 
d]])/f + ((a - I*b)*b*(A + I*B - C)*Sqrt[c + I*d]*ArcTan[Tan[e + f*x]/Sqrt 
[c + I*d]])/f)/(a^2 + b^2) - (2*(A*b^2 - a*(b*B - a*C))*Sqrt[b*c - a*d]*Ar 
cTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + 
 b^2)*f))/b + (2*C*Sqrt[c + d*Tan[e + f*x]])/(b*f)
 

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 4117

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

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])))
 

rule 4136

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3575\) vs. \(2(200)=400\).

Time = 0.21 (sec) , antiderivative size = 3576, normalized size of antiderivative = 15.28

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3576\)
default	\(\text {Expression too large to display}\)	\(3576\)

Input:

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

Output:

-2/f*b/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d 
-b*c)*b)^(1/2))*A*a*d+1/4/f/(a^2+b^2)/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e)) 
^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*C*(2*(c^2+d^2)^(1/2) 
+2*c)^(1/2)*(c^2+d^2)^(1/2)*a-1/4/f/(a^2+b^2)/d*ln(d*tan(f*x+e)+c+(c+d*tan 
(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*C*(2*(c^2+d^ 
2)^(1/2)+2*c)^(1/2)*a*c+1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e)-c+(c+d*tan(f*x+ 
e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))*A*(2*(c^2+d^2)^(1 
/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a-1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e)-c+(c+d 
*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))*A*(2*(c^ 
2+d^2)^(1/2)+2*c)^(1/2)*a*c+1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e)-c+(c+d*tan( 
f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))*B*(2*(c^2+d^2 
)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b-1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e)-c+ 
(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))*B*(2 
*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c-1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e)-c+(c+d* 
tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))*C*(2*(c^2 
+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+1/4/f/(a^2+b^2)/d*ln(-d*tan(f*x+e 
)-c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-(c^2+d^2)^(1/2))* 
C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-2/f*b/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*ar 
ctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2))*B*a*c-2/f/b/(a^2+b^2)/( 
(a*d-b*c)*b)^(1/2)*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2))...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 30.40 (sec) , antiderivative size = 62245, normalized size of antiderivative = 266.00 \[ \int \frac {\sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((((32*(4*C*a*b^8*d^11*f^4 - 4*C*b^9*c*d^10*f^4 + 8*C*a^3*b^6*d^11* 
f^4 + 4*C*a^5*b^4*d^11*f^4 - 4*C*b^9*c^3*d^8*f^4 + 4*C*a*b^8*c^2*d^9*f^4 - 
 8*C*a^2*b^7*c*d^10*f^4 - 4*C*a^4*b^5*c*d^10*f^4 - 8*C*a^2*b^7*c^3*d^8*f^4 
 + 8*C*a^3*b^6*c^2*d^9*f^4 - 4*C*a^4*b^5*c^3*d^8*f^4 + 4*C*a^5*b^4*c^2*d^9 
*f^4))/(b*f^5) - (32*(c + d*tan(e + f*x))^(1/2)*((((8*C^2*a^2*c*f^2 - 8*C^ 
2*b^2*c*f^2 + 16*C^2*a*b*d*f^2)^2/4 - (C^4*c^2 + C^4*d^2)*(16*a^4*f^4 + 16 
*b^4*f^4 + 32*a^2*b^2*f^4))^(1/2) - 4*C^2*a^2*c*f^2 + 4*C^2*b^2*c*f^2 - 8* 
C^2*a*b*d*f^2)/(16*(a^4*f^4 + b^4*f^4 + 2*a^2*b^2*f^4)))^(1/2)*(16*b^10*d^ 
10*f^4 + 16*a^2*b^8*d^10*f^4 - 16*a^4*b^6*d^10*f^4 - 16*a^6*b^4*d^10*f^4 + 
 24*b^10*c^2*d^8*f^4 + 40*a^2*b^8*c^2*d^8*f^4 + 8*a^4*b^6*c^2*d^8*f^4 - 8* 
a^6*b^4*c^2*d^8*f^4 + 8*a*b^9*c*d^9*f^4 + 24*a^3*b^7*c*d^9*f^4 + 24*a^5*b^ 
5*c*d^9*f^4 + 8*a^7*b^3*c*d^9*f^4))/(b*f^4))*((((8*C^2*a^2*c*f^2 - 8*C^2*b 
^2*c*f^2 + 16*C^2*a*b*d*f^2)^2/4 - (C^4*c^2 + C^4*d^2)*(16*a^4*f^4 + 16*b^ 
4*f^4 + 32*a^2*b^2*f^4))^(1/2) - 4*C^2*a^2*c*f^2 + 4*C^2*b^2*c*f^2 - 8*C^2 
*a*b*d*f^2)/(16*(a^4*f^4 + b^4*f^4 + 2*a^2*b^2*f^4)))^(1/2) - (32*(c + d*t 
an(e + f*x))^(1/2)*(14*C^2*a*b^7*d^11*f^2 - 2*C^2*a^5*b^3*d^11*f^2 - 10*C^ 
2*b^8*c^3*d^8*f^2 - 4*C^2*a^3*b^5*d^11*f^2 - 16*C^2*a^7*b*d^11*f^2 + 8*C^2 
*a^8*c*d^10*f^2 - 6*C^2*b^8*c*d^10*f^2 + 18*C^2*a*b^7*c^2*d^9*f^2 + 12*C^2 
*a^2*b^6*c*d^10*f^2 + 2*C^2*a^4*b^4*c*d^10*f^2 + 24*C^2*a^6*b^2*c*d^10*f^2 
 - 16*C^2*a^7*b*c^2*d^9*f^2 + 4*C^2*a^2*b^6*c^3*d^8*f^2 + 4*C^2*a^3*b^5...
 

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x)*d + c)/(tan(e + f*x)*b + a),x)*a + int((sqrt(tan(e + 
 f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*b + a),x)*c + int((sqrt(tan(e 
+ f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*b + a),x)*b