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

Optimal. Leaf size=74 \[ \frac{a \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)} \]

[Out]

-Log[1 - Sin[c + d*x]]/(2*(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)

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Rubi [A]  time = 0.0661064, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2721, 801} \[ \frac{a \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - Sin[c + d*x]]/(2*(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 2721

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]

Rule 801

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]

Rubi steps

\begin{align*} \int \frac{\tan (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{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]  time = 0.0863228, size = 87, normalized size = 1.18 \[ \frac{a \log (a+b \sin (c+d x))+(b-a) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-(a+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 76, normalized size = 1. \begin{align*}{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.55148, size = 88, normalized size = 1.19 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(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

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Fricas [A]  time = 1.61071, size = 157, normalized size = 2.12 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3001, size = 96, normalized size = 1.3 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(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