3.84 \(\int \frac{\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=105 \[ -\frac{1}{16 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{1}{16 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{16 a^4 d}-\frac{1}{12 a d (a \sin (c+d x)+a)^3}+\frac{1}{8 d (a \sin (c+d x)+a)^4} \]

[Out]

ArcTanh[Sin[c + d*x]]/(16*a^4*d) + 1/(8*d*(a + a*Sin[c + d*x])^4) - 1/(12*a*d*(a + a*Sin[c + d*x])^3) - 1/(16*
d*(a^2 + a^2*Sin[c + d*x])^2) - 1/(16*d*(a^4 + a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.0653541, antiderivative size = 105, 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} \[ -\frac{1}{16 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{1}{16 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{16 a^4 d}-\frac{1}{12 a d (a \sin (c+d x)+a)^3}+\frac{1}{8 d (a \sin (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + a*Sin[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(16*a^4*d) + 1/(8*d*(a + a*Sin[c + d*x])^4) - 1/(12*a*d*(a + a*Sin[c + d*x])^3) - 1/(16*
d*(a^2 + a^2*Sin[c + d*x])^2) - 1/(16*d*(a^4 + a^4*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))^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(a-x) (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{2 (a+x)^5}+\frac{1}{4 a (a+x)^4}+\frac{1}{8 a^2 (a+x)^3}+\frac{1}{16 a^3 (a+x)^2}+\frac{1}{16 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{1}{8 d (a+a \sin (c+d x))^4}-\frac{1}{12 a d (a+a \sin (c+d x))^3}-\frac{1}{16 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac{1}{16 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 a^3 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{16 a^4 d}+\frac{1}{8 d (a+a \sin (c+d x))^4}-\frac{1}{12 a d (a+a \sin (c+d x))^3}-\frac{1}{16 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac{1}{16 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.25388, size = 62, normalized size = 0.59 \[ \frac{3 \tanh ^{-1}(\sin (c+d x))-\frac{3 \sin ^3(c+d x)+12 \sin ^2(c+d x)+19 \sin (c+d x)+4}{(\sin (c+d x)+1)^4}}{48 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x])^4,x]

[Out]

(3*ArcTanh[Sin[c + d*x]] - (4 + 19*Sin[c + d*x] + 12*Sin[c + d*x]^2 + 3*Sin[c + d*x]^3)/(1 + Sin[c + d*x])^4)/
(48*a^4*d)

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Maple [A]  time = 0.092, size = 108, normalized size = 1. \begin{align*} -{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{32\,d{a}^{4}}}+{\frac{1}{8\,d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{12\,d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{16\,d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{16\,d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{32\,d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+a*sin(d*x+c))^4,x)

[Out]

-1/32/d/a^4*ln(sin(d*x+c)-1)+1/8/d/a^4/(1+sin(d*x+c))^4-1/12/d/a^4/(1+sin(d*x+c))^3-1/16/d/a^4/(1+sin(d*x+c))^
2-1/16/d/a^4/(1+sin(d*x+c))+1/32*ln(1+sin(d*x+c))/a^4/d

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Maxima [A]  time = 3.1573, size = 163, normalized size = 1.55 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} + 12 \, \sin \left (d x + c\right )^{2} + 19 \, \sin \left (d x + c\right ) + 4\right )}}{a^{4} \sin \left (d x + c\right )^{4} + 4 \, a^{4} \sin \left (d x + c\right )^{3} + 6 \, a^{4} \sin \left (d x + c\right )^{2} + 4 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4}}}{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))^4,x, algorithm="maxima")

[Out]

-1/96*(2*(3*sin(d*x + c)^3 + 12*sin(d*x + c)^2 + 19*sin(d*x + c) + 4)/(a^4*sin(d*x + c)^4 + 4*a^4*sin(d*x + c)
^3 + 6*a^4*sin(d*x + c)^2 + 4*a^4*sin(d*x + c) + a^4) - 3*log(sin(d*x + c) + 1)/a^4 + 3*log(sin(d*x + c) - 1)/
a^4)/d

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Fricas [B]  time = 1.5909, size = 525, normalized size = 5. \begin{align*} \frac{24 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 22\right )} \sin \left (d x + c\right ) - 32}{96 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} - 8 \, a^{4} d \cos \left (d x + c\right )^{2} + 8 \, a^{4} d - 4 \,{\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/96*(24*cos(d*x + c)^2 + 3*(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)*log(
sin(d*x + c) + 1) - 3*(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)*log(-sin(d
*x + c) + 1) + 2*(3*cos(d*x + c)^2 - 22)*sin(d*x + c) - 32)/(a^4*d*cos(d*x + c)^4 - 8*a^4*d*cos(d*x + c)^2 + 8
*a^4*d - 4*(a^4*d*cos(d*x + c)^2 - 2*a^4*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 ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin{\left (c + d x \right )} + 1}\, dx}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(tan(c + d*x)/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a**4

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Giac [A]  time = 2.14057, size = 123, normalized size = 1.17 \begin{align*} \frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4}} - \frac{25 \, \sin \left (d x + c\right )^{4} + 124 \, \sin \left (d x + c\right )^{3} + 246 \, \sin \left (d x + c\right )^{2} + 252 \, \sin \left (d x + c\right ) + 57}{a^{4}{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{384 \, 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))^4,x, algorithm="giac")

[Out]

1/384*(12*log(abs(sin(d*x + c) + 1))/a^4 - 12*log(abs(sin(d*x + c) - 1))/a^4 - (25*sin(d*x + c)^4 + 124*sin(d*
x + c)^3 + 246*sin(d*x + c)^2 + 252*sin(d*x + c) + 57)/(a^4*(sin(d*x + c) + 1)^4))/d