\(\int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\) [116]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {\log (\cos (c+d x))}{b d}+\frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3181, 3556, 3212} \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d}-\frac {\log (\cos (c+d x))}{b d} \]

[In]

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

[Out]

-(Log[Cos[c + d*x]]/(b*d)) + Log[a*Cos[c + d*x] + b*Sin[c + d*x]]/(b*d)

Rule 3181

Int[1/(cos[(c_.) + (d_.)*(x_)]*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])), x_Symbol] :>
Dist[1/b, Int[Tan[c + d*x], x], x] + Dist[1/b, Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3212

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + Simp[(c*B - b*C)*(L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 + c^2))), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 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}& = \frac {\int \frac {b \cos (c+d x)-a \sin (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b}+\frac {\int \tan (c+d x) \, dx}{b} \\ & = -\frac {\log (\cos (c+d x))}{b d}+\frac {\log (a \cos (c+d x)+b \sin (c+d x))}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.44 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\log (a+b \tan (c+d x))}{b d} \]

[In]

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

[Out]

Log[a + b*Tan[c + d*x]]/(b*d)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.46

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{d b}\)	\(19\)
default	\(\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{d b}\)	\(19\)
risch	\(-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b d}\)	\(58\)
parallelrisch	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b d}\)	\(67\)
norman	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{b d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b d}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.44 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, b d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (41) = 82\).

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.51 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\frac {\log \left (-a - \frac {2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.46 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b d} \]

[In]

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

[Out]

log(abs(b*tan(d*x + c) + a))/(b*d)

Mupad [B] (verification not implemented)

Time = 23.46 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.51 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {b\,\left (b\,\cos \left (c+d\,x\right )-a\,\sin \left (c+d\,x\right )\right )}{2\,\cos \left (c+d\,x\right )\,a^2+\sin \left (c+d\,x\right )\,a\,b+\cos \left (c+d\,x\right )\,b^2}\right )}{b\,d} \]

[In]

int(1/(cos(c + d*x)*(a*cos(c + d*x) + b*sin(c + d*x))),x)

[Out]

-(2*atanh((b*(b*cos(c + d*x) - a*sin(c + d*x)))/(2*a^2*cos(c + d*x) + b^2*cos(c + d*x) + a*b*sin(c + d*x))))/(
b*d)