3.63 \(\int \frac{\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=104 \[ \frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{3}{16 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\tanh ^{-1}(\sin (c+d x))}{8 a^2 d}+\frac{a}{12 d (a \sin (c+d x)+a)^3}-\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.085784, antiderivative size = 104, 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} \[ \frac{1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{3}{16 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\tanh ^{-1}(\sin (c+d x))}{8 a^2 d}+\frac{a}{12 d (a \sin (c+d x)+a)^3}-\frac{1}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.310755, size = 70, normalized size = 0.67 \[ -\frac{-\frac{3}{1-\sin (c+d x)}-\frac{9}{\sin (c+d x)+1}+\frac{12}{(\sin (c+d x)+1)^2}-\frac{4}{(\sin (c+d x)+1)^3}+6 \tanh ^{-1}(\sin (c+d x))}{48 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]

[Out]

-(6*ArcTanh[Sin[c + d*x]] - 3/(1 - Sin[c + d*x]) - 4/(1 + Sin[c + d*x])^3 + 12/(1 + Sin[c + d*x])^2 - 9/(1 + S
in[c + d*x]))/(48*a^2*d)

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Maple [A]  time = 0.083, size = 108, normalized size = 1. \begin{align*} -{\frac{1}{16\,d{a}^{2} \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{16\,d{a}^{2}}}+{\frac{1}{12\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{4\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3}{16\,d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{16\,d{a}^{2}}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.05494, size = 149, normalized size = 1.43 \begin{align*} \frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) - 2\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{48 \, 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))^2,x, algorithm="maxima")

[Out]

1/48*(2*(3*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 7*sin(d*x + c) - 2)/(a^2*sin(d*x + c)^4 + 2*a^2*sin(d*x + c)^3
- 2*a^2*sin(d*x + c) - a^2) - 3*log(sin(d*x + c) + 1)/a^2 + 3*log(sin(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.53143, size = 467, normalized size = 4.49 \begin{align*} \frac{12 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) - 16}{48 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\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))^2,x, algorithm="fricas")

[Out]

1/48*(12*cos(d*x + c)^2 - 3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) + 3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(3*
cos(d*x + c)^2 + 4)*sin(d*x + c) - 16)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*c
os(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{3}{\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)**3/(a+a*sin(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 2.99776, size = 138, normalized size = 1.33 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \, \sin \left (d x + c\right )}{a^{2}{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{11 \, \sin \left (d x + c\right )^{3} + 51 \, \sin \left (d x + c\right )^{2} + 45 \, \sin \left (d x + c\right ) + 13}{a^{2}{\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)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/96*(6*log(abs(sin(d*x + c) + 1))/a^2 - 6*log(abs(sin(d*x + c) - 1))/a^2 + 6*sin(d*x + c)/(a^2*(sin(d*x + c)
- 1)) - (11*sin(d*x + c)^3 + 51*sin(d*x + c)^2 + 45*sin(d*x + c) + 13)/(a^2*(sin(d*x + c) + 1)^3))/d