Integrand size = 21, antiderivative size = 146 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-a^3 x-\frac {9}{2} a b^2 x+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {2 b^3 \cos (c+d x)}{d}-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d} \]
-a^3*x-9/2*a*b^2*x+3*a^2*b*cos(d*x+c)/d+2*b^3*cos(d*x+c)/d-1/3*b^3*cos(d*x +c)^3/d+3*a^2*b*sec(d*x+c)/d+b^3*sec(d*x+c)/d+a^3*tan(d*x+c)/d+9/2*a*b^2*t an(d*x+c)/d-3/2*a*b^2*sin(d*x+c)^2*tan(d*x+c)/d
Time = 0.47 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.77 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {b \sec (c+d x) \left (108 a^2+45 b^2+4 \left (9 a^2+5 b^2\right ) \cos (2 (c+d x))-b^2 \cos (4 (c+d x))+9 a b \sin (3 (c+d x))\right )+3 a \left (-4 \left (2 a^2+9 b^2\right ) (c+d x)+\left (8 a^2+27 b^2\right ) \tan (c+d x)\right )}{24 d} \]
(b*Sec[c + d*x]*(108*a^2 + 45*b^2 + 4*(9*a^2 + 5*b^2)*Cos[2*(c + d*x)] - b ^2*Cos[4*(c + d*x)] + 9*a*b*Sin[3*(c + d*x)]) + 3*a*(-4*(2*a^2 + 9*b^2)*(c + d*x) + (8*a^2 + 27*b^2)*Tan[c + d*x]))/(24*d)
Time = 0.41 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3201, 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.
| \(\displaystyle \int \tan ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^2 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3201 |
| \(\displaystyle \int \left (a^3 \tan ^2(c+d x)+3 a^2 b \sin (c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^2(c+d x) \tan ^2(c+d x)+b^3 \sin ^3(c+d x) \tan ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \tan (c+d x)}{d}+a^3 (-x)+\frac {3 a^2 b \cos (c+d x)}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {9 a b^2 \tan (c+d x)}{2 d}-\frac {3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a b^2 x-\frac {b^3 \cos ^3(c+d x)}{3 d}+\frac {2 b^3 \cos (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}\) |
-(a^3*x) - (9*a*b^2*x)/2 + (3*a^2*b*Cos[c + d*x])/d + (2*b^3*Cos[c + d*x]) /d - (b^3*Cos[c + d*x]^3)/(3*d) + (3*a^2*b*Sec[c + d*x])/d + (b^3*Sec[c + d*x])/d + (a^3*Tan[c + d*x])/d + (9*a*b^2*Tan[c + d*x])/(2*d) - (3*a*b^2*S in[c + d*x]^2*Tan[c + d*x])/(2*d)
3.15.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Time = 0.88 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(\frac {\left (36 a^{2} b +20 b^{3}\right ) \cos \left (2 d x +2 c \right )-b^{3} \cos \left (4 d x +4 c \right )+9 \sin \left (3 d x +3 c \right ) a \,b^{2}+\left (-24 a^{3} d x -108 a \,b^{2} d x +144 a^{2} b +64 b^{3}\right ) \cos \left (d x +c \right )+\left (24 a^{3}+81 a \,b^{2}\right ) \sin \left (d x +c \right )+108 a^{2} b +45 b^{3}}{24 d \cos \left (d x +c \right )}\) | \(130\) |
| derivativedivides | \(\frac {a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{2} b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(169\) |
| default | \(\frac {a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{2} b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a \,b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(169\) |
| risch | \(-a^{3} x -\frac {9 a \,b^{2} x}{2}-\frac {3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}+\frac {7 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {3 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}+\frac {7 b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3}+6 i a \,b^{2}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{12 d}\) | \(200\) |
| norman | \(\frac {a \left (2 a^{2}+9 b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {36 a^{2} b +16 b^{3}}{3 d}+\frac {a \left (2 a^{2}+9 b^{2}\right ) x}{2}-\frac {2 \left (36 a^{2} b +16 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {3 a \left (2 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (2 a^{2}+5 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (2 a^{2}+9 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (2 a^{2}+9 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a \left (2 a^{2}+9 b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a \left (2 a^{2}+9 b^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {12 a^{2} b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(300\) |
1/24*((36*a^2*b+20*b^3)*cos(2*d*x+2*c)-b^3*cos(4*d*x+4*c)+9*sin(3*d*x+3*c) *a*b^2+(-24*a^3*d*x-108*a*b^2*d*x+144*a^2*b+64*b^3)*cos(d*x+c)+(24*a^3+81* a*b^2)*sin(d*x+c)+108*a^2*b+45*b^3)/d/cos(d*x+c)
Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.79 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {2 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} d x \cos \left (d x + c\right ) - 18 \, a^{2} b - 6 \, b^{3} - 6 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
-1/6*(2*b^3*cos(d*x + c)^4 + 3*(2*a^3 + 9*a*b^2)*d*x*cos(d*x + c) - 18*a^2 *b - 6*b^3 - 6*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 - 3*(3*a*b^2*cos(d*x + c)^ 2 + 2*a^3 + 6*a*b^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.82 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} + 9 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{3} - 18 \, a^{2} b {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{6 \, d} \]
-1/6*(6*(d*x + c - tan(d*x + c))*a^3 + 9*(3*d*x + 3*c - tan(d*x + c)/(tan( d*x + c)^2 + 1) - 2*tan(d*x + c))*a*b^2 + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b^3 - 18*a^2*b*(1/cos(d*x + c) + cos(d*x + c)))/d
Time = 0.46 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.42 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {3 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} {\left (d x + c\right )} + \frac {12 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b + b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 18 \, a^{2} b - 10 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
-1/6*(3*(2*a^3 + 9*a*b^2)*(d*x + c) + 12*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b ^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b + b^3)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2* (9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 6*b^3* tan(1/2*d*x + 1/2*c)^4 - 36*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 24*b^3*tan(1/2* d*x + 1/2*c)^2 - 9*a*b^2*tan(1/2*d*x + 1/2*c) - 18*a^2*b - 10*b^3)/(tan(1/ 2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 16.22 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.71 \[ \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3+9\,a\,b^2\right )+12\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^3+9\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^3+15\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (24\,a^2\,b+\frac {32\,b^3}{3}\right )+\frac {16\,b^3}{3}+12\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+9\,b^2\right )}{2\,a^3+9\,a\,b^2}\right )\,\left (2\,a^2+9\,b^2\right )}{d} \]
(tan(c/2 + (d*x)/2)*(9*a*b^2 + 2*a^3) + 12*a^2*b + tan(c/2 + (d*x)/2)^7*(9 *a*b^2 + 2*a^3) + tan(c/2 + (d*x)/2)^3*(15*a*b^2 + 6*a^3) + tan(c/2 + (d*x )/2)^5*(15*a*b^2 + 6*a^3) + tan(c/2 + (d*x)/2)^2*(24*a^2*b + (32*b^3)/3) + (16*b^3)/3 + 12*a^2*b*tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^8 + 1)) - (a*atan((a*tan(c/2 + (d*x)/2)*(2*a^2 + 9*b^2))/(9*a*b^2 + 2*a^3))*(2*a^2 + 9*b^2))/d