3.2.14 \(\int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx\) [114]

3.2.14.1 Optimal result

Integrand size = 47, antiderivative size = 210 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b) \sqrt {c-i d} f}-\frac {(A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) \sqrt {c+i d} f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right ) \sqrt {b c-a d} f} \]

output

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

3.2.14.2 Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {(-i a+b) (A-i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {(i a+b) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}-\frac {2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}}}{\left (a^2+b^2\right ) f} \]

input

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

output

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

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

Time = 1.23 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.255, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.2.14.3.1 Defintions of rubi rules used

rule 25

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

3.2.14.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(13473\) vs. \(2(179)=358\).

Time = 0.14 (sec) , antiderivative size = 13474, normalized size of antiderivative = 64.16

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(13474\)
default	\(\text {Expression too large to display}\)	\(13474\)

input

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

output

result too large to display
 

3.2.14.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.2.14.6 Sympy [F]

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

input

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

output

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

3.2.14.7 Maxima [F(-2)]

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

input

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

3.2.14.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.2.14.9 Mupad [B] (verification not implemented)

Time = 65.48 (sec) , antiderivative size = 25341, normalized size of antiderivative = 120.67 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

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

output

(log((((((((((128*C*b^2*d^8*(a*d + b*c)^2*(a^2 + b^2)^2)/f - 64*b^2*d^8*(a 
^2 + b^2)^2*(c + d*tan(e + f*x))^(1/2)*((4*(-C^4*f^4*(a^2*d - b^2*d + 2*a* 
b*c)^2)^(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)/(f^4* 
(a^2 + b^2)^2*(c^2 + d^2)))^(1/2)*(3*b^3*c^2 + 2*b^3*d^2 - a^2*b*c^2 - 2*a 
^2*b*d^2 + a^3*c*d + a*b^2*c*d))*((4*(-C^4*f^4*(a^2*d - b^2*d + 2*a*b*c)^2 
)^(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)/(f^4*(a^2 + 
 b^2)^2*(c^2 + d^2)))^(1/2))/4 + (64*C^2*b*d^8*(c + d*tan(e + f*x))^(1/2)* 
(5*b^6*c - 4*a^6*c - 2*a^2*b^4*c + 5*a^4*b^2*c - 2*a^3*b^3*d + 7*a*b^5*d + 
 7*a^5*b*d))/f^2)*((4*(-C^4*f^4*(a^2*d - b^2*d + 2*a*b*c)^2)^(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)/(f^4*(a^2 + b^2)^2*(c^2 + 
d^2)))^(1/2))/4 + (32*C^3*b*d^8*(4*a^5*d - b^5*c - 9*a^2*b^3*c - 15*a^3*b^ 
2*d + 12*a^4*b*c + a*b^4*d))/f^3)*((4*(-C^4*f^4*(a^2*d - b^2*d + 2*a*b*c)^ 
2)^(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)/(f^4*(a^2 
+ b^2)^2*(c^2 + d^2)))^(1/2))/4 - (32*C^4*b*d^8*(2*a^4 + b^4)*(c + d*tan(e 
 + f*x))^(1/2))/f^4)*((4*(-C^4*f^4*(a^2*d - b^2*d + 2*a*b*c)^2)^(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)/(f^4*(a^2 + b^2)^2*(c^2 
 + d^2)))^(1/2))/4 + (32*C^5*a^2*b^2*d^8)/f^5)*(((32*C^4*a^2*b^2*d^2*f^4 - 
 16*C^4*b^4*d^2*f^4 - 64*C^4*a^2*b^2*c^2*f^4 - 16*C^4*a^4*d^2*f^4 + 64*C^4 
*a*b^3*c*d*f^4 - 64*C^4*a^3*b*c*d*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)/(a^4*c^2*f^4 + a^4*d^2*f^4 + b^4*c^2*f^4 + b^...