Integrand size = 27, antiderivative size = 101 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (1+\sin (c+d x))}{8 d}-\frac {a^2 \sin (c+d x)}{d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {7 a^3}{4 d (a-a \sin (c+d x))} \]
-17/8*a^2*ln(1-sin(d*x+c))/d+1/8*a^2*ln(1+sin(d*x+c))/d-a^2*sin(d*x+c)/d+1 /4*a^4/d/(a-a*sin(d*x+c))^2-7/4*a^3/d/(a-a*sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {a^2 \left (17 \log (1-\sin (c+d x))-\log (1+\sin (c+d x))-\frac {2}{(-1+\sin (c+d x))^2}-\frac {14}{-1+\sin (c+d x)}+8 \sin (c+d x)\right )}{8 d} \]
-1/8*(a^2*(17*Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]] - 2/(-1 + Sin[ c + d*x])^2 - 14/(-1 + Sin[c + d*x]) + 8*Sin[c + d*x]))/d
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 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.
| \(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a \sin (c+d x)+a)^2}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\sin ^4(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {a^4 \sin ^4(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (\frac {a^3}{2 (a-a \sin (c+d x))^3}-\frac {7 a^2}{4 (a-a \sin (c+d x))^2}+\frac {17 a}{8 (a-a \sin (c+d x))}+\frac {a}{8 (\sin (c+d x) a+a)}-1\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {a^3}{4 (a-a \sin (c+d x))^2}-\frac {7 a^2}{4 (a-a \sin (c+d x))}-a \sin (c+d x)-\frac {17}{8} a \log (a-a \sin (c+d x))+\frac {1}{8} a \log (a \sin (c+d x)+a)\right )}{d}\) |
(a*((-17*a*Log[a - a*Sin[c + d*x]])/8 + (a*Log[a + a*Sin[c + d*x]])/8 - a* Sin[c + d*x] + a^3/(4*(a - a*Sin[c + d*x])^2) - (7*a^2)/(4*(a - a*Sin[c + d*x]))))/d
3.9.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.53
| method | result | size |
| risch | \(2 i a^{2} x +\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {4 i a^{2} c}{d}+\frac {i a^{2} \left (-12 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-7 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}-\frac {17 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\) | \(155\) |
| parallelrisch | \(\frac {2 \left (-\frac {7}{4}+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {17 \left (\frac {3}{4}-\frac {\cos \left (2 d x +2 c \right )}{4}-\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {\left (-\frac {3}{4}+\frac {\cos \left (2 d x +2 c \right )}{4}+\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {7 \cos \left (2 d x +2 c \right )}{4}+3 \sin \left (d x +c \right )-\frac {\sin \left (3 d x +3 c \right )}{4}\right ) a^{2}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(162\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(201\) |
| default | \(\frac {a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(201\) |
| norman | \(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {15 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {9 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {24 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {17 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}+\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(303\) |
2*I*a^2*x+1/2*I*a^2/d*exp(I*(d*x+c))-1/2*I*a^2/d*exp(-I*(d*x+c))+4*I*a^2/d *c+1/2*I*a^2/d/(exp(I*(d*x+c))-I)^4*(-12*I*exp(2*I*(d*x+c))+7*exp(3*I*(d*x +c))-7*exp(I*(d*x+c)))-17/4*a^2/d*ln(exp(I*(d*x+c))-I)+1/4*a^2/d*ln(exp(I* (d*x+c))+I)
Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {16 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
1/8*(16*a^2*cos(d*x + c)^2 - 4*a^2 + (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(sin(d*x + c) + 1) - 17*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d* x + c) - 2*a^2)*log(-sin(d*x + c) + 1) - 2*(4*a^2*cos(d*x + c)^2 - a^2)*si n(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, a^{2} \sin \left (d x + c\right ) + \frac {2 \, {\left (7 \, a^{2} \sin \left (d x + c\right ) - 6 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]
1/8*(a^2*log(sin(d*x + c) + 1) - 17*a^2*log(sin(d*x + c) - 1) - 8*a^2*sin( d*x + c) + 2*(7*a^2*sin(d*x + c) - 6*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 34 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a^{2} \sin \left (d x + c\right ) + \frac {51 \, a^{2} \sin \left (d x + c\right )^{2} - 74 \, a^{2} \sin \left (d x + c\right ) + 27 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]
1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 34*a^2*log(abs(sin(d*x + c) - 1)) - 16*a^2*sin(d*x + c) + (51*a^2*sin(d*x + c)^2 - 74*a^2*sin(d*x + c) + 27 *a^2)/(sin(d*x + c) - 1)^2)/d
Time = 9.85 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.23 \[ \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{4\,d}-\frac {17\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{4\,d}-\frac {\frac {9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}+\frac {2\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
(a^2*log(tan(c/2 + (d*x)/2) + 1))/(4*d) - (17*a^2*log(tan(c/2 + (d*x)/2) - 1))/(4*d) - (17*a^2*tan(c/2 + (d*x)/2)^3 - 14*a^2*tan(c/2 + (d*x)/2)^2 - 14*a^2*tan(c/2 + (d*x)/2)^4 + (9*a^2*tan(c/2 + (d*x)/2)^5)/2 + (9*a^2*tan( c/2 + (d*x)/2))/2)/(d*(7*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 8*t an(c/2 + (d*x)/2)^3 + 7*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^5 + ta n(c/2 + (d*x)/2)^6 + 1)) + (2*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d