\(\int (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [5]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3093, 3855} \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{d}+\frac {C \sin (c+d x)}{d} \]

[In]

Int[(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]

[Out]

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

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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}& = \frac {C \sin (c+d x)}{d}+A \int \sec (c+d x) \, dx \\ & = \frac {A \text {arctanh}(\sin (c+d x))}{d}+\frac {C \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{d}+\frac {C \cos (d x) \sin (c)}{d}+\frac {C \cos (c) \sin (d x)}{d} \]

[In]

Integrate[(A + C*Cos[c + d*x]^2)*Sec[c + d*x],x]

[Out]

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

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\sin \left (d x +c \right ) C}{d}\)	\(30\)
default	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+\sin \left (d x +c \right ) C}{d}\)	\(30\)
parts	\(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \sin \left (d x +c \right )}{d}\)	\(32\)
parallelrisch	\(\frac {-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right ) C}{d}\)	\(43\)
risch	\(-\frac {i C \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(71\)
norman	\(\frac {\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A \log \left (\sin \left (d x + c\right ) + 1\right ) - A \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A \log \left (\sin \left (d x + c\right ) + 1\right ) - A \log \left (\sin \left (d x + c\right ) - 1\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - A \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {C\,\sin \left (c+d\,x\right )+A\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{d} \]

[In]

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

[Out]

(C*sin(c + d*x) + A*atanh(sin(c + d*x)))/d