3.4.35 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx\) [335]

3.4.35.1 Optimal result

Integrand size = 32, antiderivative size = 44 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a d}+\frac {(B-C) \tan (c+d x)}{a d (1+\sec (c+d x))} \]

output

C*arctanh(sin(d*x+c))/a/d+(B-C)*tan(d*x+c)/a/d/(1+sec(d*x+c))
 

3.4.35.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(106\) vs. \(2(44)=88\).

Time = 0.62 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.41 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (C \cos \left (\frac {1}{2} (c+d x)\right ) \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(B-C) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{a d (1+\cos (c+d x))} \]

input

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x]),x]
 

output

(2*Cos[(c + d*x)/2]*(C*Cos[(c + d*x)/2]*(-Log[Cos[(c + d*x)/2] - Sin[(c + 
d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (B - C)*Sin[(c + d* 
x)/2]))/(a*d*(1 + Cos[c + d*x]))
 

3.4.35.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 4538, 27, 3042, 4257, 4281}

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.

\(\displaystyle \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4538

\(\displaystyle \frac {\int \frac {(B-C) \sec (c+d x)}{\sec (c+d x)+1}dx}{a}+\frac {C \int \sec (c+d x)dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(B-C) \int \frac {\sec (c+d x)}{\sec (c+d x)+1}dx}{a}+\frac {C \int \sec (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )+1}dx}{a}+\frac {C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {(B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )+1}dx}{a}+\frac {C \text {arctanh}(\sin (c+d x))}{a d}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {C \text {arctanh}(\sin (c+d x))}{a d}+\frac {(B-C) \tan (c+d x)}{a d (\sec (c+d x)+1)}\)

input

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x]),x]
 

output

(C*ArcTanh[Sin[c + d*x]])/(a*d) + ((B - C)*Tan[c + d*x])/(a*d*(1 + Sec[c + 
 d*x]))
 

3.4.35.3.1 Defintions of rubi rules used

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 3042

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4538

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[C/b   Int[Csc[ 
e + f*x], x], x] + Simp[1/b   Int[(A*b + (b*B - a*C)*Csc[e + f*x])/(a + b*C 
sc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]
 

3.4.35.4 Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {-C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (B -C \right )}{a d}\)	\(53\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(61\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(61\)
risch	\(\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\)	\(91\)
norman	\(\frac {\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(105\)

input

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE 
)
 

output

(-C*ln(tan(1/2*d*x+1/2*c)-1)+C*ln(tan(1/2*d*x+1/2*c)+1)+tan(1/2*d*x+1/2*c) 
*(B-C))/a/d
 

3.4.35.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {{\left (C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B - C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

input

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fri 
cas")
 

output

1/2*((C*cos(d*x + c) + C)*log(sin(d*x + c) + 1) - (C*cos(d*x + c) + C)*log 
(-sin(d*x + c) + 1) + 2*(B - C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
 

3.4.35.6 Sympy [F]

\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {B \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

input

integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
 

output

(Integral(B*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)* 
*2/(sec(c + d*x) + 1), x))/a
 

3.4.35.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (44) = 88\).

Time = 0.23 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.25 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {B \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

input

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="max 
ima")
 

output

(C*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x 
 + c) + 1) - 1)/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))) + B*sin(d*x + c)/ 
(a*(cos(d*x + c) + 1)))/d
 

3.4.35.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.59 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

input

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="gia 
c")
 

output

(C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - C*log(abs(tan(1/2*d*x + 1/2*c) - 
 1))/a + (B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a)/d
 

3.4.35.9 Mupad [B] (verification not implemented)

Time = 15.85 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]

input

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x)),x)
 

output

(2*C*atanh(tan(c/2 + (d*x)/2)))/(a*d) + (tan(c/2 + (d*x)/2)*(B - C))/(a*d)