Integrand size = 19, antiderivative size = 82 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{8 a^3 d}+\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {1}{8 a d (a+a \sin (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )} \] Output:
1/8*arctanh(sin(d*x+c))/a^3/d+1/6/d/(a+a*sin(d*x+c))^3-1/8/a/d/(a+a*sin(d* x+c))^2-1/8/d/(a^3+a^3*sin(d*x+c))
Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \text {arctanh}(\sin (c+d x))-\frac {2+9 \sin (c+d x)+3 \sin ^2(c+d x)}{(1+\sin (c+d x))^3}}{24 a^3 d} \] Input:
Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x])^3,x]
Output:
(3*ArcTanh[Sin[c + d*x]] - (2 + 9*Sin[c + d*x] + 3*Sin[c + d*x]^2)/(1 + Si n[c + d*x])^3)/(24*a^3*d)
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 86, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a \sin (c+d x)}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-\frac {1}{2 (\sin (c+d x) a+a)^4}+\frac {1}{4 (\sin (c+d x) a+a)^3 a}+\frac {1}{8 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^2}+\frac {1}{8 (\sin (c+d x) a+a)^2 a^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\text {arctanh}(\sin (c+d x))}{8 a^3}-\frac {1}{8 a^2 (a \sin (c+d x)+a)}-\frac {1}{8 a (a \sin (c+d x)+a)^2}+\frac {1}{6 (a \sin (c+d x)+a)^3}}{d}\) |
Input:
Int[Tan[c + d*x]/(a + a*Sin[c + d*x])^3,x]
Output:
(ArcTanh[Sin[c + d*x]]/(8*a^3) + 1/(6*(a + a*Sin[c + d*x])^3) - 1/(8*a*(a + a*Sin[c + d*x])^2) - 1/(8*a^2*(a + a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.83 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\) | \(67\) |
default | \(\frac {\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\) | \(67\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{i \left (d x +c \right )}-14 \,{\mathrm e}^{3 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+18 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{12 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d \,a^{3}}\) | \(125\) |
Input:
int(tan(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/6/(1+sin(d*x+c))^3-1/8/(1+sin(d*x+c))^2-1/8/(1+sin(d*x+c))+1/16 *ln(1+sin(d*x+c))-1/16*ln(sin(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (74) = 148\).
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.88 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, \sin \left (d x + c\right ) - 10}{48 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/48*(6*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d *x + c) - 4)*log(sin(d*x + c) + 1) + 3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(-sin(d*x + c) + 1) - 18*sin(d*x + c) - 10)/(3* a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 4*a^3*d)*sin(d*x + c))
\[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\tan {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))**3,x)
Output:
Integral(tan(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x ) + 1), x)/a**3
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.20 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2\right )}}{a^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/48*(2*(3*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^3 + 3*a ^3*sin(d*x + c)^2 + 3*a^3*sin(d*x + c) + a^3) - 3*log(sin(d*x + c) + 1)/a^ 3 + 3*log(sin(d*x + c) - 1)/a^3)/d
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, a^{3} d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, a^{3} d} - \frac {3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2}{24 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \] Input:
integrate(tan(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
1/16*log(abs(sin(d*x + c) + 1))/(a^3*d) - 1/16*log(abs(sin(d*x + c) - 1))/ (a^3*d) - 1/24*(3*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)/(a^3*d*(sin(d*x + c ) + 1)^3)
Time = 19.52 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.27 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^3\,d}+\frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )} \] Input:
int(tan(c + d*x)/(a + a*sin(c + d*x))^3,x)
Output:
atanh(tan(c/2 + (d*x)/2))/(4*a^3*d) + (tan(c/2 + (d*x)/2)^2/2 - tan(c/2 + (d*x)/2)/4 + tan(c/2 + (d*x)/2)^3/6 + tan(c/2 + (d*x)/2)^4/2 - tan(c/2 + ( d*x)/2)^5/4)/(d*(15*a^3*tan(c/2 + (d*x)/2)^2 + 20*a^3*tan(c/2 + (d*x)/2)^3 + 15*a^3*tan(c/2 + (d*x)/2)^4 + 6*a^3*tan(c/2 + (d*x)/2)^5 + a^3*tan(c/2 + (d*x)/2)^6 + a^3 + 6*a^3*tan(c/2 + (d*x)/2)))
Time = 0.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.63 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+1}{24 a^{3} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(tan(d*x+c)/(a+a*sin(d*x+c))^3,x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 3*log(tan ((c + d*x)/2) - 1) + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 9*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 3*log(tan((c + d*x)/2) + 1) + 3*sin(c + d*x)**3 + 6*sin(c + d*x)** 2 + 1)/(24*a**3*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))