\(\int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx\) [57]

Optimal result

Integrand size = 19, antiderivative size = 37 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 a d}-\frac {1}{2 d (a+a \sin (c+d x))} \]

[Out]

1/2*arctanh(sin(d*x+c))/a/d-1/2/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2746, 46, 212} \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{2 a d}-\frac {1}{2 d (a \sin (c+d x)+a)} \]

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(2*a*d) - 1/(2*d*(a + a*Sin[c + d*x]))

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))-\frac {1}{1+\sin (c+d x)}}{2 a d} \]

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]] - (1 + Sin[c + d*x])^(-1))/(2*a*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\)	\(43\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\)	\(43\)
parallelrisch	\(\frac {\left (-1-\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right )}{2 a \left (1+\sin \left (d x +c \right )\right ) d}\)	\(70\)
norman	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(71\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}\)	\(76\)

[In]

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

[Out]

1/d/a*(-1/4*ln(sin(d*x+c)-1)-1/2/(1+sin(d*x+c))+1/4*ln(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2}{4 \, {\left (a d \sin \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/4*((sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (sin(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2)/(a*d*sin(d*x +
c) + a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2}{a \sin \left (d x + c\right ) + a}}{4 \, d} \]

[In]

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

[Out]

1/4*(log(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a - 2/(a*sin(d*x + c) + a))/d

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {\sin \left (d x + c\right ) + 3}{a {\left (\sin \left (d x + c\right ) + 1\right )}}}{4 \, d} \]

[In]

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

[Out]

1/4*(log(abs(sin(d*x + c) + 1))/a - log(abs(sin(d*x + c) - 1))/a - (sin(d*x + c) + 3)/(a*(sin(d*x + c) + 1)))/
d

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{2\,a\,d}-\frac {1}{2\,d\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

atanh(sin(c + d*x))/(2*a*d) - 1/(2*d*(a + a*sin(c + d*x)))