3.74 \(\int \frac{\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=82 \[ -\frac{1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac{1}{8 a d (a \sin (c+d x)+a)^2}+\frac{1}{6 d (a \sin (c+d x)+a)^3} \]

[Out]

ArcTanh[Sin[c + d*x]]/(8*a^3*d) + 1/(6*d*(a + a*Sin[c + d*x])^3) - 1/(8*a*d*(a + a*Sin[c + d*x])^2) - 1/(8*d*(
a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.0571133, antiderivative size = 82, 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} \[ -\frac{1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac{1}{8 a d (a \sin (c+d x)+a)^2}+\frac{1}{6 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + a*Sin[c + d*x])^3,x]

[Out]

ArcTanh[Sin[c + d*x]]/(8*a^3*d) + 1/(6*d*(a + a*Sin[c + d*x])^3) - 1/(8*a*d*(a + a*Sin[c + d*x])^2) - 1/(8*d*(
a^3 + a^3*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))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(a-x) (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{2 (a+x)^4}+\frac{1}{4 a (a+x)^3}+\frac{1}{8 a^2 (a+x)^2}+\frac{1}{8 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{6 d (a+a \sin (c+d x))^3}-\frac{1}{8 a d (a+a \sin (c+d x))^2}-\frac{1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 a^2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac{1}{6 d (a+a \sin (c+d x))^3}-\frac{1}{8 a d (a+a \sin (c+d x))^2}-\frac{1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.145046, size = 52, normalized size = 0.63 \[ \frac{3 \tanh ^{-1}(\sin (c+d x))-\frac{3 \sin ^2(c+d x)+9 \sin (c+d x)+2}{(\sin (c+d x)+1)^3}}{24 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*ArcTanh[Sin[c + d*x]] - (2 + 9*Sin[c + d*x] + 3*Sin[c + d*x]^2)/(1 + Sin[c + d*x])^3)/(24*a^3*d)

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Maple [A]  time = 0.095, size = 90, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{16\,d{a}^{3}}}+{\frac{1}{6\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{8\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{8\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{16\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

-1/16/d/a^3*ln(sin(d*x+c)-1)+1/6/d/a^3/(1+sin(d*x+c))^3-1/8/d/a^3/(1+sin(d*x+c))^2-1/8/d/a^3/(1+sin(d*x+c))+1/
16*ln(1+sin(d*x+c))/a^3/d

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Maxima [A]  time = 2.14301, size = 132, normalized size = 1.61 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2\right )}}{a^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, 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))^3,x, algorithm="maxima")

[Out]

-1/48*(2*(3*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^3 + 3*a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x +
c) + a^3) - 3*log(sin(d*x + c) + 1)/a^3 + 3*log(sin(d*x + c) - 1)/a^3)/d

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Fricas [B]  time = 1.55306, size = 409, normalized size = 4.99 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, \sin \left (d x + c\right ) - 10}{48 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\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))^3,x, algorithm="fricas")

[Out]

-1/48*(6*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) +
3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(-sin(d*x + c) + 1) - 18*sin(d*x + c) - 10)/(
3*a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 4*a^3*d)*sin(d*x + c))

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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 ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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Giac [A]  time = 2.18984, size = 109, normalized size = 1.33 \begin{align*} \frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} - \frac{11 \, \sin \left (d x + c\right )^{3} + 45 \, \sin \left (d x + c\right )^{2} + 69 \, \sin \left (d x + c\right ) + 19}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{96 \, 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))^3,x, algorithm="giac")

[Out]

1/96*(6*log(abs(sin(d*x + c) + 1))/a^3 - 6*log(abs(sin(d*x + c) - 1))/a^3 - (11*sin(d*x + c)^3 + 45*sin(d*x +
c)^2 + 69*sin(d*x + c) + 19)/(a^3*(sin(d*x + c) + 1)^3))/d