\(\int (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [40]

Optimal result

Integrand size = 19, antiderivative size = 24 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3855, 3852, 8} \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \text {arctanh}(\sin (c+d x))}{d}+\frac {C \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \tan \left (d x +c \right )}{d}\)	\(30\)
default	\(\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \tan \left (d x +c \right )}{d}\)	\(32\)
parts	\(\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \tan \left (d x +c \right )}{d}\)	\(32\)
risch	\(-\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 i C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(59\)
parallelrisch	\(\frac {-B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+C \sin \left (d x +c \right )}{d \cos \left (d x +c \right )}\)	\(63\)
norman	\(-\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(67\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/2*(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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac {C \tan \left (d x + c\right )}{d} \]

[In]

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

[Out]

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. 57 vs. \(2 (24) = 48\).

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B {\left (\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right ) - \log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )\right )}}{4 \, d} + \frac {C \tan \left (d x + c\right )}{d} \]

[In]

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

[Out]

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 = 15.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(2*B*atanh(tan(c/2 + (d*x)/2)))/d - (2*C*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 - 1))