Integrand size = 27, antiderivative size = 111 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {51 a^3 x}{8}+\frac {7 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {19 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d} \]
-51/8*a^3*x+7*a^3*cos(d*x+c)/d-a^3*cos(d*x+c)^3/d+4*a^3*cos(d*x+c)/d/(1-si n(d*x+c))+19/8*a^3*cos(d*x+c)*sin(d*x+c)/d+1/4*a^3*cos(d*x+c)*sin(d*x+c)^3 /d
Time = 5.80 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {(a+a \sin (c+d x))^3 \left (-204 (c+d x)+200 \cos (c+d x)-8 \cos (3 (c+d x))+\frac {256 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+40 \sin (2 (c+d x))-\sin (4 (c+d x))\right )}{32 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \]
((a + a*Sin[c + d*x])^3*(-204*(c + d*x) + 200*Cos[c + d*x] - 8*Cos[3*(c + d*x)] + (256*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 40* Sin[2*(c + d*x)] - Sin[4*(c + d*x)]))/(32*d*(Cos[(c + d*x)/2] + Sin[(c + d *x)/2])^6)
Time = 0.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3351, 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.
| \(\displaystyle \int \sin (c+d x) \tan ^2(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a \sin (c+d x)+a)^3}{\cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle a^2 \int \left (-a \sin ^4(c+d x)-3 a \sin ^3(c+d x)-4 a \sin ^2(c+d x)-4 a \sin (c+d x)-4 a+\frac {4 a}{1-\sin (c+d x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a^2 \left (-\frac {a \cos ^3(c+d x)}{d}+\frac {7 a \cos (c+d x)}{d}+\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {19 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {4 a \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {51 a x}{8}\right )\) |
a^2*((-51*a*x)/8 + (7*a*Cos[c + d*x])/d - (a*Cos[c + d*x]^3)/d + (4*a*Cos[ c + d*x])/(d*(1 - Sin[c + d*x])) + (19*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d))
3.8.65.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04
| method | result | size |
| risch | \(-\frac {51 a^{3} x}{8}+\frac {25 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {25 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {8 a^{3}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}-\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a^{3} \cos \left (3 d x +3 c \right )}{4 d}+\frac {5 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\) | \(115\) |
| parallelrisch | \(\frac {a^{3} \left (-408 d x \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+408 d x \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+7 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )+\cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )-160 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-32 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+\sin \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )-7 \sin \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )-32 \sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+160 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-632 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-280 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d \left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )-\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(165\) |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\) | \(212\) |
| default | \(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\) | \(212\) |
| norman | \(\frac {\frac {51 a^{3} x}{8}-\frac {20 a^{3}}{d}-\frac {51 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {34 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {69 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {34 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {51 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {153 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {51 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {51 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {153 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {51 a^{3} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {4 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {60 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {44 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(282\) |
-51/8*a^3*x+25/8*a^3/d*exp(I*(d*x+c))+25/8*a^3/d*exp(-I*(d*x+c))+8*a^3/d/( exp(I*(d*x+c))-I)-1/32*a^3/d*sin(4*d*x+4*c)-1/4*a^3/d*cos(3*d*x+3*c)+5/4*a ^3/d*sin(2*d*x+2*c)
Time = 0.28 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.60 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{3} \cos \left (d x + c\right )^{4} - 15 \, a^{3} \cos \left (d x + c\right )^{3} + 51 \, a^{3} d x - 56 \, a^{3} \cos \left (d x + c\right )^{2} - 32 \, a^{3} + {\left (51 \, a^{3} d x - 67 \, a^{3}\right )} \cos \left (d x + c\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{3} \cos \left (d x + c\right )^{3} - 51 \, a^{3} d x - 21 \, a^{3} \cos \left (d x + c\right )^{2} + 35 \, a^{3} \cos \left (d x + c\right ) - 32 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]
-1/8*(2*a^3*cos(d*x + c)^5 + 8*a^3*cos(d*x + c)^4 - 15*a^3*cos(d*x + c)^3 + 51*a^3*d*x - 56*a^3*cos(d*x + c)^2 - 32*a^3 + (51*a^3*d*x - 67*a^3)*cos( d*x + c) + (2*a^3*cos(d*x + c)^4 - 6*a^3*cos(d*x + c)^3 - 51*a^3*d*x - 21* a^3*cos(d*x + c)^2 + 35*a^3*cos(d*x + c) - 32*a^3)*sin(d*x + c))/(d*cos(d* x + c) - d*sin(d*x + c) + d)
Timed out. \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.46 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {8 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{3} + {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{3} + 12 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} - 8 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \]
-1/8*(8*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^3 + (15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^3 + 12*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^3 - 8*a^3*(1/cos(d*x + c) + cos(d*x + c))) /d
Time = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.50 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {51 \, {\left (d x + c\right )} a^{3} + \frac {64 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 32 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 144 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
-1/8*(51*(d*x + c)*a^3 + 64*a^3/(tan(1/2*d*x + 1/2*c) - 1) + 2*(19*a^3*tan (1/2*d*x + 1/2*c)^7 - 32*a^3*tan(1/2*d*x + 1/2*c)^6 + 27*a^3*tan(1/2*d*x + 1/2*c)^5 - 144*a^3*tan(1/2*d*x + 1/2*c)^4 - 27*a^3*tan(1/2*d*x + 1/2*c)^3 - 160*a^3*tan(1/2*d*x + 1/2*c)^2 - 19*a^3*tan(1/2*d*x + 1/2*c) - 48*a^3)/ (tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 16.69 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.27 \[ \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {51\,a^3\,x}{8}-\frac {\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\frac {a^3\,\left (51\,c+51\,d\,x-58\right )}{8}\right )-\frac {a^3\,\left (51\,c+51\,d\,x-160\right )}{8}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\frac {a^3\,\left (51\,c+51\,d\,x-102\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-102\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-266\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-374\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-538\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{4}-\frac {a^3\,\left (306\,c+306\,d\,x-342\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{4}-\frac {a^3\,\left (306\,c+306\,d\,x-618\right )}{8}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
- (51*a^3*x)/8 - ((51*a^3*(c + d*x))/8 - tan(c/2 + (d*x)/2)*((51*a^3*(c + d*x))/8 - (a^3*(51*c + 51*d*x - 58))/8) - (a^3*(51*c + 51*d*x - 160))/8 + tan(c/2 + (d*x)/2)^8*((51*a^3*(c + d*x))/8 - (a^3*(51*c + 51*d*x - 102))/8 ) - tan(c/2 + (d*x)/2)^7*((51*a^3*(c + d*x))/2 - (a^3*(204*c + 204*d*x - 1 02))/8) - tan(c/2 + (d*x)/2)^3*((51*a^3*(c + d*x))/2 - (a^3*(204*c + 204*d *x - 266))/8) + tan(c/2 + (d*x)/2)^6*((51*a^3*(c + d*x))/2 - (a^3*(204*c + 204*d*x - 374))/8) + tan(c/2 + (d*x)/2)^2*((51*a^3*(c + d*x))/2 - (a^3*(2 04*c + 204*d*x - 538))/8) - tan(c/2 + (d*x)/2)^5*((153*a^3*(c + d*x))/4 - (a^3*(306*c + 306*d*x - 342))/8) + tan(c/2 + (d*x)/2)^4*((153*a^3*(c + d*x ))/4 - (a^3*(306*c + 306*d*x - 618))/8))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan( c/2 + (d*x)/2)^2 + 1)^4)