3.56 \(\int \frac {\cos (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {\log (\sin (c+d x)+1)}{a d} \]

[Out]

ln(1+sin(d*x+c))/a/d

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2667, 31} \[ \frac {\log (\sin (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + Sin[c + d*x]]/(a*d)

Rule 31

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

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {\log (\sin (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + Sin[c + d*x]]/(a*d)

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fricas [A]  time = 0.74, size = 16, normalized size = 1.00 \[ \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(d*x + c) + 1)/(a*d)

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giac [A]  time = 0.40, size = 19, normalized size = 1.19 \[ \frac {\log \left ({\left | a \sin \left (d x + c\right ) + a \right |}\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(a*sin(d*x + c) + a))/(a*d)

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maple [A]  time = 0.06, size = 19, normalized size = 1.19 \[ \frac {\ln \left (a +a \sin \left (d x +c \right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*ln(a+a*sin(d*x+c))/a

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maxima [A]  time = 0.31, size = 18, normalized size = 1.12 \[ \frac {\log \left (a \sin \left (d x + c\right ) + a\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*sin(d*x + c) + a)/(a*d)

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mupad [B]  time = 0.04, size = 16, normalized size = 1.00 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(c + d*x) + 1)/(a*d)

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sympy [A]  time = 0.50, size = 24, normalized size = 1.50 \[ \begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \cos {\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(sin(c + d*x) + 1)/(a*d), Ne(d, 0)), (x*cos(c)/(a*sin(c) + a), True))

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