\(\int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx\) [1334]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2800, 815} \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \]

[In]

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

[Out]

-1/2*Log[1 - Sin[c + d*x]]/((a + b)*d) - Log[1 + Sin[c + d*x]]/(2*(a - b)*d) + (a*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 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {(-a+b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 a \log (a+b \sin (c+d x))}{2 (a-b) (a+b) d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\)	\(71\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}+\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right )}}{d}\)	\(71\)
risch	\(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {2 i a x}{a^{2}-b^{2}}-\frac {2 i a c}{d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{2}-b^{2}\right )}\)	\(175\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sin(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 = 65, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.55 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int \frac {\tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )} \]

[In]

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

[Out]

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