\(\int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [1335]

Optimal result

Integrand size = 25, antiderivative size = 93 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {b^2 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2916, 12, 908} \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^2 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)}+\frac {\log (\sin (c+d x))}{a d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {\log (1-\sin (c+d x))}{a+b}-\frac {2 \log (\sin (c+d x))}{a}+\frac {\log (1+\sin (c+d x))}{a-b}-\frac {2 b^2 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )}}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(87\)
default	\(\frac {-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(87\)
parallelrisch	\(\frac {\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{2}-a \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (a +b \right ) \left (a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (a -b \right )\right )}{d \left (a^{3}-a \,b^{2}\right )}\)	\(106\)
norman	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {b^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a d \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}\)	\(114\)
risch	\(\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}-\frac {2 i b^{2} x}{a \left (a^{2}-b^{2}\right )}-\frac {2 i b^{2} c}{a d \left (a^{2}-b^{2}\right )}-\frac {2 i x}{a}-\frac {2 i c}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(226\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, b^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3} - a b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b - a b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int \frac {\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}+\frac {b^2\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a\,d\,\left (a^2-b^2\right )} \]

[In]

int(1/(cos(c + d*x)*sin(c + d*x)*(a + b*sin(c + d*x))),x)

[Out]

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