\(\int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx\) [115]

Optimal result

Integrand size = 19, antiderivative size = 47 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {\text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, 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 = {3153, 212} \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {\text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \sqrt {a^2+b^2}} \]

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-1),x]

[Out]

-(ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]]/(Sqrt[a^2 + b^2]*d))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3153

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{d} \\ & = -\frac {\text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {2 \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} d} \]

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-1),x]

[Out]

(2*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(Sqrt[a^2 + b^2]*d)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {2 \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}}\)	\(43\)
default	\(\frac {2 \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}}\)	\(43\)
risch	\(\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}\, d}\)	\(88\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (43) = 86\).

Time = 0.25 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.79 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\frac {\log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{2 \, \sqrt {a^{2} + b^{2}} d} \]

[In]

integrate(1/(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 - 2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*co
s(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2))/(sqrt(a^2
+ b^2)*d)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.90 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.47 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x}{\sin {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{b d} & \text {for}\: a = 0 \\\frac {x}{a \cos {\left (c \right )} + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {1}{i b d \sin {\left (c + d x \right )} + b d \cos {\left (c + d x \right )}} & \text {for}\: a = - i b \\- \frac {1}{- i b d \sin {\left (c + d x \right )} + b d \cos {\left (c + d x \right )}} & \text {for}\: a = i b \\- \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{d \sqrt {a^{2} + b^{2}}} + \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{d \sqrt {a^{2} + b^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x/sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(tan(c/2 + d*x/2))/(b*d), Eq(a, 0)), (x/(a*cos(c
) + b*sin(c)), Eq(d, 0)), (-1/(I*b*d*sin(c + d*x) + b*d*cos(c + d*x)), Eq(a, -I*b)), (-1/(-I*b*d*sin(c + d*x)
+ b*d*cos(c + d*x)), Eq(a, I*b)), (-log(tan(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)/(d*sqrt(a**2 + b**2)) +
log(tan(c/2 + d*x/2) - b/a + sqrt(a**2 + b**2)/a)/(d*sqrt(a**2 + b**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.70 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {\log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} d} \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {\log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} d} \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.56 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {b-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{d\,\sqrt {a^2+b^2}} \]

[In]

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

[Out]

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