\(\int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx\) [359]

Optimal result

Integrand size = 30, antiderivative size = 49 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=-\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {3193} \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=-\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

[In]

Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-1),x]

[Out]

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

Rule 3193

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

Rubi steps \begin{align*} \text {integral}& = -\frac {c-\sqrt {b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=\frac {-c+\sqrt {b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

[In]

Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(-\frac {2 \left (\sqrt {b^{2}+c^{2}}+b \right )}{e \,c^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )}\)	\(50\)
default	\(-\frac {2 \left (\sqrt {b^{2}+c^{2}}+b \right )}{e \,c^{2} \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+\frac {\sqrt {b^{2}+c^{2}}}{c}+\frac {b}{c}\right )}\)	\(50\)
risch	\(\frac {2 i b}{\left (i \sqrt {b^{2}+c^{2}}\, c +b^{2} {\mathrm e}^{i \left (e x +d \right )}+c^{2} {\mathrm e}^{i \left (e x +d \right )}+\sqrt {b^{2}+c^{2}}\, b \right ) e}-\frac {2 c}{\left (i \sqrt {b^{2}+c^{2}}\, c +b^{2} {\mathrm e}^{i \left (e x +d \right )}+c^{2} {\mathrm e}^{i \left (e x +d \right )}+\sqrt {b^{2}+c^{2}}\, b \right ) e}\)	\(121\)

[In]

int(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=-\frac {b^{2} + c^{2} - \sqrt {b^{2} + c^{2}} {\left (b \cos \left (e x + d\right ) + c \sin \left (e x + d\right )\right )}}{{\left (b^{2} c + c^{3}\right )} e \cos \left (e x + d\right ) - {\left (b^{3} + b c^{2}\right )} e \sin \left (e x + d\right )} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2)),x, algorithm="fricas")

[Out]

-(b^2 + c^2 - sqrt(b^2 + c^2)*(b*cos(e*x + d) + c*sin(e*x + d)))/((b^2*c + c^3)*e*cos(e*x + d) - (b^3 + b*c^2)
*e*sin(e*x + d))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b**2+c**2)**(1/2)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=-\frac {2}{{\left (c - \frac {{\left (b - \sqrt {b^{2} + c^{2}}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )} e} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=-\frac {2 \, {\left (b + \sqrt {b^{2} + c^{2}}\right )}}{{\left (c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b + \sqrt {b^{2} + c^{2}}\right )} c e} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2)),x, algorithm="giac")

[Out]

-2*(b + sqrt(b^2 + c^2))/((c*tan(1/2*e*x + 1/2*d) + b + sqrt(b^2 + c^2))*c*e)

Mupad [B] (verification not implemented)

Time = 27.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)} \, dx=\frac {2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e\,\left (b+\sqrt {b^2+c^2}+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )} \]

[In]

int(1/(b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2)),x)

[Out]

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