Integrand size = 21, antiderivative size = 132 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {a}{20 d (a+a \sin (c+d x))^5}-\frac {1}{8 d (a+a \sin (c+d x))^4}+\frac {1}{16 a d (a+a \sin (c+d x))^3}+\frac {1}{32 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {1}{64 d \left (a^4-a^4 \sin (c+d x)\right )}+\frac {1}{64 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
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))
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.38 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {1+4 \sin (c+d x)+5 \sin ^2(c+d x)}{20 a^4 d (-1+\sin (c+d x)) (1+\sin (c+d x))^5} \] Input:
Integrate[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]
Output:
-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)
Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3}{(a \sin (c+d x)+a)^4}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {a}{4 (\sin (c+d x) a+a)^6}+\frac {1}{2 (\sin (c+d x) a+a)^5}-\frac {3}{16 (\sin (c+d x) a+a)^4 a}-\frac {1}{16 (\sin (c+d x) a+a)^3 a^2}+\frac {1}{64 (a-a \sin (c+d x))^2 a^3}-\frac {1}{64 (\sin (c+d x) a+a)^2 a^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{64 a^3 (a-a \sin (c+d x))}+\frac {1}{64 a^3 (a \sin (c+d x)+a)}+\frac {1}{32 a^2 (a \sin (c+d x)+a)^2}+\frac {a}{20 (a \sin (c+d x)+a)^5}-\frac {1}{8 (a \sin (c+d x)+a)^4}+\frac {1}{16 a (a \sin (c+d x)+a)^3}}{d}\) |
Input:
Int[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]
Output:
(1/(64*a^3*(a - a*Sin[c + d*x])) + a/(20*(a + a*Sin[c + d*x])^5) - 1/(8*(a + a*Sin[c + d*x])^4) + 1/(16*a*(a + a*Sin[c + d*x])^3) + 1/(32*a^2*(a + a *Sin[c + d*x])^2) + 1/(64*a^3*(a + a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(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]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[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] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 5.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61
method | result | size |
derivativedivides | \(\frac {-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )}+\frac {1}{20 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{64+64 \sin \left (d x +c \right )}}{d \,a^{4}}\) | \(81\) |
default | \(\frac {-\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )}+\frac {1}{20 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{64+64 \sin \left (d x +c \right )}}{d \,a^{4}}\) | \(81\) |
risch | \(-\frac {4 \left (8 i {\mathrm e}^{7 i \left (d x +c \right )}+5 \,{\mathrm e}^{8 i \left (d x +c \right )}-8 i {\mathrm e}^{5 i \left (d x +c \right )}-14 \,{\mathrm e}^{6 i \left (d x +c \right )}+5 \,{\mathrm e}^{4 i \left (d x +c \right )}\right )}{5 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d \,a^{4}}\) | \(95\) |
Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
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)))
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {5 \, \cos \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) - 6}{20 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 8 \, a^{4} d \cos \left (d x + c\right )^{4} + 8 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
-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))
\[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(tan(d*x+c)**3/(a+a*sin(d*x+c))**4,x)
Output:
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
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) + 1}{20 \, {\left (a^{4} \sin \left (d x + c\right )^{6} + 4 \, a^{4} \sin \left (d x + c\right )^{5} + 5 \, a^{4} \sin \left (d x + c\right )^{4} - 5 \, a^{4} \sin \left (d x + c\right )^{2} - 4 \, a^{4} \sin \left (d x + c\right ) - a^{4}\right )} d} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
-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)
Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.58 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {1}{64 \, a^{4} d {\left (\sin \left (d x + c\right ) - 1\right )}} + \frac {5 \, \sin \left (d x + c\right )^{4} + 30 \, \sin \left (d x + c\right )^{3} + 80 \, \sin \left (d x + c\right )^{2} + 50 \, \sin \left (d x + c\right ) + 11}{320 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{5}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-1/64/(a^4*d*(sin(d*x + c) - 1)) + 1/320*(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*d*(sin(d*x + c) + 1) ^5)
Time = 18.03 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {56\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^2\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^{10}} \] Input:
int(tan(c + d*x)^3/(a + a*sin(c + d*x))^4,x)
Output:
(4*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 + (32*cos(c/2 + (d*x)/2)^5*si n(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)
Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin \left (d x +c \right )^{4} \left (-\sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )-5\right )}{20 a^{4} d \left (\sin \left (d x +c \right )^{6}+4 \sin \left (d x +c \right )^{5}+5 \sin \left (d x +c \right )^{4}-5 \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )-1\right )} \] Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c))^4,x)
Output:
(sin(c + d*x)**4*( - sin(c + d*x)**2 - 4*sin(c + d*x) - 5))/(20*a**4*d*(si n(c + d*x)**6 + 4*sin(c + d*x)**5 + 5*sin(c + d*x)**4 - 5*sin(c + d*x)**2 - 4*sin(c + d*x) - 1))