\(\int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx\) [61]

Optimal result

Integrand size = 19, antiderivative size = 31 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos (c+d x)}{a d}+\frac {\log (1+\cos (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2912, 12, 45} \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\log (\cos (c+d x)+1)}{a d}-\frac {\cos (c+d x)}{a d} \]

[In]

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

[Out]

-(Cos[c + d*x]/(a*d)) + Log[1 + Cos[c + d*x]]/(a*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (c+d x) \sin (c+d x)}{-a-a \cos (c+d x)} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x}{a (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x}{-a+x} \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (1-\frac {a}{a-x}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = -\frac {\cos (c+d x)}{a d}+\frac {\log (1+\cos (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos (c+d x)-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \]

[In]

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

[Out]

-((Cos[c + d*x] - 2*Log[Cos[(c + d*x)/2]])/(a*d))

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {-\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+1-\cos \left (d x +c \right )}{a d}\)	\(32\)
derivativedivides	\(\frac {\ln \left (1+\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}-\ln \left (\sec \left (d x +c \right )\right )}{d a}\)	\(37\)
default	\(\frac {\ln \left (1+\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}-\ln \left (\sec \left (d x +c \right )\right )}{d a}\)	\(37\)
norman	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {\ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a d}\)	\(58\)
risch	\(-\frac {i x}{a}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d a}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d a}-\frac {2 i c}{d a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}\)	\(73\)

[In]

int(sin(d*x+c)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos \left (d x + c\right ) - \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{a d} \]

[In]

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

[Out]

-(cos(d*x + c) - log(1/2*cos(d*x + c) + 1/2))/(a*d)

Sympy [F]

\[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(sin(c + d*x)/(sec(c + d*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {\cos \left (d x + c\right )}{a} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a}}{d} \]

[In]

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

[Out]

-(cos(d*x + c)/a - log(cos(d*x + c) + 1)/a)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos \left (d x + c\right )}{a d} + \frac {\log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a d} \]

[In]

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

[Out]

-cos(d*x + c)/(a*d) + log(abs(-cos(d*x + c) - 1))/(a*d)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {\sin (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )-\cos \left (c+d\,x\right )}{a\,d} \]

[In]

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

[Out]

(log(cos(c + d*x) + 1) - cos(c + d*x))/(a*d)