3.217 \(\int \frac {\cos (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {\log \left (\sin ^2(c+d x)+1\right )}{2 d} \]

[Out]

1/2*ln(1+sin(d*x+c)^2)/d

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4334, 260} \[ \frac {\log \left (\sin ^2(c+d x)+1\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]

[Out]

Log[1 + Sin[c + d*x]^2]/(2*d)

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4334

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[d/(b
*c), Subst[Int[SubstFor[1, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a +
b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{\csc (c+d x)+\sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\log \left (1+\sin ^2(c+d x)\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 20, normalized size = 1.11 \[ \frac {\log (3-\cos (2 (c+d x)))}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(Csc[c + d*x] + Sin[c + d*x]),x]

[Out]

Log[3 - Cos[2*(c + d*x)]]/(2*d)

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fricas [A]  time = 0.50, size = 18, normalized size = 1.00 \[ \frac {\log \left (-\cos \left (d x + c\right )^{2} + 2\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(-cos(d*x + c)^2 + 2)/d

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giac [A]  time = 0.26, size = 16, normalized size = 0.89 \[ \frac {\log \left (\sin \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(sin(d*x + c)^2 + 1)/d

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maple [A]  time = 0.06, size = 17, normalized size = 0.94 \[ \frac {\ln \left (\cos ^{2}\left (d x +c \right )-2\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x)

[Out]

1/2/d*ln(cos(d*x+c)^2-2)

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maxima [A]  time = 0.47, size = 16, normalized size = 0.89 \[ \frac {\log \left (\sin \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(sin(d*x + c)^2 + 1)/d

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mupad [B]  time = 0.06, size = 16, normalized size = 0.89 \[ \frac {\ln \left ({\sin \left (c+d\,x\right )}^2+1\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(c + d*x)^2 + 1)/(2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(csc(d*x+c)+sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)/(sin(c + d*x) + csc(c + d*x)), x)

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