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

Optimal. Leaf size=126 \[ \frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{1}{16 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac{a}{16 d (a \sin (c+d x)+a)^4}-\frac{1}{6 d (a \sin (c+d x)+a)^3}+\frac{3}{32 a d (a \sin (c+d x)+a)^2} \]

[Out]

-ArcTanh[Sin[c + d*x]]/(32*a^3*d) + a/(16*d*(a + a*Sin[c + d*x])^4) - 1/(6*d*(a + a*Sin[c + d*x])^3) + 3/(32*a
*d*(a + a*Sin[c + d*x])^2) + 1/(32*d*(a^3 - a^3*Sin[c + d*x])) + 1/(16*d*(a^3 + a^3*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.090512, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2707, 88, 206} \[ \frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{1}{16 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac{a}{16 d (a \sin (c+d x)+a)^4}-\frac{1}{6 d (a \sin (c+d x)+a)^3}+\frac{3}{32 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sin[c + d*x]]/(32*a^3*d) + a/(16*d*(a + a*Sin[c + d*x])^4) - 1/(6*d*(a + a*Sin[c + d*x])^3) + 3/(32*a
*d*(a + a*Sin[c + d*x])^2) + 1/(32*d*(a^3 - a^3*Sin[c + d*x])) + 1/(16*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(a-x)^2 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{32 a^2 (a-x)^2}-\frac{a}{4 (a+x)^5}+\frac{1}{2 (a+x)^4}-\frac{3}{16 a (a+x)^3}-\frac{1}{16 a^2 (a+x)^2}-\frac{1}{32 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a}{16 d (a+a \sin (c+d x))^4}-\frac{1}{6 d (a+a \sin (c+d x))^3}+\frac{3}{32 a d (a+a \sin (c+d x))^2}+\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{1}{16 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 )}{32 a^2 d}\\ &=-\frac{\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac{a}{16 d (a+a \sin (c+d x))^4}-\frac{1}{6 d (a+a \sin (c+d x))^3}+\frac{3}{32 a d (a+a \sin (c+d x))^2}+\frac{1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{1}{16 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.363215, size = 82, normalized size = 0.65 \[ -\frac{-\frac{3}{1-\sin (c+d x)}-\frac{6}{\sin (c+d x)+1}-\frac{9}{(\sin (c+d x)+1)^2}+\frac{16}{(\sin (c+d x)+1)^3}-\frac{6}{(\sin (c+d x)+1)^4}+3 \tanh ^{-1}(\sin (c+d x))}{96 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*ArcTanh[Sin[c + d*x]] - 3/(1 - Sin[c + d*x]) - 6/(1 + Sin[c + d*x])^4 + 16/(1 + Sin[c + d*x])^3 - 9/(1 + S
in[c + d*x])^2 - 6/(1 + Sin[c + d*x]))/(96*a^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.092, size = 126, normalized size = 1. \begin{align*} -{\frac{1}{32\,d{a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{64\,d{a}^{3}}}+{\frac{1}{16\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{1}{6\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{3}{32\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{16\,d{a}^{3} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{64\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/d/a^3/(sin(d*x+c)-1)+1/64/d/a^3*ln(sin(d*x+c)-1)+1/16/d/a^3/(1+sin(d*x+c))^4-1/6/d/a^3/(1+sin(d*x+c))^3+
3/32/d/a^3/(1+sin(d*x+c))^2+1/16/d/a^3/(1+sin(d*x+c))-1/64*ln(1+sin(d*x+c))/a^3/d

________________________________________________________________________________________

Maxima [A]  time = 1.93841, size = 197, normalized size = 1.56 \begin{align*} \frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 27 \, \sin \left (d x + c\right ) - 8\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, 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}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/192*(2*(3*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 27*sin(d*x + c) - 8)/(a^3*sin(d*x + c)^5 +
3*a^3*sin(d*x + c)^4 + 2*a^3*sin(d*x + c)^3 - 2*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

________________________________________________________________________________________

Fricas [A]  time = 1.58082, size = 586, normalized size = 4.65 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{4} + 38 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \,{\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 60}{192 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} +{\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\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)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/192*(6*cos(d*x + c)^4 + 38*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4*cos
(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) + 3*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 -
4*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 18*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 60)/(3*a^3*d*c
os(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{3}{\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)**3/(a+a*sin(d*x+c))**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 2.55233, size = 154, normalized size = 1.22 \begin{align*} -\frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac{12 \,{\left (\sin \left (d x + c\right ) + 1\right )}}{a^{3}{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{25 \, \sin \left (d x + c\right )^{4} + 148 \, \sin \left (d x + c\right )^{3} + 366 \, \sin \left (d x + c\right )^{2} + 260 \, \sin \left (d x + c\right ) + 65}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/768*(12*log(abs(sin(d*x + c) + 1))/a^3 - 12*log(abs(sin(d*x + c) - 1))/a^3 + 12*(sin(d*x + c) + 1)/(a^3*(si
n(d*x + c) - 1)) - (25*sin(d*x + c)^4 + 148*sin(d*x + c)^3 + 366*sin(d*x + c)^2 + 260*sin(d*x + c) + 65)/(a^3*
(sin(d*x + c) + 1)^4))/d