Integrand size = 19, antiderivative size = 71 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {5 a \log (1-\sin (c+d x))}{4 d}-\frac {a \log (1+\sin (c+d x))}{4 d}+\frac {a \sin (c+d x)}{d}+\frac {a^2}{2 d (a-a \sin (c+d x))} \]
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {3 a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{2 d}-\frac {a \sin (c+d x) \tan ^2(c+d x)}{d}+\frac {a \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d} \]
(-3*a*ArcTanh[Sin[c + d*x]])/(2*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(2*d) - (a*Sin[c + d*x]*Tan[c + d*x]^2)/d + (a*(2*Log[Cos[c + d*x]] + Tan[c + d *x]^2))/(2*d)
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.
\(\displaystyle \int \tan ^3(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {a^2}{2 (a-a \sin (c+d x))^2}-\frac {5 a}{4 (a-a \sin (c+d x))}-\frac {a}{4 (\sin (c+d x) a+a)}+1\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2}{2 (a-a \sin (c+d x))}+a \sin (c+d x)+\frac {5}{4} a \log (a-a \sin (c+d x))-\frac {1}{4} a \log (a \sin (c+d x)+a)}{d}\) |
((5*a*Log[a - a*Sin[c + d*x]])/4 - (a*Log[a + a*Sin[c + d*x]])/4 + a*Sin[c + d*x] + a^2/(2*(a - a*Sin[c + d*x])))/d
3.1.2.3.1 Defintions of rubi rules used
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 = 0.85 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(81\) |
default | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(81\) |
parts | \(\frac {a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(89\) |
risch | \(-i a x -\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{\left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {5 a \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}\) | \(115\) |
1/d*(a*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2* ln(sec(d*x+c)+tan(d*x+c)))+a*(1/2*tan(d*x+c)^2+ln(cos(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} + {\left (a \sin \left (d x + c\right ) - a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, {\left (a \sin \left (d x + c\right ) - a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a \sin \left (d x + c\right ) - 2 \, a}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]
-1/4*(4*a*cos(d*x + c)^2 + (a*sin(d*x + c) - a)*log(sin(d*x + c) + 1) - 5* (a*sin(d*x + c) - a)*log(-sin(d*x + c) + 1) + 4*a*sin(d*x + c) - 2*a)/(d*s in(d*x + c) - d)
\[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \]
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {a \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \]
-1/4*(a*log(sin(d*x + c) + 1) - 5*a*log(sin(d*x + c) - 1) - 4*a*sin(d*x + c) + 2*a/(sin(d*x + c) - 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 22388 vs. \(2 (66) = 132\).
Time = 105.66 (sec) , antiderivative size = 22388, normalized size of antiderivative = 315.32 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\text {Too large to display} \]
1/4*(3*a*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2* d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + t an(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 - 3*a*lo g(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2 *d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*ta n(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 + 2*a*log(4*(tan(d* x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 + 2*a*tan( d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 - 6*a*log(2*(tan(1/2*d*x)^2*ta n(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + t an(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2 *d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)*tan(1/ 2*d*x)^6*tan(1/2*c)^6*tan(c) + 6*a*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2* tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2 *c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(d*x)*tan(1/2*d*x)^6*tan(1/ 2*c)^6*tan(c) - 4*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(t an(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(1/2*d*x)^...
Time = 6.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.17 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{2\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{2\,d}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
(5*a*log(tan(c/2 + (d*x)/2) - 1))/(2*d) - (a*log(tan(c/2 + (d*x)/2) + 1))/ (2*d) + (3*a*tan(c/2 + (d*x)/2) - 4*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^3)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 + 1)) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d