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

Optimal. Leaf size=37 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{1}{2 d (a \sin (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.051378, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2667, 44, 206} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{1}{2 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[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 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])

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\sec (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{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{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{1}{2 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0374701, size = 30, normalized size = 0.81 \[ \frac{\tanh ^{-1}(\sin (c+d x))-\frac{1}{\sin (c+d x)+1}}{2 a d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.052, size = 54, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{4\,da}}-{\frac{1}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{4\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.981991, size = 63, normalized size = 1.7 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.69868, size = 163, normalized size = 4.41 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.20812, size = 78, normalized size = 2.11 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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