\(\int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx\) [372]

Optimal result

Integrand size = 12, antiderivative size = 97 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {a x}{a^2+b^2}+\frac {2 a c \text {arctanh}\left (\frac {b-(a-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt {a^2+b^2-c^2}}+\frac {b \log (c+a \cos (x)+b \sin (x))}{a^2+b^2} \] Output:

a*x/(a^2+b^2)+2*a*c*arctanh((b-(a-c)*tan(1/2*x))/(a^2+b^2-c^2)^(1/2))/(a^2 
+b^2)/(a^2+b^2-c^2)^(1/2)+b*ln(c+a*cos(x)+b*sin(x))/(a^2+b^2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {a x+\frac {2 a c \text {arctanh}\left (\frac {b+(-a+c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}}+b \log (c+a \cos (x)+b \sin (x))}{a^2+b^2} \] Input:

Integrate[(a + c*Sec[x] + b*Tan[x])^(-1),x]
 

Output:

(a*x + (2*a*c*ArcTanh[(b + (-a + c)*Tan[x/2])/Sqrt[a^2 + b^2 - c^2]])/Sqrt 
[a^2 + b^2 - c^2] + b*Log[c + a*Cos[x] + b*Sin[x]])/(a^2 + b^2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3638, 3042, 3617, 3042, 3603, 1083, 219}

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.

\(\displaystyle \int \frac {1}{a+b \tan (x)+c \sec (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{a+b \tan (x)+c \sec (x)}dx\)

\(\Big \downarrow \) 3638

\(\displaystyle \int \frac {\cos (x)}{a \cos (x)+b \sin (x)+c}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (x)}{a \cos (x)+b \sin (x)+c}dx\)

\(\Big \downarrow \) 3617

\(\displaystyle -\frac {a c \int \frac {1}{c+a \cos (x)+b \sin (x)}dx}{a^2+b^2}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a c \int \frac {1}{c+a \cos (x)+b \sin (x)}dx}{a^2+b^2}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2}\)

\(\Big \downarrow \) 3603

\(\displaystyle -\frac {2 a c \int \frac {1}{-\left ((a-c) \tan ^2\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a+c}d\tan \left (\frac {x}{2}\right )}{a^2+b^2}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {4 a c \int \frac {1}{4 \left (a^2+b^2-c^2\right )-\left (2 b-2 (a-c) \tan \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 (a-c) \tan \left (\frac {x}{2}\right )\right )}{a^2+b^2}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 a c \text {arctanh}\left (\frac {2 b-2 (a-c) \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt {a^2+b^2-c^2}}+\frac {b \log (a \cos (x)+b \sin (x)+c)}{a^2+b^2}+\frac {a x}{a^2+b^2}\)

Input:

Int[(a + c*Sec[x] + b*Tan[x])^(-1),x]
 

Output:

(a*x)/(a^2 + b^2) + (2*a*c*ArcTanh[(2*b - 2*(a - c)*Tan[x/2])/(2*Sqrt[a^2 
+ b^2 - c^2])])/((a^2 + b^2)*Sqrt[a^2 + b^2 - c^2]) + (b*Log[c + a*Cos[x] 
+ b*Sin[x]])/(a^2 + b^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3603

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f 
/e)   Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) 
/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
 

rule 3617

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_) 
]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[b*B*((d + e*x)/ 
(e*(b^2 + c^2))), x] + (Simp[c*B*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/ 
(e*(b^2 + c^2))), x] + Simp[(A*(b^2 + c^2) - a*b*B)/(b^2 + c^2)   Int[1/(a 
+ b*Cos[d + e*x] + c*Sin[d + e*x]), x], x]) /; FreeQ[{a, b, c, d, e, A, B}, 
 x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*b*B, 0]
 

rule 3638

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)]) 
^(-1), x_Symbol] :> Int[Cos[d + e*x]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), 
 x] /; FreeQ[{a, b, c, d, e}, x]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.76

method	result	size
default	\(\frac {-b \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+2 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}+b^{2}}+\frac {\frac {2 \left (b a -c b \right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -c \tan \left (\frac {x}{2}\right )^{2}-2 b \tan \left (\frac {x}{2}\right )-a -c \right )}{2 a -2 c}+\frac {2 \left (a c -b^{2}+\frac {\left (b a -c b \right ) b}{a -c}\right ) \arctan \left (\frac {2 \left (a -c \right ) \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}-b^{2}+c^{2}}}}{a^{2}+b^{2}}\)	\(171\)
risch	\(\text {Expression too large to display}\)	\(1314\)

Input:

int(1/(a+c*sec(x)+b*tan(x)),x,method=_RETURNVERBOSE)
 

Output:

2/(a^2+b^2)*(-1/2*b*ln(1+tan(1/2*x)^2)+a*arctan(tan(1/2*x)))+2/(a^2+b^2)*( 
1/2*(a*b-b*c)/(a-c)*ln(tan(1/2*x)^2*a-c*tan(1/2*x)^2-2*b*tan(1/2*x)-a-c)+( 
a*c-b^2+(a*b-b*c)*b/(a-c))/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(a-c)*tan(1/ 
2*x)-2*b)/(-a^2-b^2+c^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (93) = 186\).

Time = 0.15 (sec) , antiderivative size = 553, normalized size of antiderivative = 5.70 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\left [\frac {\sqrt {a^{2} + b^{2} - c^{2}} a c \log \left (\frac {2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (a^{2} - b^{2}\right )} c^{2} + 2 \, {\left (a^{3} + a b^{2}\right )} c \cos \left (x\right ) - {\left (a^{4} - b^{4} - 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \left (x\right )^{2} + 2 \, {\left ({\left (a^{2} b + b^{3}\right )} c - {\left (a^{3} b + a b^{3} - 2 \, a b c^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) + 2 \, {\left (2 \, a b c \cos \left (x\right )^{2} - a b c + {\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) - {\left (a^{3} + a b^{2} + {\left (a^{2} - b^{2}\right )} c \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt {a^{2} + b^{2} - c^{2}}}{2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )}\right ) + 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x + {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}, -\frac {2 \, \sqrt {-a^{2} - b^{2} + c^{2}} a c \arctan \left (\frac {{\left (a c \cos \left (x\right ) + b c \sin \left (x\right ) + a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}}{{\left (a^{2} b + b^{3} - b c^{2}\right )} \cos \left (x\right ) - {\left (a^{3} + a b^{2} - a c^{2}\right )} \sin \left (x\right )}\right ) - 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x - {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, a c \cos \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2} + c^{2} + 2 \, {\left (a b \cos \left (x\right ) + b c\right )} \sin \left (x\right )\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}\right ] \] Input:

integrate(1/(a+c*sec(x)+b*tan(x)),x, algorithm="fricas")
 

Output:

[1/2*(sqrt(a^2 + b^2 - c^2)*a*c*log((2*a^4 + 3*a^2*b^2 + b^4 - (a^2 - b^2) 
*c^2 + 2*(a^3 + a*b^2)*c*cos(x) - (a^4 - b^4 - 2*(a^2 - b^2)*c^2)*cos(x)^2 
 + 2*((a^2*b + b^3)*c - (a^3*b + a*b^3 - 2*a*b*c^2)*cos(x))*sin(x) + 2*(2* 
a*b*c*cos(x)^2 - a*b*c + (a^2*b + b^3)*cos(x) - (a^3 + a*b^2 + (a^2 - b^2) 
*c*cos(x))*sin(x))*sqrt(a^2 + b^2 - c^2))/(2*a*c*cos(x) + (a^2 - b^2)*cos( 
x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b*c)*sin(x))) + 2*(a^3 + a*b^2 - a*c^2) 
*x + (a^2*b + b^3 - b*c^2)*log(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + 
 c^2 + 2*(a*b*cos(x) + b*c)*sin(x)))/(a^4 + 2*a^2*b^2 + b^4 - (a^2 + b^2)* 
c^2), -1/2*(2*sqrt(-a^2 - b^2 + c^2)*a*c*arctan((a*c*cos(x) + b*c*sin(x) + 
 a^2 + b^2)*sqrt(-a^2 - b^2 + c^2)/((a^2*b + b^3 - b*c^2)*cos(x) - (a^3 + 
a*b^2 - a*c^2)*sin(x))) - 2*(a^3 + a*b^2 - a*c^2)*x - (a^2*b + b^3 - b*c^2 
)*log(2*a*c*cos(x) + (a^2 - b^2)*cos(x)^2 + b^2 + c^2 + 2*(a*b*cos(x) + b* 
c)*sin(x)))/(a^4 + 2*a^2*b^2 + b^4 - (a^2 + b^2)*c^2)]
 

Sympy [F]

\[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\int \frac {1}{a + b \tan {\left (x \right )} + c \sec {\left (x \right )}}\, dx \] Input:

integrate(1/(a+c*sec(x)+b*tan(x)),x)
 

Output:

Integral(1/(a + b*tan(x) + c*sec(x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+c*sec(x)+b*tan(x)),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(c^2-b^2-a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, c\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - c \tan \left (\frac {1}{2} \, x\right ) - b}{\sqrt {-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{{\left (a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}} + \frac {a x}{a^{2} + b^{2}} + \frac {b \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a + c\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{a^{2} + b^{2}} \] Input:

integrate(1/(a+c*sec(x)+b*tan(x)),x, algorithm="giac")
 

Output:

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*c) + arctan(-(a*tan(1/2*x) - c*t 
an(1/2*x) - b)/sqrt(-a^2 - b^2 + c^2)))*a*c/((a^2 + b^2)*sqrt(-a^2 - b^2 + 
 c^2)) + a*x/(a^2 + b^2) + b*log(-a*tan(1/2*x)^2 + c*tan(1/2*x)^2 + 2*b*ta 
n(1/2*x) + a + c)/(a^2 + b^2) - b*log(tan(1/2*x)^2 + 1)/(a^2 + b^2)
 

Mupad [B] (verification not implemented)

Time = 26.66 (sec) , antiderivative size = 988, normalized size of antiderivative = 10.19 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx =\text {Too large to display} \] Input:

int(1/(a + b*tan(x) + c/cos(x)),x)
 

Output:

(log(32*a*c - 32*c^2 + 32*b*tan(x/2)*(a - c) + ((32*a^2*b - 32*b*c^2 + 32* 
tan(x/2)*(a - c)*(2*a*c - a^2 + 3*b^2 - 2*c^2) - ((a^2*b - b*c^2 + b^3 + a 
*c*(a^2 + b^2 - c^2)^(1/2))*(32*a^4 - 64*a^3*c - 64*a^2*b^2 + 32*a^2*c^2 - 
 32*b^2*c^2 + 96*a*b^2*c + 32*b*tan(x/2)*(a - c)*(4*a^2 - 4*a*c + b^2) + ( 
32*(a - c)*(a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2))*(3*b^4*tan( 
x/2) + 3*a*b^3 + 3*a^3*b + b^3*c + 3*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(x/2) 
 - 2*b^2*c^2*tan(x/2) - 2*a^3*c*tan(x/2) - 4*a*b*c^2 + a^2*b*c - 2*a*b^2*c 
*tan(x/2)))/((a^2 + b^2)*(a^2 + b^2 - c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^ 
2)))*(a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2)))/((a^2 + b^2)*(a^ 
2 + b^2 - c^2)))*(b*(a^2 - c^2) + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2)))/(c^2 
*(a^2 + b^2 - c^2) + (a^2 + b^2 - c^2)^2) - log(tan(x/2) + 1i)/(a*1i + b) 
- (log(tan(x/2) - 1i)*1i)/(a + b*1i) + (log(32*a*c - 32*c^2 + 32*b*tan(x/2 
)*(a - c) + ((32*a^2*b - 32*b*c^2 + 32*tan(x/2)*(a - c)*(2*a*c - a^2 + 3*b 
^2 - 2*c^2) - ((a^2*b - b*c^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a^4 
 - 64*a^3*c - 64*a^2*b^2 + 32*a^2*c^2 - 32*b^2*c^2 + 96*a*b^2*c + 32*b*tan 
(x/2)*(a - c)*(4*a^2 - 4*a*c + b^2) + (32*(a - c)*(a^2*b - b*c^2 + b^3 - a 
*c*(a^2 + b^2 - c^2)^(1/2))*(3*b^4*tan(x/2) + 3*a*b^3 + 3*a^3*b + b^3*c + 
3*a^2*b^2*tan(x/2) + 2*a^2*c^2*tan(x/2) - 2*b^2*c^2*tan(x/2) - 2*a^3*c*tan 
(x/2) - 4*a*b*c^2 + a^2*b*c - 2*a*b^2*c*tan(x/2)))/((a^2 + b^2)*(a^2 + b^2 
 - c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^2)))*(a^2*b - b*c^2 + b^3 - a*c*...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.70 \[ \int \frac {1}{a+c \sec (x)+b \tan (x)} \, dx=\frac {-2 \sqrt {-a^{2}-b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) c -b}{\sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a c -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) b \,c^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} c -2 \tan \left (\frac {x}{2}\right ) b -a -c \right ) b \,c^{2}+a^{3} x +a \,b^{2} x -a \,c^{2} x}{a^{4}+2 a^{2} b^{2}-a^{2} c^{2}+b^{4}-b^{2} c^{2}} \] Input:

int(1/(a+c*sec(x)+b*tan(x)),x)
 

Output:

( - 2*sqrt( - a**2 - b**2 + c**2)*atan((tan(x/2)*a - tan(x/2)*c - b)/sqrt( 
 - a**2 - b**2 + c**2))*a*c - log(tan(x/2)**2 + 1)*a**2*b - log(tan(x/2)** 
2 + 1)*b**3 + log(tan(x/2)**2 + 1)*b*c**2 + log(tan(x/2)**2*a - tan(x/2)** 
2*c - 2*tan(x/2)*b - a - c)*a**2*b + log(tan(x/2)**2*a - tan(x/2)**2*c - 2 
*tan(x/2)*b - a - c)*b**3 - log(tan(x/2)**2*a - tan(x/2)**2*c - 2*tan(x/2) 
*b - a - c)*b*c**2 + a**3*x + a*b**2*x - a*c**2*x)/(a**4 + 2*a**2*b**2 - a 
**2*c**2 + b**4 - b**2*c**2)