3.227 \(\int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=32 \[ \frac {\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {3075} \[ \frac {\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3075

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

Rubi steps

\begin {align*} \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx &=\frac {\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 32, normalized size = 1.00 \[ \frac {\sin (c+d x)}{a d (a \cos (c+d x)+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 57, normalized size = 1.78 \[ -\frac {b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )}{{\left (a^{3} + a b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{2} b + b^{3}\right )} d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 20, normalized size = 0.62 \[ -\frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((b*tan(d*x + c) + a)*b*d)

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maple [A]  time = 0.49, size = 21, normalized size = 0.66 \[ -\frac {1}{d b \left (a +b \tan \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 21, normalized size = 0.66 \[ -\frac {1}{{\left (b^{2} \tan \left (d x + c\right ) + a b\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((b^2*tan(d*x + c) + a*b)*d)

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mupad [B]  time = 2.34, size = 47, normalized size = 1.47 \[ \frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))**(-2), x)

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