3.83 \(\int \frac{\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=132 \[ \frac{1}{64 d \left (a^4-a^4 \sin (c+d x)\right )}+\frac{1}{64 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac{1}{32 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac{a}{20 d (a \sin (c+d x)+a)^5}-\frac{1}{8 d (a \sin (c+d x)+a)^4}+\frac{1}{16 a d (a \sin (c+d x)+a)^3} \]

[Out]

a/(20*d*(a + a*Sin[c + d*x])^5) - 1/(8*d*(a + a*Sin[c + d*x])^4) + 1/(16*a*d*(a + a*Sin[c + d*x])^3) + 1/(32*d
*(a^2 + a^2*Sin[c + d*x])^2) + 1/(64*d*(a^4 - a^4*Sin[c + d*x])) + 1/(64*d*(a^4 + a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.0887882, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2707, 88} \[ \frac{1}{64 d \left (a^4-a^4 \sin (c+d x)\right )}+\frac{1}{64 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac{1}{32 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac{a}{20 d (a \sin (c+d x)+a)^5}-\frac{1}{8 d (a \sin (c+d x)+a)^4}+\frac{1}{16 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]

[Out]

a/(20*d*(a + a*Sin[c + d*x])^5) - 1/(8*d*(a + a*Sin[c + d*x])^4) + 1/(16*a*d*(a + a*Sin[c + d*x])^3) + 1/(32*d
*(a^2 + a^2*Sin[c + d*x])^2) + 1/(64*d*(a^4 - a^4*Sin[c + d*x])) + 1/(64*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 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]))

Rubi steps

\begin{align*} \int \frac{\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(a-x)^2 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{64 a^3 (a-x)^2}-\frac{a}{4 (a+x)^6}+\frac{1}{2 (a+x)^5}-\frac{3}{16 a (a+x)^4}-\frac{1}{16 a^2 (a+x)^3}-\frac{1}{64 a^3 (a+x)^2}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a}{20 d (a+a \sin (c+d x))^5}-\frac{1}{8 d (a+a \sin (c+d x))^4}+\frac{1}{16 a d (a+a \sin (c+d x))^3}+\frac{1}{32 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac{1}{64 d \left (a^4-a^4 \sin (c+d x)\right )}+\frac{1}{64 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.115064, size = 50, normalized size = 0.38 \[ -\frac{5 \sin ^2(c+d x)+4 \sin (c+d x)+1}{20 a^4 d (\sin (c+d x)-1) (\sin (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]

[Out]

-(1 + 4*Sin[c + d*x] + 5*Sin[c + d*x]^2)/(20*a^4*d*(-1 + Sin[c + d*x])*(1 + Sin[c + d*x])^5)

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Maple [A]  time = 0.088, size = 81, normalized size = 0.6 \begin{align*}{\frac{1}{d{a}^{4}} \left ( -{\frac{1}{64\,\sin \left ( dx+c \right ) -64}}+{\frac{1}{20\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{8\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{16\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{1}{32\, \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{1}{64+64\,\sin \left ( dx+c \right ) }} \right ) } \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))^4,x)

[Out]

1/d/a^4*(-1/64/(sin(d*x+c)-1)+1/20/(1+sin(d*x+c))^5-1/8/(1+sin(d*x+c))^4+1/16/(1+sin(d*x+c))^3+1/32/(1+sin(d*x
+c))^2+1/64/(1+sin(d*x+c)))

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Maxima [A]  time = 2.93891, size = 128, normalized size = 0.97 \begin{align*} -\frac{5 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) + 1}{20 \,{\left (a^{4} \sin \left (d x + c\right )^{6} + 4 \, a^{4} \sin \left (d x + c\right )^{5} + 5 \, a^{4} \sin \left (d x + c\right )^{4} - 5 \, a^{4} \sin \left (d x + c\right )^{2} - 4 \, a^{4} \sin \left (d x + c\right ) - a^{4}\right )} 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))^4,x, algorithm="maxima")

[Out]

-1/20*(5*sin(d*x + c)^2 + 4*sin(d*x + c) + 1)/((a^4*sin(d*x + c)^6 + 4*a^4*sin(d*x + c)^5 + 5*a^4*sin(d*x + c)
^4 - 5*a^4*sin(d*x + c)^2 - 4*a^4*sin(d*x + c) - a^4)*d)

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Fricas [A]  time = 1.35404, size = 250, normalized size = 1.89 \begin{align*} -\frac{5 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 6}{20 \,{\left (a^{4} d \cos \left (d x + c\right )^{6} - 8 \, a^{4} d \cos \left (d x + c\right )^{4} + 8 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/20*(5*cos(d*x + c)^2 - 4*sin(d*x + c) - 6)/(a^4*d*cos(d*x + c)^6 - 8*a^4*d*cos(d*x + c)^4 + 8*a^4*d*cos(d*x
+ c)^2 - 4*(a^4*d*cos(d*x + c)^4 - 2*a^4*d*cos(d*x + c)^2)*sin(d*x + c))

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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 ^{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)**3/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(tan(c + d*x)**3/(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 = 2.77507, size = 103, normalized size = 0.78 \begin{align*} -\frac{\frac{5}{a^{4}{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{5 \, \sin \left (d x + c\right )^{4} + 30 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} + 50 \, \sin \left (d x + c\right ) + 11}{a^{4}{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{320 \, 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))^4,x, algorithm="giac")

[Out]

-1/320*(5/(a^4*(sin(d*x + c) - 1)) - (5*sin(d*x + c)^4 + 30*sin(d*x + c)^3 + 80*sin(d*x + c)^2 + 50*sin(d*x +
c) + 11)/(a^4*(sin(d*x + c) + 1)^5))/d