\(\int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) [254]

Optimal result

Integrand size = 26, antiderivative size = 75 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {a \log (b+a \cos (c+d x))}{\left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4482, 2747, 720, 31, 647} \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {a \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 720

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {(a-b) \log (1-\cos (c+d x))+(a+b) \log (1+\cos (c+d x))-2 a \log (b+a \cos (c+d x))}{2 (a-b) (a+b) d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {a \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(70\)
default	\(\frac {-\frac {a \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(70\)
risch	\(-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}+\frac {2 i a x}{a^{2}-b^{2}}+\frac {2 i a c}{\left (a^{2}-b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{\left (a^{2}-b^{2}\right ) d}\)	\(171\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, a \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a - b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {a \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} - b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.35 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, a \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{2} - b^{2}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {a\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^2-b^2\right )} \]

[In]

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

[Out]

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