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

Optimal. Leaf size=49 \[ -\frac{c-\sqrt{b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.036152, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.033, Rules used = {3114} \[ -\frac{c-\sqrt{b^2+c^2} \sin (d+e x)}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

Antiderivative was successfully verified.

[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 3114

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*} \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))}\\ \end{align*}

Mathematica [A]  time = 0.0995726, size = 49, normalized size = 1. \[ \frac{\sqrt{b^2+c^2} \sin (d+e x)-c}{c e (c \cos (d+e x)-b \sin (d+e x))} \]

Antiderivative was successfully verified.

[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]))

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Maple [A]  time = 0.074, size = 50, normalized size = 1. \begin{align*} -2\,{\frac{\sqrt{{b}^{2}+{c}^{2}}+b}{{c}^{2}e} \left ( \tan \left ( d/2+1/2\,ex \right ) +{\frac{\sqrt{{b}^{2}+{c}^{2}}}{c}}+{\frac{b}{c}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.999555, size = 54, normalized size = 1.1 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.15614, size = 173, normalized size = 3.53 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1434, size = 58, normalized size = 1.18 \begin{align*} -\frac{2 \,{\left (b + \sqrt{b^{2} + c^{2}}\right )} e^{\left (-1\right )}}{{\left (c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + b + \sqrt{b^{2} + c^{2}}\right )} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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