\(\int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\) [1015]

Optimal result

Integrand size = 29, antiderivative size = 71 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {(A+B) \text {arctanh}(\sin (c+d x))}{4 a^2 d}-\frac {A-B}{4 d (a+a \sin (c+d x))^2}-\frac {A+B}{4 d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 78, 212} \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {(A+B) \text {arctanh}(\sin (c+d x))}{4 a^2 d}-\frac {A+B}{4 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {A-B}{4 d (a \sin (c+d x)+a)^2} \]

[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 78

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 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 2915

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/b)*x
)^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]

Rubi steps \begin{align*} \text {integral}& = \frac {a \text {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 \text {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) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 a d} \\ & = \frac {(A+B) \text {arctanh}(\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] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {a \left (\frac {(A+B) \text {arctanh}(\sin (c+d x))}{4 a^3}-\frac {A-B}{4 a (a+a \sin (c+d x))^2}-\frac {A+B}{4 a^2 (a+a \sin (c+d x))}\right )}{d} \]

[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 [A] (verified)

Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {-\frac {\frac {A}{2}-\frac {B}{2}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {\frac {B}{4}+\frac {A}{4}}{1+\sin \left (d x +c \right )}+\left (\frac {A}{8}+\frac {B}{8}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {A}{8}-\frac {B}{8}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d \,a^{2}}\)	\(81\)
default	\(\frac {-\frac {\frac {A}{2}-\frac {B}{2}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {\frac {B}{4}+\frac {A}{4}}{1+\sin \left (d x +c \right )}+\left (\frac {A}{8}+\frac {B}{8}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {A}{8}-\frac {B}{8}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{d \,a^{2}}\)	\(81\)
parallelrisch	\(\frac {-\left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \cos \left (2 d x +2 c \right )+\left (-6 A +2 B \right ) \sin \left (d x +c \right )-2 A}{4 d \,a^{2} \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\)	\(131\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (4 i A \,{\mathrm e}^{i \left (d x +c \right )}+A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}-A -B \right )}{2 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{4 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{4 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 a^{2} d}\)	\(168\)
norman	\(\frac {\frac {2 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (3 A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (3 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (3 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{2} d}+\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{2} d}\)	\(196\)

[In]

int(sec(d*x+c)*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (65) = 130\).

Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.89 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left ({\left (A + B\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 2 \, B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 2 \, B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A + B\right )} \sin \left (d x + c\right ) + 4 \, A}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \]

[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]

\[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left ({\left (A + B\right )} \sin \left (d x + c\right ) + 2 \, A\right )}}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.46 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} - \frac {3 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} + 10 \, A \sin \left (d x + c\right ) + 10 \, B \sin \left (d x + c\right ) + 11 \, A + 3 \, B}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{16 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A+B\right )}{4\,a^2\,d}-\frac {\frac {A}{2}+\sin \left (c+d\,x\right )\,\left (\frac {A}{4}+\frac {B}{4}\right )}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]

[In]

int((A + B*sin(c + d*x))/(cos(c + d*x)*(a + a*sin(c + d*x))^2),x)

[Out]

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