\(\int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [36]

Optimal result

Integrand size = 24, antiderivative size = 17 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=a x-\frac {b \log (\cos (c+d x))}{d} \]

[Out]

a*x-b*ln(cos(d*x+c))/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3165, 3556} \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=a x-\frac {b \log (\cos (c+d x))}{d} \]

[In]

Int[Sec[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

a*x - (b*Log[Cos[c + d*x]])/d

Rule 3165

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3556

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

Rubi steps \begin{align*} \text {integral}& = \int (a+b \tan (c+d x)) \, dx \\ & = a x+b \int \tan (c+d x) \, dx \\ & = a x-\frac {b \log (\cos (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=a x-\frac {b \log (\cos (c+d x))}{d} \]

[In]

Integrate[Sec[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

a*x - (b*Log[Cos[c + d*x]])/d

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-b \ln \left (\cos \left (d x +c \right )\right )+a \left (d x +c \right )}{d}\)	\(23\)
default	\(\frac {-b \ln \left (\cos \left (d x +c \right )\right )+a \left (d x +c \right )}{d}\)	\(23\)
parts	\(\frac {a \left (d x +c \right )}{d}+\frac {b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(24\)
risch	\(i b x +a x +\frac {2 i b c}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(36\)
parallelrisch	\(\frac {a x d -b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(54\)
norman	\(\frac {a x +a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(91\)

[In]

int(sec(d*x+c)*(cos(d*x+c)*a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-b*ln(cos(d*x+c))+a*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a d x - b \log \left (-\cos \left (d x + c\right )\right )}{d} \]

[In]

integrate(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

(a*d*x - b*log(-cos(d*x + c)))/d

Sympy [F]

\[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))*sec(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {2 \, {\left (d x + c\right )} a - b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

[In]

integrate(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(2*(d*x + c)*a - b*log(-sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {2 \, {\left (d x + c\right )} a + b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

[In]

integrate(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*(d*x + c)*a + b*log(tan(d*x + c)^2 + 1))/d

Mupad [B] (verification not implemented)

Time = 20.92 (sec) , antiderivative size = 70, normalized size of antiderivative = 4.12 \[ \int \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {b\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {2\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {b\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )}{d} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))/cos(c + d*x),x)

[Out]

(b*log(1/cos(c/2 + (d*x)/2)^2))/d + (2*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (b*log(cos(c + d*x)/
(cos(c + d*x) + 1)))/d