3.1007 \(\int \frac{\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=45 \[ \frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{A-B}{2 d (a \sin (c+d x)+a)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0941355, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 77, 206} \[ \frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{A-B}{2 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.060957, size = 44, normalized size = 0.98 \[ \frac{(A+B) (\sin (c+d x)+1) \tanh ^{-1}(\sin (c+d x))-A+B}{2 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.092, size = 112, normalized size = 2.5 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{4\,da}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{4\,da}}-{\frac{A}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{B}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) B}{4\,da}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{4\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.017, size = 78, normalized size = 1.73 \begin{align*} \frac{\frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac{2 \,{\left (A - B\right )}}{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+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.71091, size = 207, normalized size = 4.6 \begin{align*} \frac{{\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, A + 2 \, B}{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+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.3048, size = 107, normalized size = 2.38 \begin{align*} \frac{\frac{{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac{A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) + 3 \, A - B}{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+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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