\(\int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) [1548]

Optimal result

Integrand size = 29, antiderivative size = 90 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {(A+B) \log (1-\sin (c+d x))}{2 (a+b) d}+\frac {(A-B) \log (1+\sin (c+d x))}{2 (a-b) d}-\frac {(A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2916, 815} \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {(A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {(A+B) \log (1-\sin (c+d x))}{2 d (a+b)}+\frac {(A-B) \log (\sin (c+d x)+1)}{2 d (a-b)} \]

[In]

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

[Out]

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {(a-b) (A+B) \log (1-\sin (c+d x))-(a+b) (A-B) \log (1+\sin (c+d x))+2 (A b-a B) \log (a+b \sin (c+d x))}{2 (-a+b) (a+b) d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\)	\(89\)
default	\(\frac {\frac {\left (-A -B \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\left (A -B \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\)	\(89\)
parallelrisch	\(\frac {\left (-A b +B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a -b \right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a +b \right ) \left (A -B \right )}{d \left (a^{2}-b^{2}\right )}\)	\(96\)
norman	\(\frac {\left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}-\frac {\left (A b -B a \right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{2}-b^{2}\right )}\)	\(107\)
risch	\(-\frac {i A x}{a -b}+\frac {i B c}{\left (a +b \right ) d}+\frac {i B c}{d \left (a -b \right )}+\frac {2 i A b x}{a^{2}-b^{2}}+\frac {2 i A b c}{d \left (a^{2}-b^{2}\right )}+\frac {i B x}{a +b}+\frac {i A x}{a +b}+\frac {i A c}{\left (a +b \right ) d}-\frac {2 i B a x}{a^{2}-b^{2}}-\frac {2 i B a c}{d \left (a^{2}-b^{2}\right )}+\frac {i B x}{a -b}-\frac {i A c}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A b}{d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B a}{d \left (a^{2}-b^{2}\right )}\)	\(366\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (A - B\right )} a + {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]

[In]

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} + \frac {{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]

[In]

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (B a b - A b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} + \frac {{\left (A - B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]

[In]

integrate(sec(d*x+c)*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {A}{2}-\frac {B}{2}\right )}{d\,\left (a-b\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {A}{2}+\frac {B}{2}\right )}{d\,\left (a+b\right )} \]

[In]

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

[Out]

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