\(\int (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [58]

Optimal result

Integrand size = 20, antiderivative size = 27 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=A x+\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[Out]

A*x+B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3855, 3852, 8} \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=A x+\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[In]

Int[A + B*Sec[c + d*x] + C*Sec[c + d*x]^2,x]

[Out]

A*x + (B*ArcTanh[Sin[c + d*x]])/d + (C*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = A x+B \int \sec (c+d x) \, dx+C \int \sec ^2(c+d x) \, dx \\ & = A x+\frac {B \text {arctanh}(\sin (c+d x))}{d}-\frac {C \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = A x+\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=A x+\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[In]

Integrate[A + B*Sec[c + d*x] + C*Sec[c + d*x]^2,x]

[Out]

A*x + (B*ArcTanh[Sin[c + d*x]])/d + (C*Tan[c + d*x])/d

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30

method	result	size
default	\(A x +\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \tan \left (d x +c \right )}{d}\)	\(35\)
parts	\(A x +\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \tan \left (d x +c \right )}{d}\)	\(35\)
derivativedivides	\(\frac {\left (d x +c \right ) A +B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \tan \left (d x +c \right )}{d}\)	\(37\)
risch	\(A x -\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 i C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(62\)
parallelrisch	\(\frac {-B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+C \sin \left (d x +c \right )}{d \cos \left (d x +c \right )}+A x\)	\(67\)
norman	\(\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A x -\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(87\)

[In]

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

[Out]

A*x+B/d*ln(sec(d*x+c)+tan(d*x+c))+C*tan(d*x+c)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (27) = 54\).

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.63 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, A d x \cos \left (d x + c\right ) + B \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

integrate(A+B*sec(d*x+c)+C*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(A + B*sec(c + d*x) + C*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=A x + \frac {B \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac {C \tan \left (d x + c\right )}{d} \]

[In]

integrate(A+B*sec(d*x+c)+C*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

A*x + B*log(sec(d*x + c) + tan(d*x + c))/d + C*tan(d*x + c)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=A x + \frac {B {\left (\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right ) - \log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )\right )}}{4 \, d} + \frac {C \tan \left (d x + c\right )}{d} \]

[In]

integrate(A+B*sec(d*x+c)+C*sec(d*x+c)^2,x, algorithm="giac")

[Out]

A*x + 1/4*B*(log(abs(1/sin(d*x + c) + sin(d*x + c) + 2)) - log(abs(1/sin(d*x + c) + sin(d*x + c) - 2)))/d + C*
tan(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 14.95 (sec) , antiderivative size = 161, normalized size of antiderivative = 5.96 \[ \int \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,\mathrm {atan}\left (\frac {64\,A^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^3+64\,A\,B^2}+\frac {64\,A\,B^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^3+64\,A\,B^2}\right )}{d}+\frac {2\,B\,\mathrm {atanh}\left (\frac {64\,B^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,B+64\,B^3}+\frac {64\,A^2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,A^2\,B+64\,B^3}\right )}{d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(2*A*atan((64*A^3*tan(c/2 + (d*x)/2))/(64*A*B^2 + 64*A^3) + (64*A*B^2*tan(c/2 + (d*x)/2))/(64*A*B^2 + 64*A^3))
)/d + (2*B*atanh((64*B^3*tan(c/2 + (d*x)/2))/(64*A^2*B + 64*B^3) + (64*A^2*B*tan(c/2 + (d*x)/2))/(64*A^2*B + 6
4*B^3)))/d - (2*C*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 - 1))