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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 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 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]

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 = 0.10, 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]

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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 102, normalized size = 0.77 \[ -\frac {5 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 6}{20 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 8 \, a^{4} d \cos \left (d x + c\right )^{4} + 8 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

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))

________________________________________________________________________________________

giac [A]  time = 2.34, size = 76, normalized size = 0.58 \[ -\frac {\frac {5}{a^{4} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {5 \, \sin \left (d x + c\right )^{4} + 30 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} + 50 \, \sin \left (d x + c\right ) + 11}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{320 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.24, size = 81, normalized size = 0.61 \[ \frac {-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )}+\frac {1}{20 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{64+64 \sin \left (d x +c \right )}}{d \,a^{4}} \]

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)))

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 95, normalized size = 0.72 \[ -\frac {5 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) + 1}{20 \, {\left (a^{4} \sin \left (d x + c\right )^{6} + 4 \, a^{4} \sin \left (d x + c\right )^{5} + 5 \, a^{4} \sin \left (d x + c\right )^{4} - 5 \, a^{4} \sin \left (d x + c\right )^{2} - 4 \, a^{4} \sin \left (d x + c\right ) - a^{4}\right )} d} \]

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)

________________________________________________________________________________________

mupad [B]  time = 7.51, size = 172, normalized size = 1.30 \[ \frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {56\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^2\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 + (32*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7)/5 + (56*cos(c/2
+ (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/5 + (32*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5)/5 + 4*cos(c/2 + (d*x)/2)
^8*sin(c/2 + (d*x)/2)^4)/(a^4*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^2*(cos(c/2 + (d*x)/2) + sin(c/2 + (d
*x)/2))^10)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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

________________________________________________________________________________________