3.1015 \(\int \frac{\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=71 \[ -\frac{A+B}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{A-B}{4 d (a \sin (c+d x)+a)^2} \]

[Out]

((A + B)*ArcTanh[Sin[c + d*x]])/(4*a^2*d) - (A - B)/(4*d*(a + a*Sin[c + d*x])^2) - (A + B)/(4*d*(a^2 + a^2*Sin
[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.1075, antiderivative size = 71, 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} \[ -\frac{A+B}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{A-B}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

((A + B)*ArcTanh[Sin[c + d*x]])/(4*a^2*d) - (A - B)/(4*d*(a + a*Sin[c + d*x])^2) - (A + B)/(4*d*(a^2 + a^2*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))^2} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{A+\frac{B x}{a}}{(a-x) (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (\frac{A-B}{2 a (a+x)^3}+\frac{A+B}{4 a^2 (a+x)^2}+\frac{A+B}{4 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{A-B}{4 d (a+a \sin (c+d x))^2}-\frac{A+B}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{(A+B) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 a d}\\ &=\frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac{A-B}{4 d (a+a \sin (c+d x))^2}-\frac{A+B}{4 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.119526, size = 69, normalized size = 0.97 \[ \frac{a \left (-\frac{A+B}{4 a^2 (a \sin (c+d x)+a)}+\frac{(A+B) \tanh ^{-1}(\sin (c+d x))}{4 a^3}-\frac{A-B}{4 a (a \sin (c+d x)+a)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

(a*(((A + B)*ArcTanh[Sin[c + d*x]])/(4*a^3) - (A - B)/(4*a*(a + a*Sin[c + d*x])^2) - (A + B)/(4*a^2*(a + a*Sin
[c + d*x]))))/d

________________________________________________________________________________________

Maple [B]  time = 0.115, size = 150, normalized size = 2.1 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{8\,d{a}^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{8\,d{a}^{2}}}-{\frac{A}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{A}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{B}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{8\,d{a}^{2}}}+{\frac{B\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{8\,d{a}^{2}}} \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))^2,x)

[Out]

-1/8/d/a^2*ln(sin(d*x+c)-1)*A-1/8/d/a^2*ln(sin(d*x+c)-1)*B-1/4/d/a^2/(1+sin(d*x+c))^2*A+1/4/d/a^2/(1+sin(d*x+c
))^2*B-1/4/d/a^2/(1+sin(d*x+c))*A-1/4/d/a^2/(1+sin(d*x+c))*B+1/8/d/a^2*ln(1+sin(d*x+c))*A+1/8*B*ln(1+sin(d*x+c
))/a^2/d

________________________________________________________________________________________

Maxima [A]  time = 1.04265, size = 113, normalized size = 1.59 \begin{align*} -\frac{\frac{2 \,{\left ({\left (A + B\right )} \sin \left (d x + c\right ) + 2 \, A\right )}}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, 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))^2,x, algorithm="maxima")

[Out]

-1/8*(2*((A + B)*sin(d*x + c) + 2*A)/(a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - (A + B)*log(sin(d*x + c
) + 1)/a^2 + (A + B)*log(sin(d*x + c) - 1)/a^2)/d

________________________________________________________________________________________

Fricas [B]  time = 1.53666, size = 358, normalized size = 5.04 \begin{align*} \frac{{\left ({\left (A + B\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 2 \, B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left ({\left (A + B\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 2 \, B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A + B\right )} \sin \left (d x + c\right ) + 4 \, A}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/8*(((A + B)*cos(d*x + c)^2 - 2*(A + B)*sin(d*x + c) - 2*A - 2*B)*log(sin(d*x + c) + 1) - ((A + B)*cos(d*x +
c)^2 - 2*(A + B)*sin(d*x + c) - 2*A - 2*B)*log(-sin(d*x + c) + 1) + 2*(A + B)*sin(d*x + c) + 4*A)/(a^2*d*cos(d
*x + c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sin{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

(Integral(A*sec(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x) + Integral(B*sin(c + d*x)*sec(c + d*x)/(si
n(c + d*x)**2 + 2*sin(c + d*x) + 1), x))/a**2

________________________________________________________________________________________

Giac [A]  time = 1.38443, size = 140, normalized size = 1.97 \begin{align*} \frac{\frac{2 \,{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{2 \,{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} - \frac{3 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} + 10 \, A \sin \left (d x + c\right ) + 10 \, B \sin \left (d x + c\right ) + 11 \, A + 3 \, B}{a^{2}{\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{16 \, 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))^2,x, algorithm="giac")

[Out]

1/16*(2*(A + B)*log(abs(sin(d*x + c) + 1))/a^2 - 2*(A + B)*log(abs(sin(d*x + c) - 1))/a^2 - (3*A*sin(d*x + c)^
2 + 3*B*sin(d*x + c)^2 + 10*A*sin(d*x + c) + 10*B*sin(d*x + c) + 11*A + 3*B)/(a^2*(sin(d*x + c) + 1)^2))/d