Integrand size = 29, antiderivative size = 87 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {5 a^3}{4 d (a-a \sin (c+d x))} \]
-7/8*a^2*ln(1-sin(d*x+c))/d-1/8*a^2*ln(1+sin(d*x+c))/d+1/4*a^4/d/(a-a*sin( d*x+c))^2-5/4*a^3/d/(a-a*sin(d*x+c))
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2 \left (3 \text {arctanh}(\sin (c+d x))-6 \sec ^3(c+d x) \tan (c+d x)+\sec (c+d x) \tan (c+d x) \left (3+8 \tan ^2(c+d x)\right )-2 \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)-\tan ^4(c+d x)\right )\right )}{4 d} \]
(a^2*(3*ArcTanh[Sin[c + d*x]] - 6*Sec[c + d*x]^3*Tan[c + d*x] + Sec[c + d* x]*Tan[c + d*x]*(3 + 8*Tan[c + d*x]^2) - 2*(2*Log[Cos[c + d*x]] + Tan[c + d*x]^2 - Tan[c + d*x]^4)))/(4*d)
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a \sin (c+d x)+a)^2}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\sin ^3(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^2 \int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a^2 \int \left (\frac {a^2}{2 (a-a \sin (c+d x))^3}-\frac {5 a}{4 (a-a \sin (c+d x))^2}+\frac {7}{8 (a-a \sin (c+d x))}-\frac {1}{8 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \left (\frac {a^2}{4 (a-a \sin (c+d x))^2}-\frac {5 a}{4 (a-a \sin (c+d x))}-\frac {7}{8} \log (a-a \sin (c+d x))-\frac {1}{8} \log (a \sin (c+d x)+a)\right )}{d}\) |
(a^2*((-7*Log[a - a*Sin[c + d*x]])/8 - Log[a + a*Sin[c + d*x]]/8 + a^2/(4* (a - a*Sin[c + d*x])^2) - (5*a)/(4*(a - a*Sin[c + d*x]))))/d
3.9.63.3.1 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 = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44
| method | result | size |
| risch | \(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {i \left (-8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2} {\mathrm e}^{i \left (d x +c \right )}+5 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\) | \(125\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(137\) |
| default | \(\frac {a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(137\) |
| parallelrisch | \(-\frac {7 \left (\frac {4}{7}+\frac {4 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{7}-\frac {4 \cos \left (2 d x +2 c \right )}{7}-\frac {6 \sin \left (d x +c \right )}{7}\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(151\) |
| norman | \(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {5 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {2 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(302\) |
I*a^2*x+2*I*a^2/d*c+1/2*I*(-8*I*a^2*exp(2*I*(d*x+c))-5*a^2*exp(I*(d*x+c))+ 5*a^2*exp(3*I*(d*x+c)))/(exp(I*(d*x+c))-I)^4/d-1/4*a^2/d*ln(exp(I*(d*x+c)) +I)-7/4*a^2/d*ln(exp(I*(d*x+c))-I)
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {10 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
-1/8*(10*a^2*sin(d*x + c) - 8*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) + 7*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]
-1/8*(a^2*log(sin(d*x + c) + 1) + 7*a^2*log(sin(d*x + c) - 1) - 2*(5*a^2*s in(d*x + c) - 4*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 14 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {21 \, a^{2} \sin \left (d x + c\right )^{2} - 22 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]
-1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) + 14*a^2*log(abs(sin(d*x + c) - 1) ) - (21*a^2*sin(d*x + c)^2 - 22*a^2*sin(d*x + c) + 5*a^2)/(sin(d*x + c) - 1)^2)/d
Time = 10.50 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.91 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {7\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{4\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{4\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]
(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (7*a^2*log(tan(c/2 + (d*x)/2) - 1) )/(4*d) - (a^2*log(tan(c/2 + (d*x)/2) + 1))/(4*d) - ((3*a^2*tan(c/2 + (d*x )/2)^3)/2 - 4*a^2*tan(c/2 + (d*x)/2)^2 + (3*a^2*tan(c/2 + (d*x)/2))/2)/(d* (6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 + 1))