3.64 \(\int \frac{\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=60 \[ -\frac{1}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}+\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

[Out]

ArcTanh[Sin[c + d*x]]/(4*a^2*d) + 1/(4*d*(a + a*Sin[c + d*x])^2) - 1/(4*d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.0483769, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2707, 77, 206} \[ -\frac{1}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}+\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + a*Sin[c + d*x])^2,x]

[Out]

ArcTanh[Sin[c + d*x]]/(4*a^2*d) + 1/(4*d*(a + a*Sin[c + d*x])^2) - 1/(4*d*(a^2 + a^2*Sin[c + d*x]))

Rule 2707

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(a-x) (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{2 (a+x)^3}+\frac{1}{4 a (a+x)^2}+\frac{1}{4 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{4 d (a+a \sin (c+d x))^2}-\frac{1}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 a d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}+\frac{1}{4 d (a+a \sin (c+d x))^2}-\frac{1}{4 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.0807824, size = 36, normalized size = 0.6 \[ \frac{\tanh ^{-1}(\sin (c+d x))-\frac{\sin (c+d x)}{(\sin (c+d x)+1)^2}}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x])^2,x]

[Out]

(ArcTanh[Sin[c + d*x]] - Sin[c + d*x]/(1 + Sin[c + d*x])^2)/(4*a^2*d)

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Maple [A]  time = 0.082, size = 72, normalized size = 1.2 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{8\,d{a}^{2}}}+{\frac{1}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{8\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

-1/8/d/a^2*ln(sin(d*x+c)-1)+1/4/d/a^2/(1+sin(d*x+c))^2-1/4/d/a^2/(1+sin(d*x+c))+1/8*ln(1+sin(d*x+c))/a^2/d

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Maxima [A]  time = 1.02483, size = 95, normalized size = 1.58 \begin{align*} -\frac{\frac{2 \, \sin \left (d x + c\right )}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*(2*sin(d*x + c)/(a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - log(sin(d*x + c) + 1)/a^2 + log(sin(d*x
+ c) - 1)/a^2)/d

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Fricas [A]  time = 1.45633, size = 274, normalized size = 4.57 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, \sin \left (d x + c\right )}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8*((cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) - (cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(-
sin(d*x + c) + 1) + 2*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.84808, size = 122, normalized size = 2.03 \begin{align*} \frac{\frac{\log \left ({\left | \frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right )}{a^{2}} - \frac{\log \left ({\left | \frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )}{a^{2}} - \frac{\frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 6}{a^{2}{\left (\frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2\right )}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(log(abs(1/sin(d*x + c) + sin(d*x + c) + 2))/a^2 - log(abs(1/sin(d*x + c) + sin(d*x + c) - 2))/a^2 - (1/s
in(d*x + c) + sin(d*x + c) + 6)/(a^2*(1/sin(d*x + c) + sin(d*x + c) + 2)))/d