\(\int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\) [1007]

Optimal result

Integrand size = 29, antiderivative size = 45 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(A+B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {A-B}{2 d (a+a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, 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} \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(A+B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {A-B}{2 d (a \sin (c+d x)+a)} \]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {-A+B+(A+B) \text {arctanh}(\sin (c+d x)) (1+\sin (c+d x))}{2 a d (1+\sin (c+d x))} \]

[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 [A] (verified)

Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\left (-\frac {A}{4}-\frac {B}{4}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {\frac {A}{2}-\frac {B}{2}}{1+\sin \left (d x +c \right )}+\left (\frac {B}{4}+\frac {A}{4}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(62\)
default	\(\frac {\left (-\frac {A}{4}-\frac {B}{4}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {\frac {A}{2}-\frac {B}{2}}{1+\sin \left (d x +c \right )}+\left (\frac {B}{4}+\frac {A}{4}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(62\)
parallelrisch	\(\frac {-\left (1+\sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\sin \left (d x +c \right )\right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right ) \left (A -B \right )}{2 a d \left (1+\sin \left (d x +c \right )\right )}\)	\(81\)
norman	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(122\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A -B \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a d}\)	\(127\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {{\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} \sin \left (d x + c\right ) + A + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, A + 2 \, B}{4 \, {\left (a d \sin \left (d x + c\right ) + a d\right )}} \]

[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]

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

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (A - B\right )}}{a \sin \left (d x + c\right ) + a}}{4 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.76 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) + 3 \, A - B}{a {\left (\sin \left (d x + c\right ) + 1\right )}}}{4 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A+B\right )}{2\,a\,d}-\frac {\frac {A}{2}-\frac {B}{2}}{d\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

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