Integrand size = 21, antiderivative size = 104 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{8 a^2 d}+\frac {a}{12 d (a+a \sin (c+d x))^3}-\frac {1}{4 d (a+a \sin (c+d x))^2}+\frac {1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {3}{16 d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
-1/8*arctanh(sin(d*x+c))/a^2/d+1/12*a/d/(a+a*sin(d*x+c))^3-1/4/d/(a+a*sin( d*x+c))^2+1/16/d/(a^2-a^2*sin(d*x+c))+3/16/d/(a^2+a^2*sin(d*x+c))
Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {6 \text {arctanh}(\sin (c+d x))-\frac {3}{1-\sin (c+d x)}-\frac {4}{(1+\sin (c+d x))^3}+\frac {12}{(1+\sin (c+d x))^2}-\frac {9}{1+\sin (c+d x)}}{48 a^2 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]
Output:
-1/48*(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 + Sin[c + d*x]))/(a^2*d)
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a}{4 (\sin (c+d x) a+a)^4}+\frac {1}{2 (\sin (c+d x) a+a)^3}-\frac {1}{8 \left (a^2-a^2 \sin ^2(c+d x)\right ) a}+\frac {1}{16 (a-a \sin (c+d x))^2 a}-\frac {3}{16 (\sin (c+d x) a+a)^2 a}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\text {arctanh}(\sin (c+d x))}{8 a^2}+\frac {a}{12 (a \sin (c+d x)+a)^3}-\frac {1}{4 (a \sin (c+d x)+a)^2}+\frac {1}{16 a (a-a \sin (c+d x))}+\frac {3}{16 a (a \sin (c+d x)+a)}}{d}\) |
Input:
Int[Tan[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]
Output:
(-1/8*ArcTanh[Sin[c + d*x]]/a^2 + 1/(16*a*(a - a*Sin[c + d*x])) + a/(12*(a + a*Sin[c + d*x])^3) - 1/(4*(a + a*Sin[c + d*x])^2) + 3/(16*a*(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 = 0.88 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16}+\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16}}{d \,a^{2}}\) | \(79\) |
| default | \(\frac {-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16}+\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16}}{d \,a^{2}}\) | \(79\) |
| risch | \(\frac {i \left (-19 \,{\mathrm e}^{3 i \left (d x +c \right )}-12 i {\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}+19 \,{\mathrm e}^{5 i \left (d x +c \right )}+40 i {\mathrm e}^{4 i \left (d x +c \right )}-12 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d \,a^{2}}\) | \(162\) |
Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(-1/16/(sin(d*x+c)-1)+1/16*ln(sin(d*x+c)-1)+1/12/(1+sin(d*x+c))^3- 1/4/(1+sin(d*x+c))^2+3/16/(1+sin(d*x+c))-1/16*ln(1+sin(d*x+c)))
Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.71 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {12 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 4\right )} \sin \left (d x + c\right ) - 16}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
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*co s(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*cos(d*x + c)^2)
\[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(tan(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) - 2\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{48 \, d} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/48*(2*(3*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 7*sin(d*x + c) - 2)/(a^2*si n(d*x + c)^4 + 2*a^2*sin(d*x + c)^3 - 2*a^2*sin(d*x + c) - a^2) - 3*log(si n(d*x + c) + 1)/a^2 + 3*log(sin(d*x + c) - 1)/a^2)/d
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, a^{2} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, a^{2} d} + \frac {3 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) - 2}{24 \, a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/16*log(abs(sin(d*x + c) + 1))/(a^2*d) + 1/16*log(abs(sin(d*x + c) - 1)) /(a^2*d) + 1/24*(3*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 7*sin(d*x + c) - 2) /(a^2*d*(sin(d*x + c) + 1)^3*(sin(d*x + c) - 1))
Time = 19.98 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.31 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-10\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^2\,d} \] Input:
int(tan(c + d*x)^3/(a + a*sin(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)/4 + tan(c/2 + (d*x)/2)^2 + (13*tan(c/2 + (d*x)/2)^3)/1 2 + (10*tan(c/2 + (d*x)/2)^4)/3 + (13*tan(c/2 + (d*x)/2)^5)/12 + tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7/4)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 - 4*a ^2*tan(c/2 + (d*x)/2)^3 - 10*a^2*tan(c/2 + (d*x)/2)^4 - 4*a^2*tan(c/2 + (d *x)/2)^5 + 4*a^2*tan(c/2 + (d*x)/2)^6 + 4*a^2*tan(c/2 + (d*x)/2)^7 + a^2*t an(c/2 + (d*x)/2)^8 + a^2 + 4*a^2*tan(c/2 + (d*x)/2))) - atanh(tan(c/2 + ( d*x)/2))/(4*a^2*d)
Time = 0.18 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.17 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-7 \sin \left (d x +c \right )^{4}-8 \sin \left (d x +c \right )^{3}-12 \sin \left (d x +c \right )^{2}+3}{48 a^{2} d \left (\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )-1\right )} \] Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
Output:
(6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 12*log(tan((c + d*x)/2) - 1 )*sin(c + d*x)**3 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 6*log(tan( (c + d*x)/2) - 1) - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 12*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**3 + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 6*log(tan((c + d*x)/2) + 1) - 7*sin(c + d*x)**4 - 8*sin(c + d*x)* *3 - 12*sin(c + d*x)**2 + 3)/(48*a**2*d*(sin(c + d*x)**4 + 2*sin(c + d*x)* *3 - 2*sin(c + d*x) - 1))